Problem 1 Answer

Problem 1 Solution

I have had a lot of mail about this problem over the years. Some people question the answer, and some admit the answer is right, but that the solution is wrong. Here is my original solution:

For n points choose any given point and evaluate the probability that the other n-1 lie within a semicircle going clockwise. This probability is (1/2)^n-1.

To prove at least the answer is correct, I did ten million trials for various numbers of points on the circle. I assumed the circle to have 32,767 units or degrees. The first point was defined to be at a halfway point, or 16,383. Then a maximum and minimum were taken for the rest of the points. If the difference between the maximum and minimum were less than 16,383 then the trial was considered to be a success. Note that 16,383 is half of 32,767, rounded down. Because of the rounding we should expect a slightly lower number of successes. Here are the results:

Several people have written, saying my answer is right, but the solution is wrong. Here is what David Beim of Columbia University wrote to me about it:

There is no doubt that your answer is right, but your reasoning is unsettling because it confounds ex ante with ex post information. We cannot construct the semicircles until the n points have been selected, and then it is too late to determine ex ante probabilities. This kind of mixing can lead to all kinds of paradoxes in probability, as I am sure you know better than I. For one simple example, select a point on a line. Ex ante, the probability of selecting that point is zero; ex post, the probability must have been greater than zero because it happened.

I attach my own solution (PDF 33K) to the problem, in case it is of interest. Incidentally, I would give this problem four stars rather than three --took me days to get it clear!

Per his comment, I did add a star to the difficulty level, it was 3 starts until November, 2008.

This problem has been published in book form in "Contests in Higher Mathematics" by Gabor Szekely, which is a collection of math problems given to the best mathematicians of Poland over the last several decades. The problem (1963 problem 10) is stated as, "Select n points on a circle independently with uniform distribution. Let P_n be the probability that the center of the circle is in the interior of the convex hull of these n points. Calculate the probabilities P₃ and P₄. I'm told the answer agrees with mine.

Problem 2 Answer

Problem 2 Solution

The distance covered in the first hour is the integral from x to x+1 of 1/t dt. The antiderivative of 1/t is ln(t) so the total distance covered in the first hour is ln((x+1)/x).

By the same reasoning the distance covered in the second hour in ln((x+2)/(x+1)).

Using the fact that it the plow traveled twice as far in the first hour as the second: ln((x+1)/x) = ln((x+2)/(x+1))²

Exp both sides and you have (x+1)/x = ((x+2)/(x+1))².

Solving for x you get x=(5^1/2-1)/2, which is the number of hours that elapsed between the time it started snowing and the snow plow left.

This problem was taken from the Actuarial Review, although I heard it somewhere else before.

Problem 3 Answer

Problem 3 Solution

So the total arc length is the integral from 0 to 2pi of ( (dx/dt)² + (dy/dt)² ) ^1/2.

After a few steps this integral becomes:

r×2^1/2 × (1-cos(t))^1/2.

Using the hint:

r×2^1/2 × 2^1/2 × integral of sin(t/2) dt from 0 to 2×pi

= 2×r × (-2×cos(t/w) from 2×pi to 0)
= 8r

Reference: Example 3, page 550, Calculus and Analytic Geometry, 1982 edition.

Problem 4 Answer

If the spider started at a corner diagonally on the same face as the ant the answer would be 9, and if the spider started at an adjacent corner the answer would be 7.

Here are answers for other figures:

Square: 4
Octahedron: 6
Dodecahedron: 35
Icosahedon: 15

Problem 5 Answer

Problem 5 Solution

Each wise man initially assumes that he isn't painted and can plainly see the other two are.

However, if only two were painted it would not take long for either of the painted wise men to realize they were painted. For either of the painted ones would see the other painted one laughing and know he is laughing at them, since the other guy is not painted. In other words if you saw a wise man laughing and the other guy wasn't he painted he must be laughing at you, thus you must be painted.

After a sufficient period of time if nobody stops laughing then it can no longer be assumed that you are not painted, thus you must be painted as well. When you realize this you stop laughing.

Problem 6 Answer

Problem 6 Solution

I get a lot of Emails and forum posts about this problem. Based on feedback given, I've changed my "solution" to this problem numerous times. Suffice it to say, I've never been entirely satisfied with my explanation of why there isn't a 25% expected gain in switching. My readers have never been satisfied with my explanations either.

As of this writing, I feel comfortable that I understand the answer, but I don't know the proper way to explain it. I think there are two ways you can attack the 25% argument but I think one of them is a red herring and there needs be only one.

Suffice it to say, that I recently took down my latest explanation because it was getting too much negative feedback. So, come back later and maybe I'll have another solution to present.

Problem 7 Answer

Problem 7 Solution

The height of a face is 3^.5/2.

The area of any face is 3^.5/4.

The distance from any corner to the center of a joining face is 1/3^.5.

The height of the tetrahedron is (2/3)^.5

The area is 1/3*base*height =

(1/3) * 3^.5/4 * (2/3)^.5 = 2^1/2/12.

Solution 2

Provided by
Mircea Petrache, of "Liceo scientifico Varano" in Camerino, Italy

The above cube has sides of length 1/sqr(2). The area formed by the diagonals along four of the faces as shown in the diagram is a tetrahedrom. Left over are four pyramids. The base of each pyramid is half of the side of the cube and the height is the height of the cube. Thus the area of each pyramid is: 1/2*(1/sqr(2))² * 1/sqr(2) * 1/3 = (12*sqr(2))^-1

The total area of the cube is (1/sqr(2))³ = (2*sqr(2))^-1

The area of the tetrahedrom is thus (2*sqr(2))^-1 - 4*(12*sqr(2))^-1 =

3/(6*sqr(2)) - 2/(6*sqr(2)) = 1/(6*sqr(2)) = sqr(2)/12

Problem 8 Answer

Problem 8 Solution

Problem 9 Answer

If the number is not divisible by 3 then use one 20 pack. If the remaining number is divisible by 3 then use the above method for the rest.

If the number still isn't divisible by 3 use a second 20 pack. The remainder must be divisible by 3, in which case use the 6 and 9 packs as above.

The largest impossible number would be such that you would have to subtract 20 twice to get a remainder divisible by 3. However, you can't make 3 itself with 6 and 9 packs. So the largest impossible number is 2*20+3=43.

Problem 11 Answer

Problem 11 Solution

Think of this problem instead as a random walk along the number line, starting at zero. If the walker ever returns to zero, then success is achieved. Let x_i be the probability of success from point i on the number line. Let p be the probability that the walker moves to the right. Let's also let p be the lesser of the probabilities of flipping heads and tails.

I hope I won't have to prove that x_-1=1. In other words, the gravity of the greater probability of q will pull the walker to the right, over the long run. So if his first move is to the left, he will eventually make it back to point 0. For the same reason, given a casino game with a house advantage, the house always wins in the long run. If the player wins his first bet, if he plays indefinitely, he will eventually lose it back.

That said, we can set up the following equations:

x₀ = p + q×x₁
x₁ = p + q×x₂
x₂ = p×x₁ + q×x₃
x₃ = p×x₂ + q×x₄
.
.
.

Next, take the sum of both sides.

Let E = x₀ + x₁ + x₂ + ...

E = 2p + pE -p×x₀ + qE - q×x₀
E = 2p + E×(p+q) - x₀×(p+q)
E = 2p + E - x₀ (because p+q=1)
0 = 2p - x₀
x₀ = 2p So the answer is 2p.

Below are the results of computer trials which have verified the answer above. For each probability p 100,000 trials were conducted, in which a trial ended when either the number of heads equalled the number of tails, or the difference was more than 25. It seems reasonable to assume that for p<=.4 if the difference ever equals 25 it is very unlikely that the gap ever be closed. A 'win' means the number of heads was ever equal to the number of tails. The number of losses would be 100,000 minus the number of wins.

  p    wins 
-----  ------
 0.4   80058
 0.3   59799
 0.2   39957
 0.1   19803
 

If you don't buy my solution, I have posted altnernate solutions by Richard Tucker and David Beim.

Thanks to Guy de Kindler for providing the problem and Richard Tucker and David Beim for the alternate solutions.

Problem 12 Answer

Problem 12 Solution

The total money paid is $9 * 3 = $27.

The total money received is $25 for the room and $2 to the bellboy = $27.

There simply is no extra dollar. The $2 to the belloy is included in the $27 the three men paid. By adding it to the $27 it is being counted twice.

Problem 13 Answer

Problem 13 Solution

U=w²*c.

3*w + 2*c = 100.

c=(100-3w)/2.

U=w²*(100-3w)/2.

dU/dw=2w*(100-3w)/2 + w²*-3/2=0.

Solving for w yields w=200/9.

With wine costing $3 a glass this will cost 3*200/9=$66.67, leaving $33.33 for cigars, for which he can buy 16.67.

Problem 14 Answer

  3     5
pint  pint
glass glass
----- -----
  0     5
  3     2
  0     2
  2     0
  2     5
  3     4

  3     5
pint  pint
glass glass
----- -----
  3     0
  0     3
  3     3
  1     5
  1     0
  0     1
  3     1
  0     4

In other words, fill the 3 pint glass and pour it into the 5 pint glass. Then fill it again and from it top off the 5 pint glass, leaving 1 pint in the 3 pint glass. Then empty the 5 pint glass and move the 1 pint to the 5 pint glass. Then fill the 3 pint glass and pour that into the 5 pint glass. A problem like this appeared in the movie Die Hard III.

Problem 14 Solution

  3     5
pint  pint
glass glass
----- -----
  0     5
  3     2
  0     2
  2     0
  2     5
  3     4

At this point there are four pints in the five pint glass.

A problem like this appeared in the movie Die Hard III.

Problem 15 Answer

The maximum probability that can be achieved of winning is 2/3. See the solution for how to do it.

Problem 15 Solution

The maximum probability that can be achieved of winning is 2/3. Here is way to do it.

1. Red opens door 1. If first door is blue, he opens 2. If first door is green, he opens 3.
2. Blue opens door 2. If first door is red, he opens 1. If first door is green, he opens 3.
3. Green opens door 3. If first door is red, he opens 1. If first door is blue, he opens 2.

This should work for these combinations, where the colors are ordered from door 1 first to door 3 last:

• RBG (all three guess correctly on first guess)
• RGB (red is correct first time, blue sees green so opens door 3, green sees blue so opens 2.)
• BRG (red sees blue so opens 2, blue sees red so opens 1, green is correct on first guess).
• GBR (red sees green so opens 3, blue is correct on first guess, green sees red so opens 1).

Problem 16 Answer

Problem 16 Solution

To illustrate why the 1/2 answer is wrong let propose a similar problem. Suppose a soup company canned 2,000,000 cans of tomato soup. Suppose further that they put the wrong label on half of the cans. The company then sent 999,999 correct cans and 1 incorrect can to one store and 999,999 incorrect cans and 1 correct can to another store. To determine which store has the correct cans you go to one store at random and open one can. You see it does indeed have tomato soup inside. What is the probability you went to the store with 999,999 correct cans? It is not reasonable to assume the probability is 50/50 that you are in either store. The odds that you would pick the one correct can in the store with 999,999 incorrect cans are extremely remote. It is true that before opening a can the odds were 50/50 but once you do open a can that information gives you a clue that can not be ignorred. In fact the probability that you are in the store with the correctly labled cans is 99.9999%.

Now let me introduce Bayes' theorem on conditional probability to explain why the answer is 2/3:

Let the events A₁, ...,A_k for a partition of the space S such that Pr(A_j > 0 for j=1,...,k and let B be any event such that Pr(B) > 0. Then, for i=1,...,k,

Pr(A_i | B) = Pr(A_i) * Pr(B|A_i) / [ $\Sigma$ (for j=1 to k) Pr(A_j)*Pr(B|A_j) ] .

For the problem in question we can reword this to: Pr(two headed coin|head chosen) = Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) / [Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) + Pr(choosing one headed coin) * Pr(choosing heads given that one headed coin was chosen)] =

1/2 * 1 / [ (1/2 * 1) + (1/2 * 1/2) ] =

1/2 / (1/2 + 1/4) = (1/2)/(3/4) = (1/2)*(4/3) = 2/3.

If this all went over your head think if it another way. There were 3 heads to begin with. 2 of them have another head on the other side. You could have chosen any head with equal probability and since 2 out of 3 have a head on the reverse side the answer is 2/3. Of course this answer would probably not merit very much in partial credit in probability class.

Some other people have questioned my exact wording of this problem, mainly objecting over the tense of the verb I use for when the coin is chosen. What appears now was taken almost exactly from a similar problem in Probability and Statistics (second edition) by Morris H. Degroot on page 63, problem number 5. I changed cards to coins and eliminated what would be the coin with two tails. The book is a commonly used college text on the subject and should be above reproach. Here is how they stated a similar problem:

A box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability taht the other side of the card is also green?

Essentially the same problem, but worded using sunken boats, secret compartments, and bars of gold, is asked on page 603 of Statistics for Business and Economics by Edwin Mansfield.

A reader sent in the following explanation that he thought might help:

You have probably received similar e-mails to the one I am now sending, but
just in case, ...

Perhaps the best way to demonstrate that 2/3 is the correct answer to this
problem is to approach it in a simple intuitive manner, such as:

Denote the double headed coin A and the regular coin B.  Further, denote
the sides of the double headed coin A1 (a head) and A2 (another head), and
the sides of the regular coin B1 (a head) and B2 (a tail).

We then have the following possible outcomes:

coin selected           side shown              other side
         A                 A1 (H)                   A2 (H)
         A                 A2 (H)                   A1 (H)
         B                 B1 (H)                   B2 (T)
         B                 B2 (T)                   B1 (H)

We exclude the last possibility from consideration since it is not a head.
Three possibilities remain, all equally likely, two of which have a head on
the other side.  The required probability is thus 2/3.

Gary.

Here is my favotite e-mail I received on this problem. I x'd out some private information.

Dear Michael,

I am a biology teacher at Xxxxxxx High School in Xxxxxxx, MA. I recently was searching for an extra credit question to put on a test for tenth graders. I used the three coins in a bag problem (#16). Unfortunately, none of my students got the correct answer. They all answered 1/2. I have done my best to explain the problem and some of the students now understand why the correct answer is 2/3. The debate has spread throughout the whole school. Most teachers and students beleive the answer to be 1/2. So I started an organization called Team 2/3 - The Probability Masters. Students take an oath to become official members of the team. The mission of the team is to recruit as many new members as possible. Team members get a membership card that has your original question on the back which helps them recruit new team members. Many people think I have gone off the deep end, but I find it amazing that so many people including most teachers and the principal still do not see the truth. If fact, the day after I started the Team 2/3, the principal got on the morning announcements and stated the the answer is 1/2 and the group that thinks otherwise should rethink the problem.

One student created a computer program that simulates 10,000 coin picks in 1 second. The data clearly show the 2/3 answer. But that same student insists that the probability is 2/3 only if many trials are conducted. He doesn't understand that the probability is the same in the first and last trial. People are stuck on the idea that you must have either the H-H or H-T coin and say it is 1 out of two. I have tried to simplify your solution without the T-T coin to help people understand.

If two coins: H-H, H-T

Conditional Prob = 1/2 divided by 3/4 = 2/3

Team 2/3 will recruit for two more days and then reveal the truth. We will try to get the principal to announce that Team 2/3 was correct all along. All team members will hold their heads high if that happens. I only have one math teacher on my side at this time. One of my student team members had a hat made for me that says "Team 2/3 Captain". I wore it on Friday and will wear it this week. We have almost everyone in the school discussing your coin problem. Many of the students are giving it a great deal of thought.

We are very confident that the answer is 2/3, however most people still disagree.

It would be most helpful if you would place a call to the school and talk to Mr. Xxxxxxx Xxxxxxxxx, the principal. This would be greatly appreciated by the team, your team, Team 2/3.

Xxxxxxx High School 1-(xxx)-xxx-xxxx

Sincerely,

Troy X. Xxxxxxx
Biology Teacher P.S. Please feel free to call me at home: 1-(xxx)-xxx-xxxx

I did call the principal and tried to explain why the answer was 2/3. He seemed pretty entertained that somebody would call all the way from Baltimore about this. After going over the problem he seemed to partially understand but added that "he didn't want to be confused with facts." Later the Biology teacher wrote me back, here is his second e-mail:

I really appreciate you talking to the principal. He did talk to me afterward and he seemed very unsure of the correct answer. Team 2/3 has 42 oath-taking official members including an English teacher and the guidance secretary. We allowed people to join unofficially by signing the unofficial membership list. That way they still get a membership card without taking the oath; it's our "chicken list". The Latin teacher, the US history techer, and the alternative education teacher are on that list. Our membership card states-"TEAM 2/3 the Probability Masters" and there is a picture of a man holding two coins. On the back of the card is your original problem. Another team formed (not as organized as our team) called Team 1/2. They also have membership cards. Just this past Thursday, a good week after your call, I reminded the principal that we needed closure to the debate. I just assumed he believe the answer was 2/3. And I told him it would be nice to recognize the courage of team 2/3 and congratulate them. He got on the announcements and stated "After consulting with several mathematical experts" (and other people heard about the controversy and contacted him with explanations for 2/3) "the answer to the coin probability problem is 'probably' 2/3, so congratulations to Team 2/3, you are probably correct. While we would have liked a stronger statement(without the probablies), we have accepted the principal's announcement as victory and feel that we have been successful in our mission.

Thanks for your help,

Troy

Problem 17 Answer

Problem 17 Solution

Integral from 0 to 1 of d*2d = 2d³/3 from 0 to 1 = 2/3.

Problem 18 Answer

Note: To convert into gallons divide cubic inches by 231 .

Problem 18 Solution

Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross section of the tank. By the pythagorean formula the length of the level of gas line is 2×(r²-x²)^1/2.

Next measure the area of the triangle with the level of gas segment as a base and the center of the circle as the opposite angle, which is the the sum of two right triangles, or x×(r²-x²)^1/2.

Next, determine the degree of the angle opposite of the base of gas line segment in the triange above. With a little trigonometry it is found that this angle is 2×cos^-1(x/r).

Next, find the area of the slice of the circle made between the two equal sides of the triangle mentioned above, which is the product of the area of the circule and the ratio of the angle of the wedge to 2×pi: pi×r² × (2×cos^-1(x/r)/2×pi) = r² × (cos^-1(x/r)) .

From this wedge subtract the area of the triangle, leaving r² × (cos^-1(x/r)) - x×(r²-x²)^1/2.

Next, multiply by the length of the tank: l × ((r² × (cos^-1(x/r)) - x×(r²-x²)^1/2).

If g is the depth of the gas then g=r-x. The solution in terms of g is l × ((r² × (cos^-1((r-g)/r)) - (r-g)×(2rg-g²)^1/2).

Thanks for Craig Katz for this solution.

First, ignore the length of the tank until the end of the problem.

Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross section of the tank. By the pythagorean formula the length of the level of gas line is 2×(r²-x²)^1/2.

Next, take the integral from 0 to the depth of the gas, measured in distance from the center of the side. For example if the level of gas is y (where y is less than x) then the answer is the integral from r-y to r of 2×(r²-x²)^1/2.

To solve this integral it is helpful to know that the integral of (r²-x²)^1/2 is 1/2×[x×(r²-x²)^1/2 + r²×sin^-1(x/r).

I'm sure if you can get this far you can do the rest yourself.

Problem 19 Answer

Problem 19 Solution

After the first turn you will have a configuration like 1,1,2,3. Call this configuration 1.

From configuration 1 if you draw a 1 on the first ball you will end up with the same or similar configuration. The probability of this happening is 1/2.

From configuration 1 if you draw a 3 or 4 first and then a 1 you will have a configuration like 1,1,1,2. The probability of this happening is 1/3. Call this new pattern configuration 2.

From configuration 1 if you draw a 3 or 4 first and then draw the other non 1 you will have a configuration like 1,1,2,2. The probability of this is 1/6. Call this new pattern configuration 3.

From configuration 2 you will end up with the same pattern with probability 1/2, configuration 3 with probability 1/4, and ending the experiment with probability 1/4.

From configuration 3 you will end up with the same thing with probability 1/3, and with configuration 2 with probability 2/3.

From this information you can draw the following equations, where a is the expected number of turns from equation 1, b from equation 2, and c from equation 3:

• a=(a+1)/2 + (b+1)/3 + (c+1)/6
• b=(b+1)/2 + (1)/4 + (c+1)/4
• c=(c+1)/3 + (2/3)*(b+1)

Problem 20 Answer

Thanks to The Grey Labyrinth for this one.

Problem 20 Solution

You can also try this problem witha a 15 minute egg and hourglasses of 7 and 11 minutes.

Thanks to The Grey Labyrinth for this one.

Problem 22 Answer

The answer is $2.

Problem 22 Solution

Let the total number of sheep be 10x+y, where y<10. The total money raised is (10x+y)² = 100x² + 20xy + y². Regardless of the values of x and y 100x² + 20xy will be divisible by 20. Because the number of $10 bills is odd y² mod 20 must be greater than 10 and less than 20. The only values of y where this is true is 4 and 6, where y² is either 16 or 36. Either way there will be 6 $1 bills left over. Before the check the first brother will have $4 more than the second brother. A $2 check will balance things out.

Thanks to Bill Feldman for suggesting this problem.

Problem 23 Answer

Answer 2: You point to either path and say, "Are you from this village?" If the person answers 'yes' then you are pointing to the truthful village, if they answer 'no' then you are pointing to the lying village.

Answer 3: Ask "Which path would a person from the other village say leads to the truthful (or untruthful) village?" Regardless of whom you speak to the answer will filter through one lie, so if you ask for the truthful village the person will point to the lying village, and vise versa.

Problem 23 Solution

This is not the only answer that works but is the simplist I have heard so far.

I continue to receive a lot of e-mail on this problem from people who don't understand or disagree with my solution. Before you question me on this answer these four questions:

1. If you point to the truthful village what will a truthful person say in answer to your question?
2. If you point to the truthful village what will a untruthful person say in answer to your question?
3. If you point to the untruthful village what will a truthful person say in answer to your question?
4. If you point to the untruthful village what will a untruthful person say in answer to your question?

Note that in the two cases where the answer is "yes" you are pointing to the truthful village and in the two cases where the answer is "no" you are pointing to the untruthful village. Thus you can tell from the reply to which village you are pointing. You do not have to know to whom you are speaking.

Another possible solution which I have heard is to ask "Which path would a person from the other village say leads to the truthful (or untruthful) village?" Regardless of whom you speak to the answer will filter through one lie, so the person will point to the lying village if you ask for the truthful village, and vise versa.

Problem 24 Answer

Problem 24 Solution

dc = 100-c dt

dc 1/(100-c) = dt

-ln(100-c) = t + K₁

t = ln(K₂) - ln(100-c) (where K₁ = ln(K₂)

t = ln(K₂/(100-c))

When t=0 c=15 so K₂ must be 85

t = ln(85/(100-c))

Solving for t when c=25:

t = ln(85/(100-25)) = ln(85/75) =~ ln(17/15) = 0.1252

Problem 25 Answer

Problem 25 Solution

1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4

From the statement that the sum equals the house number it is possible to eliminate all but two possibilities. The sums of the rest are unique and would allow for an immediate answer. For example if the house number were 16 the ages must be 1, 3, and 12. The two remaining possibilities are 2, 2, and 9; or 1, 6, and 6.

After the clue that the oldest has red hair you can eliminate 1, 6, and 6 because the oldest two have the same age thus there is no oldest son. The only remaining posibility is 2, 2, and 9.

Problem 26 Answer

Problem 26 Solution

The probability of winning is sum for i=x+1 to 100 of the probability that the highest dowry is in the ith position multiplied by the probability that you choose it if it is in that position.

This is based on the condition probability formula: Pr(A)=Sum for i=1 to n of Pr( A | B_i ) * Pr( B_i ), where the sum for i=1 to n of Pr(B_i)=1.

For all i the probability that the highest dowry is in that position is 1/100.

The probability that you will choose the dowry if it is in the ith position is equal to the probability that the highest of the first i-1 dowries belongs to one of the first x ladies. This equals x/(i-1). If the highest dowry of the first i-1 were not in the first x ladies you would choose it before getting to the largest dowry, thus losing the game.

For example if you let 30 ladies pass the probability that the highest dowry is in the 75th postion and that you will choose it equals 1/100 * 30/74.

Let A denote the overall probability of winning. The proability of winning given x is the sum for i=x+1 to 100 of x/(i-1) * 1/n. This equals:

Pr(1/n * [ 1 + x/(x+1) + x/(x+2) + ... + x/99 ] ).

Some value of x must be optimal to maximize this formula. This value of x must the first such that Pr(A|x+1) - Pr(A|x) < 0.

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(x+2) + 1/(x+3) + 1/(x+4) + ... + 1/99 - x/(x+1) ].

let y=x+1.

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - (y-1)/y ]. =

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - 1 + 1/y ]. =

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - 1 ].

If Pr(A|x+1) - Pr(A|x) = 0, then 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 = 1.

Theorem: The integral from r to n of 1/x = log(n) - log(r) = log(n/r).

The equation above the theorm is an approximation of the integral in the theorem. By applying the theorem log(100/y) = 1.

Taking e to the power of both sides: 100/y = e.

y=100/e.

Since e=~2.7182818 y=~36.79, since x=y-1 x=~35.79.

Since the integral is an underestimate of the series look at some specific values of Pr(A) for values of x close to 35.79:

 x   A
--- ---
35  37.0709%
36  37.1015%
37  37.1043%
38  37.0801%

In general the answer is going to be n/e, where n is the total number of ladies. Because the answer must be an exact integer and error in applying the theorem mentioned above you should check the few integers just above the optimal value. As n increases the error decreases. The probability of winning turns out to approach 1/e =~ 36.79% as n approaches infinity (I'll leave this proof up to you).

Here are some optimal values of x for various values of n and the probability of winning given x:

  n   x   Pr(winning)
---- ---  -----------
   3   2   50.00%
   4   2   43.83%
   5   3   43.33%
   6   3   42.78%
   7   3   41.43%
   8   4   40.98%
   9   4   40.60%
  10   4   39.87%
  15   6   38.94%
  20   8   38.42%
  30  12   37.87%
  50  19   37.42%
 100  38   37.10%
1000 369   36.82%

Here another version of my solution to this problem

Let x be the number of ladies you pass before choosing the next lady with a dowry greater than the first x ladies. Let max_x be the maximum dowry in the first x ladies.

The probability you will select the highest dowry is:

Sum for i=1 to 100-x of: ( Probability that i of the remaining ladies will have a dowry greater than max_x ) * (1/i).

This probability for any i is:

(100-x)*(99-x)*(98-x)*...*(101-x-i) * x
---------------------------------------
100*99*98*...*(100-i)

Pr(n₁) * 1 +
Pr(n₂) * 1/2 +
Pr(n₃) * 1/3 +
Pr(n₄) * 1/4 +
Pr(n₅) * 1/5 +
.
.
.
Pr(n₇₀) * 1/70.

To derive Pr(n_i) consider the probability that the i highest dowrys fall in the last 100-x ladies, multiplied the the probability that the (i+1)th highest dowry falls in the first x ladies. For example:

Pr(n₃) = 70/100 * 69/99 * 68/98 * 30/97.

70/100 = Probability that the highest dowry is not in first 30.
69/99 = Probability that the second highest dowry is not in first 30 (99 positions left, 69 not if first 30).
68/98 = Probability that the third highest dowry is not in first 30.
30/97 = Probability that fourth highest dowry is in first 30.

Personally I had to use a spreadsheet to find that the maximum probability was for x=37, which was 37.1%.

If you change the number of total ladies from 100 to any other number the strategy seems to still be to let 37% of them go by first.

Problem 27 Answer

Problem 27 Solution

Consider the two ladders two lines on a graph. Let the shorter ladder extend from (0,0) to (6,8). Let the longer ladder extend from (6,0) to (0,108^1/2). The square root of 108 can be found using the pythagorean formula. Then solve for the slope and y intercept to find the equations of the two lines:

The shorter ladder have the equation y=(4/3)*x.

The longer ladder has the equation y=(-108^1/2/6)*x + 108^1/2.

The lines meet where the ladders cross.

Use substitution to solve for y.

Acknowledgement: Thanks to Steve Lutz for suggesting this method of solution.

Scott R. Walshon pointed out that if x and y are the heights where the ladders touch the wall, and p is the height of the intersection then 1/p = 1/x + 1/y.

To prove this let x be the height on the right wall and y the height on the left wall. Call a the distance from the left edge of the alley to the point on the alley directly below the intersection point of the ladders. Call b the distance from the right edge of the alley to the point on the alley directly below the intersection point of the ladders. From similar triangles we get:

p/x = a/(a+b)
p/y = b/(a+b)

Add the two equations:

p/x + p/y = a/(a+b) + b/(a+b)
py/xy + px/xy = (a+b)/(a+b)
p(x+y)/xy = 1
p = xy/x+y
1/p = x+y/xy
1/p = x/xy + y/xy
1/p = 1/y + 1/x

Thanks Scott for this observation.

Problem 28 Answer

Problem 28 Solution

Let c be the number of chairs produced per hour and t the number of tables produced per hour.

The number of saws limit the combination of chairs and tables to 600=10c+5t.
The number of lathes limit the combination of chairs and tables to 360=5c+5t.
The number of sanding machines limit the combination of chairs and tables to 1080=5c+20t.

Next graph these three lines. It should be expected that the answer will lie on the intersection of two of these lines or to make all chairs or all tables. The intersection of the saw and sanding machine line occurs outside of how many chairs the lathe can make so this combination is not a viable answer. The saw and lathe lines cross at 48 chairs and 24 tables. The lathe and sanding machine lines cross at 24 chairs and 48 tables.

Next determine the revenue at all points of intersection.

So the optimal answer is to make 24 chairs and 48 tables for revenue of $1200 per hour.

To check it will take 24*10 + 48*5 = 480 minutes of saw time. There are 600 minutes available so the saws will be idle 20% of the time.

It will take 24*5 + 48*5 = 360 minutes of lathe time which is exactly what we have.

It will take 24*5 + 48*20 = 1080 minutes of sanding machine time which is exactly what we have.

I would like to thank Brain Storm for this problem.

Problem 29 Answer

The total distance covered is a.

Problem 29 Solution

Hard Solution

Let C be the center of the square at point (0,0). Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1. Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e^-x.

At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2^-1/2. So r(x)=2^-1/2*e^-x.

The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))²+(f^'(x))²)^1/2 =
Integral from 0 to infinity of ((2^-1/2*e^-x)² + (-2^-1/2*e^-x)²)^1/2 =
Integral from 0 to infinity of (2^-1*e^-2x + 2^-1*e^-2x)^1/2 =
Integral from 0 to infinity of e^-x =
-e^-x from 0 to infinity =
=0 - (-e⁰) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral. It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e^-2*pi/2^1/2 = 0.03055663 times it's size before the revolution.

Easy Solution

Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs. He also provided the following graphics.

For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)). So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines. In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309 units towards dog A. The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.

Thus the general formula is 1/(1+cos(t)), where t is an interior angle of the shape.

Number of sides = 3
Lengh of initial side = 1
Each angle = 180*(1-(2/3)) = 60°
Length of path = 1/(1+cos(60)) = 1/(1+0.5) = 2/3

Number of sides = 4
Lengh of initial side = 1
Each angle = 180*(1-(2/4)) = 90°
Length of path = 1/(1+cos(90)) = 1/(1+0) = 1

Number of sides = 5
Lengh of initial side = 1
Each angle = 180*(1-(2/5)) = 108°
Length of path = 1/(1+cos(108)) = 1/(1-.309) = 1.4472

Number of sides = 6
Lengh of initial side = 1
Each angle = 180*(1-(2/6)) = 120°
Length of path = 1/(1+cos(120)) = 1/(1-.5) = 2

Number of sides = 10
Lengh of initial side = 1
Each angle = 180*(1-(2/10)) = 144°
Length of path = 1/(1+cos(144)) = 1/(1-.809) = 5.2361

Number of sides = 15
Lengh of initial side = 1
Each angle = 180*(1-(2/15)) = 156°
Length of path = 1/(1+cos(156)) = 1/(1-.9135) = 11.5668

Problem 30 answer

x = 75 * [ 1/5*(y/50)^5/3 - (y/50)^1/3 ] + 60

It takes 3 hours for the chasing boat to catch the other boat.

Problem 30 Solution

Place the slower boat initially at the origin of a coordinate plane, and equate "north" with the positive y-axis, and east with the positive x-axis. The faster boat starts at the point (50,0). The "curve" shall refer to the path of the faster boat. t shall refer to time since both boats left their starting positions.

There are two ways to indicate the slope at any point along the curve. One is dy/dx. The other is (x-20t)/y, since the faster boat is always pointing towards the slower boat. Thus dy/dx=y/(x-20t).

The above equation can also be derived this way (optional):

Pick a point on the curve and label it (x₁,y₁). The x₁ and the y₁ are functions of the current time, t₁. The tangent to the curve at point (x₁,y₁) passes through the point (20t₁,0), which is the current location of the 20kph boat. The equation of the line is y=mx+b, where m is the value of dy/dx at (x₁,y₁) and b is some constant chosen to make sure that (20_t1,0) is on the line. b must be -20*t₁*y'(x₁), where I've used y'(x₁) to mean the value of dy/dx evaluated at the point (x₁,y₁). The equation of the tangent line is therefore:

y = y'(x₁)*x - 20*t₁*y'(x₁)

Rearranging that, and phrasing it as a general statement rather than a particular point (the point x₁ was arbitrary, so this applies to every point on the curve).

dy/dx * (x-20t) = y
dy/dx = y/(x-20t)

Something interesting can be found if you treat x as a function of y, instead of the other way around:

dx/dy = (x-20t)/y

This is justifiable, though it may look like abuse of Leibniz, the derivative of the inverse of a function is the reciprocal of the derivative of the function.

Then multiply both sides by y:

y * dx/dy = x - 20t

And here's the killer... differentiate with respect to y, using the product rule on the left hand side:

(1 * dx/dy) + (y * d²x/dy² ) = dx/dy - (20 * dt/dy)

There's a dx/dy term on each side of the equation, and they now happily disappear.

[1] y * d²x/dy² = -20 * dt/dy

Now it's time to turn dt/dy into something we can work with. I split it up as follows:

dt/dy = (dt/ds) * (ds/dy)

Where s is the arc length, the distance traveled so far by the chasing boat.

dt/ds is known (it's the reciprocal of the chasing boat's speed ds/dt=30kph), and ds/dy can be expressed in terms of dx and dy by applying the arc length formula:

The formula for arc length is:

s=Integral of [ (1+(dx/dy)²)^1/2 ] dy

Take the derivative of both sides:

ds/dy=(1+(dx/dy)²)^1/2, remember that the square root can be positive or negative.

Here is another way to derive this:

ds² = dx² + dy²
(ds²)/(dy²) = (dx²)/(dy²) + 1
(ds/dy)² = (dx/dy)² + 1

sqrt both sides...

|ds/dy| = sqrt((dx/dy)² + 1)
ds/dy = +/-sqrt((dx/dy)² + 1)

We need to choose a sign here. Look which way the curve goes. As t increases, y decreases. At the same time, s (the distance traveled by the chasing boat) increases. Therefore, ds/dy must be negative.

ds/dy = -sqrt((dx/dy)² + 1)

Substituting all this back into the equation marked [1], you get:

y * d²x/dy² = -20 * (1/30) * -sqrt((dx/dy)² + 1)
y * d²x/dy² = 2/3 * sqrt((dx/dy)² + 1)

That's a second-order differential equation, but not one of the extremely difficult kind. Rewrite it with a new variable u=dx/dy, and it becomes a first-order diffeq:

y * du/dy = 2/3 * sqrt(u²+1)

Breathe a sigh of relief, because that's separable. (I spent most of my solution time needlessly hacking away at this with more advanced diffeq techniques before I noticed it was separable. D'oh!)

     du         2    dy
  ----------- = - * ----
  sqrt(u^2+1)   3     y

ln(u + sqrt(u²+1)) = 2/3 * ln y + K

I've dispensed with the absolute value stuff, since u+sqrt(u²+1) can't be negative, and y isn't negative either, for the part of the curve that's relevant to the original question.

exp() both sides...

u + sqrt(u²+1) = C*y^2/3 K was an integration constant; C is exp(K).

To find C, use the point (0,50) on the curve. This is at t=0, when the target boat is due south of the chasing boat. The curve's tangent at that point is vertical, so u=dx/dy=0. Plug in u=0 and y=50:

0+1 = C*50^2/3
C = 1/50^2/3

Put that back in...

u + sqrt(u²+1) = (y/50)^2/3

We've still got some differential equations work to do.. u is a derivative! argh, a derivative in a radical. But it works itself out, just rationalize it:

sqrt(u²+1) = (y/50)^2/3 - u
u²+1 = (y/50)^4/3 - 2u(y/50)^2/3 + u²

Lucked out again. Cancel u²'s...

1 = (y/50)^4/3 - 2u(y/50)^2/3
2u(y/50)^2/3 = (y/50)^4/3 - 1
2u = (y/50)^2/3 - (y/50)^-2/3
u = 1/2*[ (y/50)^2/3 - (y/50)^-2/3 ]
dx/dy = 1/2*[ (y/50)^2/3 - (y/50)^-2/3 ]

A simple integration:

x = 1/2 * [ 3/5*(y/50)^5/3*50 - 3*(y/50)^1/3*50 ] + K

Factor out those 3's and 50's

x = 75 * [ 1/5*(y/50)^5/3 - (y/50)^1/3 ] + K

Again look at the starting point. When x is 0, y is 50.

0 = 75 * [ 1/5*1^5/3 - 1^1/3 ] + K
0 = 75 * [ 1/5 - 1 ] + K
0 = -60 + K
60 = K

x = 75 * [ 1/5*(y/50)^5/3 - (y/50)^1/3 ] + 60

That's the path the chasing boat follows. If that curve has a name, I don't know what it is.

The chasing boat intercepts the other boat when y=0.

x = 75 * [ 1/5*0 - 0 ] + 60
x = 60

The boats meet at the point (60,0). Since the slower boat is traveling due east at 20kph, it must have taken 3 hours to get there.

It takes 3 hours for the chasing boat to catch the other boat.

Below is another solution provided by Tristan Simbulan:

let t = s/30, distance traveled by first boat is 20 * s/30 = 2/3 * s dy/dx=y/( x - 2/3 * s), at t = 0 dy/dx is 50/0 is undefinable.

At t=0 dx/dy=0/50=0 is definable.

dx/dy = (x-2/3*s)/y, y * dx/dy = x - 2/3 * s, take second derivative.

Let first derivative dx/dy = p and second derivative dp/dy, ds/dy = sqrt(1 + (dx/dy)^2) = sqrt( 1+ p^2 ) dy/y = -3/2 * dp/sqrt(1+ p^2), lny + C = -3/2 * ln( p + sqrt( 1 + p^2 )), substitute initial condition y = 50, p = 0.

C = -ln 50. p = dx/dy = 0 y = { p + sqrt(1 + p^2 ) }^(-3/2) * 50 = { dx/dy + sqrt( 1 + (dx/dy)^2) }^(-3/2) * 50

Integration result: 2 * 50^(2/3)x =50^(4/3) * y^(1/3) * 3 - { y^(5/3) * 3/5 } + C C = 12/5 * 50^(5/3) at initial x = 0 y = 50 substitute value of C and you get equation of the curve.

At y = 0, x = 60 . divide distance by rate = time = 60/20 = 3 hours

Note: For the general case in which the slow boat travels at a k.p.h, the fast boat travels at b k.p.h., and the initial distance is d kilometers the formula of the curve is:

x = 1/2*[ db/(a+b)*(y/d)^(a+b)/b - db/(b-a)*(y/d)^(b-a)/b + db/(b-a) - db/(a+b) ]

The time to interception is [db/(b-a) - db/(a+b)]/2a.

Problem 31 Answer

Problem 31 Solution

Below is another approach:

The value of $1 invested for n years at interest rate i, compounded x times per year is (1+(i/x))^nx.

In this case f(x) = the value of $1 invested at 100% interest compounded n times per year = (1+(1/x))^x, where x approaches infinity.

The following table shows values of f(x) for various values of x:

     x              f(x)
-----------     ----------
          1	2.00000000
          4	2.44140625
         12	2.61303529
         52	2.69259695
        365	2.71456748
      1,000	2.71692393
     10,000	2.71814593
    100,000	2.71826824
  1,000,000	2.71828047
 10,000,000	2.71828169
100,000,000	2.71828180

As x approaches infinity it can be seen that f(x) approaches e =~ 2.71828182846.

Problem 32 Answer

The shadow over your yard is (540^2/3-36)^1/2 =~ 5.505698 feet.

The angle formed is approximately 47.46 degrees.

Problem 32 Solution

Let y be the shadow over neighbor's yard.

By similar triangles: x/(36+x²)^1/2=(x+y)/15.

y=15x/(36+x²)^1/2-x.

y^'=(((15*(36+x²)^1/2)-(15x²*(36+x²)^-1/2)) / 36+x²) - 1.

Then set y^'=0 and solve for x.

Finally subsitute x in the similar triangles formula to solve for y.

Problem 33 Answer

Problem 33 Solution

Mark Perkins suggested an alternative solution:

p = 1/6 + (5/6)*(1-p)
6p = 1 + 5 - 5p
11p = 6
p = 6/11

Problem 34 Answer

Problem 34 Solution

In this specific case this is [(.1)(.9)]/[(.1)(.9)+(.9)(.1)] = 1/2.

Problem 35 Answer

Problem 35 Solution

The density function of the shorter lived life bulbs is f(x)=1/100 * e^-x/100.

The density function of the shorter lived life bulbs is f(x)=1/200 * e^-x/200. The survival is 1 minus the integral of the density function. S(x)=e^-x/100, and S(x)=e^-x/200.

The survival function for the entire group is (e^-x/100)⁵ * (e^-x/200)⁵.

The density function for the group is the derivitive of 1 minus the survival function = e^-3x/40.

The mean for the group is the integral from 0 to infinity of x times the density function for the group = 40/3 hours.

Problem 36 Answer

Problem 36 Solution

The density function of the shorter lived life bulbs is f(x)=1/100 * e^-x/100.

The density function of the longer lived life bulbs is f(x)=1/200 * e^-x/200.

The probability of any given 100 hour bulb burning out first is the integral from 0 to infinity of (1/100)*e^-x/100*(e^-x/100>⁴*(e^-x/200)⁵ =

1/100 * integral e^-15x/200 =

(1/100)*(200/15)=2/15.

The probability that ANY of the 100 hour bulbs burn out first is five times this answer, or 10/15=2/3.

Problem 37 Answer

Problem 37 Solution

One minus this figure is the probability that there is at least one common birthday. The probability that there is at least one common birthday in 22 people is 47.57%, for 23 people it is 50.73%. Since the percentage must be at least 50% the answer is 23.

Problem 38 Answer

Problem 38 Answer

Problem 38 Answer

Problem 39 Solution

All multiples of 25 end in either 00, 25, 50, or 75. The only one of these composed of 0's and 1's is obviously 00, so the answer must end in 00. The hard part is finding a series of 0's and 1's preceeding the 00 that will make the entire number divisible by 9.

If you didn't already know the following trick then this problem would be very hard. If you did know it then the problem was was likely very easy. The trick is that if the sum of digits of a number is divisible by 9 then the number itself is also divisible by 9. Note that this is true for 3 also. For example the number 17685 is divisible by 9 because 1+7+6+8+5=27, and 27 is divisible by 9.

To prove this let's consider any five digit number, abcde. This number can be expressed as follows.

a*10000 + b*1000 + c*100 + d*10 + e =

a*(9999+1) + b*(999+1) + c*(99+1) + d*(9+1) + e*1 =

a*9999 + b*999 + c*99 + d*9 + a + b + c + d + e =

9*(a*1111 + b*111 + c*11 + d*1) + a + b + c + d + e

Thus 9*(a*1111 + b*111 + c*11 + d*1) is a multiple of 9. So if a+b+c+d+e is also a multiple of 9 then the entire number must be a multiple of 9. Note also that the remainder of abcde/9 is the same as the remainder of (a+b+c+d+e)/9.

The smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end results in the answer to the problem: 11,111,111,100.

Problem 40 Answer

Problem 40 Solution

t=(4-x)/6 + (1+x²)^1/2/2.

Set the derivative equal to 0:

dt/dx= -1/6 + 1/2 * 1/2 * 2x * (1+x²)^-1/2 = 0.

The solution is x=1/8^1/2.

y=4-x.

Problem 41 Answer

Problem 41 Solution

Let V_t = V = The volume in the tank at time t.

Let K_x = Some constant of integration.

V_t+Δt = V_t + 1*Δt - (Δt/100)*V_t.

dv/dt = (V_t - V_t+Δt) / Δt = 1 - V/100.

dv = (1 - V/100) dt

100/(100-V) dv = dt

Integrate each side:

-100*ln(100-V)=t + K₁

exponentiate each side:

(100-V)^-100 = K₂ * e^t

100-V = e^-t/100 * K₃

V = 100 - K₃*e^-t/100

We know V=0 when t=0, thus K₃=100

V = 100*(1-e^-t/100)

Substitute 50 for V and it is not difficult to solve for t=100*ln(2).

Tristan Simbulan sent in the following solution you may find better:

Let v be the volume at time t and let dv/dt the volume rate of change at time t.

Volume rate of change = rate in - rate out at time t. rate in = 1 gal. per second, rate out = v/100 * 1 gal. per sec.

Therefore dv/dt = 1 - v/100 * 1 ====> dv = 1/100 * (100 - v ) dt ======> dv/(100 - v ) = dt/100 - ln ( 100 - v ) = t/100 + C.

At t = 0 and v = 0, C = - ln 100. v = 100 * { 1 - e^(-t/100) } , let v = 50 .

Find value of t: e^(t/100) = 2, t = 100 * ln 2

Problem 42 Answer

Problem 42 Solution

Problem 43 Answer

Problem 43 Solution

Let a be the angle between the line from the yellow house to the north gate and the line from the yellow house to the blue house.

sin(a) = r/(r+3)

tan(a) = 9/(2r+3)

cos(a) = sqr(1-sin²(a)) = sqr(1-sin²(r/(r+3))) = sqr(6r+9)/(r+3)

tan(a) = sin(a)/cos(a) = (r/(r+3)) * ((r+3)/sqr(3*(2r+3))) = r/sqr(3*(2r+3))

Equating the two expressions for tan(a):

9/(2r+3) = r/sqr(3*(2r+3))

9 * sqr(3) * sqr(2r+3) = r*(2r+3)

9 * sqr(3) = r * sqr(2r+3)

243 = r²*(2r+3)

2r³ + 3r² - 243 = 0

A visit to The MathServ Calculus Toolkit will factor this to (2r-9)*(r² + 6r + 27) = 0.

The real solution is r=4.5

Here are two more solutions presented by Heng Cheng Suang:

I would like to suggest another soln to problem 43.
- just by using pythagoras thm (and/or similar triangles)
  the solution may be solved too.

Let the triangles be labelled as

        A
        |\
        | \E          It is noted that angle AED is 90 deg,
        | /\          and EC is 9 units (properties of tangents)
        |/  \
       D|    \
        |     \    
        --------
        B       C      


USING ONLY PYTHAGORAS THM
-------------------------------------------------------------------------
        
Based on triangle AED (By pyt. thm.),                           STEP 1

             AE = sqrt[(3+r)^2 - r^2)] = sqrt(9+6r)

Based on triangle ABC (By pyt. thm.),                           STEP 2

             (3+2r)^2 + 9^2 = [sqrt(9+6r) + 9]^2
Solving                                                         STEP 3
             4r^4 + 12r^3 + 9r^2 - 486r - 729 = 0
                                            r = 4.5


                                   OR 

-------------------------------------------------------------------------
USING SIM TRIANGLES PROPERTIES
-------------------------------------------------------------------------
STEP 1 same

STEP 2 is replaced by
             9 / (3 + 2r)   = r / sqrt (9 + 6r)              
            
STEP 3 same

I would like to thank Elliot Hunter for suggesting this problem which originally appeared the the November 1937 issue of Popular Science Monthly. The solution is by Dr. Anthony of the math forum.

Problem 44 Answer

Problem 44 Solution

The payoff for getting x heads followed by a tails is $2^x.

The expected return is the sum of the products of the probabilities and the payoffs is the sum for x=0 to infinity of (1/2) * (1/2)^x * 2^x.

(1/2)^x * 2^x = 1. So the answer is an infinite sum of 1/2's, or infinity.

This problem is a well known paradox in mathematics. The final result must be finite yet the expected value is infinite.

Problem 45 Answer

Problem 45 Solution

Problem 46 Answer

Problem 46 Solution

Next determine the expected number of turns given such barriers using the same method of interlocking expected values as in the ant and spider problem. The answer turns out to be 2x.

Thus, when x is infinity the expected number of turns must be 2*infinity=infinity.

Related question: What is the expected number of turns it would take a random walker on a number line to revist any point? The answer is here.

Problem 47 Answer

Problem 47 Solution

Let x be the height of the pentagon and a be the length of the extensions of the sides, as shown in the following diagram:

Using the Pythagorean formula we know x² + (a+1/2)² = (a+1)², or

(1) x² = a + 3/4.

Now consider the added two chords in the next diagram:

The isosceles triangle formed by these two chords and one side of the pentagon is congruent to the triangle formed by two extensions of length a and the same side of the pentagon.

Thus the distance between two non-adjacent corners of the pentagon is also a. Consider the chord going from either corner of the base to the top of the pentagon. We can use the Pythagorean formula, with this chord as the hypotenuse to obtain:

x² + (1/2)² = a².

Substituting for x² from above we can solve for a:

a + 3/4 = a² - (1/2)², or

a² - a - 1 = 0, or

a=(1+sqr(5))/2.

Now that we know a we can solve for x:

x² = ((1+sqr(5))/2)² - 1/4, or

x² = (5+2*sqr(5))/4, or

x = sqr(5+2*sqr(5))/2 =~ 1.538842

Old Solution

• sin(5x)=5*sin(x)-20*sin³(x)+16sin⁵(x).

• tan(x)=sin(x)/(1-sin²(x))^1/2.

• The interior angles of an n-gon are each 180*(n-2)/n.

With some simple trig we can express the height as 1/2*[1/sin(36) + 1/tan(36)].

From the formula in the hint we can express tan(36) as sin(36)/sqr(1-sin²(36)).

After a few simple steps we can express the height as (1) 1/2*[(1+sqr(1-sin²(36))/2*sin(36)].

Next lets work on expressing sin(36).

From the formula in the hint we know sin(180) = 16*sin⁵(36) - 20*sin³(36) + 5*sin(36)

= 16*sin⁴(36) - 20*sin²(36) + 5

= 16x⁴ - 20*x² + 5 (where x=sin(36)).

= 16y² - 20*y + 5 (where y=x²).

y = (5-sqr(5))/8.

x = sqr((5-sqr(5))/8) = sin(36).

Now put this expression for sin(36) into equation (1):

1/2 * [ (1+sqr(1-(5-sqr(5))/8)) / sqr((5-sqr(5))/8) ] =

1/2 * [ (1+sqr(3+sqr(5))/8)) / sqr((5-sqr(5))/8) ] =

1/2 * [ sqr(8/(5-sqr(5))) + sqr((3+sqr(5))/(5-sqr(5))) ] =

1/2 * [ (sqr(8) + sqr((3+sqr(5)))/(5-sqr(5))) ] =~ 1.5389

Problem 48 Answer

Thanks to Mark Hoyle for some help with this problem.

Problem 48 Solution

This solution is going to be hard to explain without pictures so bear with me. Call any corner of the dodecahedron a. Call the three corners surrounding a to be b, c, and d. Call q the point at the center of triangle bcd. Call p the point at the center of the dodecahedron.

From problem 123 we know the distance from two non-adjacent corners in a pentagon to be (1+sqr(5))/2. Knowing this and that cos(30)=sqr(3)/2 we can determine that BQ=(sqr(3)+sqr(15))/6.

It is given that AB=1, using the pythagorian formula we can deduce AQ from AB and BQ, which we know. After going through the math we find AQ to be sqr((3-sqr(5))/6).

Lets review what we know so far:

AQ + QP = AP
BP = AP
BQ² + QP² = BP²
BQ = (sqr(3)+sqr(15))/6
AQ = sqr((3-sqr(5))/6)

So:

PQ² + BQ² = BP², or

PQ² + ((sqr(3)+sqr(15))/6)² = BP², or

(BP - sqr((3-sqr(5))/6))² + ((sqr(3)+sqr(15))/6)² = BP² (substituting sqr((3-sqr(5))/6) + PQ = BP)

After working through the math we get BP = sqr(6*(3+sqr(5)))/4

From problem 123 we know the distance from the corner of a pentagon to the center of the pentagon is 2*sqr(5+2*sqr(5))/(5+sqr(5)) .

Lets call the center of a pentagon E, where B is one of the corners of the pentagon. We now know the distance from B to P and we know the distance from E to B from problem 123. Next we must work through the pythagorean formula again to solve for EP, which turns out to be sqr((25+11*sqr(5))/40) =~ 1.113516.

The volume of the dodecahedron is composed of 12 pyramids. Each pyramid has base sqr(25+10*sqr(5))/4 and height (25+11*sqr(5))/40. The volume of each of these pyramids is (1/3) * sqr(25+10*sqr(5))/4 * sqr((25+11*sqr(5))/40)

The volume of the entire dodecahedron is 4 * sqr(25+10*sqr(5))/4 * sqr((25+11*sqr(5))/40)

= sqr(25+10*sqr(5)) * sqr((25+11*sqr(5))/40) =

(1/sqr(40)) * sqr(625+275*sqr(5)+250*sqr(5)+550) =

(1/sqr(40)) * sqr(1175+525*sqr(5)) =

(5/sqr(40)) * sqr(47 + 21*sqr(5)) =

(sqr(10)/4) * sqr(47 + 21*sqr(5)) =

(sqr(5)/4) * sqr(94 + 42*sqr(5)) =

(sqr(5)/4) * sqr(49 + 42*sqr(5) + 45) =

(sqr(5)/4) * sqr(((7+3*sqr(5))²) =

(sqr(5)/4) * (7+3*sqr(5)) =

(15+7*sqr(5))/4 =~ 7.663119

Thanks to Mark Hoyle for some help with this problem.

Problem 49 Answer

Problem 49 Solution

However, if you were willing to not touch your investment and make the payments out of your pocket there is still a problem. You have to consider the time value of money. Lets assume inflation equal to the 6% interest rate, compounded monthly. The time value of the car payments made would be $10,364.77, however the time value of your $13,488.50 after five years would only be $10,000. In other words after five years your initial investment of $10,000 has gone down in value more than the interest savings.

Note: The payments on a loan of $x, at intrest rate i, over n periods is $xi/(1-vⁿ), where v=1/(1+i). In this case x=$10,000, i=.075/12=.00625, and n=60.

Problem 50 Answer

Problem 50 Solution

For a series to last exactly n games at least one team must have won exactly 3 out of the last (n-1). Thus:

The probability of the series lasting 4 games is 1/8.

The probability of the series lasting 5 games is 1/4.

The probability of the series lasting 6 games is 5/16.

The probability of the series lasting 7 games is 5/16.

The expected number of games played in team A's home field is: 2*(1/8) + 2*(1/4) + 3*(5/16) + 4*(5/16) = 188/64.

The expected number of games played in team B's home field is: 2*(1/8) + 3*(1/4) + 3*(5/16) + 3*(5/16) = 184/64.

Problem 51 Answer

The answer is 1/6.

Problem 51 Solution

Please see my solution to problem 51 (PDF).

Problem 52 Answer

Problem 52 Solution

1. Boy then boy
2. Boy then girl
3. Girl then boy
4. Girl then girl

Note: This is similar to the question in problem 16 which approaches the solution another way.

Problem 53 Answer

Problem 53 Solution

To find the answer we must solve I(0,1) I(0,1) I(0,1) I(0,1) sqr((x-a)²+(y-b)²) db da dy dx.

Even the first integration of this is very messy, requiring a big table of integrals, and I think at best a small book of integrals would only take you through the second one. So we need to be more creative to make the problem more managable. First we introduce a change of variables:

• u=x-a
• s=x
• v=y-b
• t=y

Or:

• x=s
• y=t
• a=s-u
• b=t-v

(1): Now we have I(0,1) I(t,t-1) I(0,1) I(s,s-1) sqr(u² + v²) * |J| du ds dv dt

Where |J| = the Jacobian =

| db/du db/ds db/dv du/dt |   |  0  0 -1  1 |
|                         |   |             |
| da/du da/ds da/dv da/dt |   | -1  1  0  0 |
|                         | = |             | = -1
| dy/du dy/ds dy/dv dy/dt |   |  0  0  0  1 |
|                         |   |             |
| dx/du dx/ds dx/dv dx/dt |   |  0  1  0  0 |

Now lets solve the following inner two integrals of equation (1):

(2): I(0,1) I(s,s-1) sqr(u² + v²) du ds.

The area of integration is bounded by the quadrilateral with vertices (0,0), (1,1), (0,1), and (-1,0), with s being the horizontal axis and u being the vertical.

Changing the order of integral we get:

I(-1,0) I(0,u+1) sqr(u² + v²) ds du + I(0,1) I(u,1) sqr(u² + v²) ds du =

I(1,0) sqr(u² + v²) * (1+u) du + I(0,1) sqr(u² + v²) * (1-u) du =

-I(1,0) sqr(p² + v²) * (1-p) dp + I(0,1) sqr(u² + v²) * (1-u) du (letting p=-u in the first integral) =

I(0,1) sqr(p² + v²) * (1-p) dp + I(0,1) sqr(u² + v²) * (1-u) du (Notice the two terms are equal)=

2* I(0,1) sqr(u² + v²) * (1-u) du.

Now put this into equation (1), and moving the constant terms to the left:

-2* I(0,1) I(t,t-1) I(0,1) sqr(u² + v²) * (1-u) du dv dt =

-2* I(0,1) I(0,1) I(t,t-1) sqr(u² + v²) * (1-u) dv dt du (changing the order of integration)=

-2* I(0,1) (1-u) I(0,1) I(t,t-1) sqr(u² + v²) dv dt du =

Note that the inner two integral are of the same form as equation (2), thus:

4* I(0,1) I(0,1) sqr(u² + v²) * (1-u) * (1-v) dv du =

4* I(0,1) I(0,1) sqr(u² + v²) * (1-u) * (1-v) du dv (changing the order of integration)

Next we make another substitution:

• u=r*sin(w)
• v=r*cos(w)

This time the Jacobian shall equal:

| du/dr du/dw |   | sin(w)  r*cos(w) |
|             | = |                  | = -r
| dv/dr dv/dw |   | cos(w) -r*sin(w) |

Thus the integral as a function of w and r is:

-8 * I(0,pi/4) I(0,1/cos(w)) -r * sqr(r*sin²(w) + r*cos²(w)) * (1-r*cos(w)) * (1-r*sin(w)) dr dw.

Note that we halved the area of integration, going from 0 to pi/4 as opposed to pi/2, and multiplying by 8 instead of the 4. This is permissible because of the symetry of the two regions.

= 8 * I(0,pi/4) I(0,1/cos(w)) r² * (1-r*cos(w)) * (1-r*sin(w)) dr dw.

= 8 * I(0,pi/4) I(0,1/cos(w)) r² - r³*(cos(w)+sin(w)) + r⁴*(cos(w)+sin(w)) dr dw.

= 8 * I(0,pi/4) r³/3 - r⁴*(cos(w)+sin(w))/4 + r⁵*(cos(w)+sin(w))/5 ( from 0 to 1/cos(w) ) dw.

= 8 * I(0,pi/4) (3*cos³(w))^-1 - (cos(w)+sin(w))/(4*cos⁴) + (cos(w)+sin(w))/(5*cos⁵(w)) dw.

= 8 * I(0,pi/4) (3*cos³(w))^-1 - (4*cos³(w))^-1 - sin(w)*(4*cos⁴(w))^-1 + sin(w)*(5*cos⁴(w))^-1 dw.

= 8 * I(0,pi/4) (12*cos³(w))^-1 - sin(w)*(20*cos⁴(w))^-1 dw.

(3) = 2/15 * I(0,pi/4) 5/cos³(w) - 3*sin(w)/cos⁴(w) dw.

Next we must dust off our table of integrals to help us with the integral of cos^-3(w).

I dx/cosⁿ(x) = (1/(n-1)) * sin(x)/cos^n-1(x) + ((n-2)/(n-1)) * I dx/cos^n-2(x).

In our case n=3 so,

I dx/cos³(x) = (1/2) * sin(x)/cos²(x) + (1/2) * I sec(x) dx.

Now lets look up the integral of sec(x):

I sec(x) dx = (1/2) * ln((1+sin(x))/(1-sin(x))).

Now we are ready to integrate cos^-3(w):

I(0,pi/4) cos^-3(w) dw = sin(w)/(2*cos²(w)) (from 0 to pi/4) + (1/2)*I(0,pi/4) sec(x) dx.

= (1/2)*(2^-1/2/2^-1 - 0) + (1/4)*ln((2+sqr(2))/(2-sqr(2))).

= (1/2)*sqr(2) + (1/4)*ln(((2+sqr(2))/2+sqr(2))/((2-sqr(2))*(2+sqr(2))).

= 1/sqr(2) + (1/4)*ln((4+4*sqr(2)+2)/(4-2)).

= 1/sqr(2) + (1/4)*ln(3+2*sqr(2)).

= 1/sqr(2) + (1/2)*ln(3+2*sqr(2))^1/2.

= 1/sqr(2) + (1/2)*ln(1+sqr(2)).

Now lets integrate sin(w)/cos⁴(w).

let m=cos(w), thus dm=-sin(w) dw.

I(0,pi/4) sin(w)/cos⁴(w).

= I(1,1/sqr(2)) -1*m^-4 dm.

= 1/(3*m³) from 0 to 1/sqr(2).

= (1/3) * (2*sqr(2) - 1).

Not it is time to plus these integrals into equation (3):

(2/15) * [ 5*(1/sqr(2) + (1/2)*ln(1+sqr(2))) + 3*((1/3) * (2*sqr(2) - 1)) ]

= (2/15) * [ 5*sqr(2)/2 + (5/2)*ln(1+sqr(2)) - 4*sqr(2)/2 + 1 ]

= (2/15) * [ sqr(2)/2 + (5/2)*ln(1+sqr(2)) + 1 ]

= (sqr(2) + 2 + 5*ln(1+sqr(2))) / 15.

Problem 54 Answer

Problem 54 Solution

Let h be the number of hours worked per week.

U = h*(80-h).

U^' = 80-2h.

Set the derivative equal to 0:

80-2h=0, thus h=40.

Problem 55 Answer

Problem 55 Solution

First reword the problem such that the coordinates of the bird's initial position are (100,0) and the nest is at (0,0). In polar coordinates the inital position of the bird is (100,0^o). Let p stand for the distance to the origin in polar coordinates, and let a stand for the angle.

Considering that the bird always flied towards the nest (p direction) and the wind blows north (y direction) the following graph shows the change in the variables with respect to a change of one infinitesimal unit of time in units of time. Note that as an angle approaches zero its arc approaches a straight line.

(1) dp/dt = -5 + 2sin(a).
(2) da/dt = 2cos(a)/p.

(3) dx/dt = -5cos(a).
(4) dy/dt = 2 - 5sin(a).

Divide eq. (3) by eq. (2):

(5) dx/da = (dx/dt)/(da/dt) = -5pcos(a)/2cos(a) = -5p/2 = -(5/2)*(x/cos(a)). (see graph above)

Rearrange:

(6) (2/x)dx = (-5/cos(a))da.

Integrate both sides:

(7) ln(x²) = -ln(sec(a)+tan(a))⁵ + C you might need an integral table for the integral of 1/cos(a).

Exp() both sides:

(8) x²*(sec(a) + tan(a))⁵ = C.

Put in the intial condition of x=100, a=0 to solve for the constant:

(9) x²*(sec(a) + tan(a)) = 10,000.

(sec(a) + tan(a))⁵ = (100/x)².

(sec(a) + tan(a)) = (100/x)^2/5.

Let b=(100/x)^2/5, then:

(10) sec(a) + tan(a) = b.

(11) (1+sin(a))/cos(a) = b (sec(a)=1/cos(a), tan(a)=sin(a)/cos(a)).

(12) (1+(1-cos²(a))^1/2) / cos(a) = b.

1+(1-cos²(a))^1/2 = b*cos(a).

(1-cos²(a))^1/2 = b*cos(a) - 1.

1-cos²(a) = b²*cos²(a) - 2*b*cos(a) + 1.

2*b*cos(a) = (1+b²)*cos²(a).

2*b = (1+b²)*cos(a).

(13) cos(a) = 2*b / (1+b²).

Substitute equation (13) into equation (3):

(14) dx/dt = -5*cos(a) = -10*b / (1+b²).

Rearranging:

dt = -(1+b²)/(10*b) dx.

-10 dt = b+(1/b) dx

(15) -10 dt = (100/x)^2/5 dx + (x/100)^2/5 dx, remember b=(100/x)^2/5

Integrating:

(16) Constant -10t = (5/3)*100^2/5*x^3/5 + (5/7)*(1/100)^2/5*x^7/5.

Put in the initial constants t=0 and x=100:

Const = (5/3)*100^2/5*100^3/5 + (5/7)*(1/100)^2/5*100^7/5.

Const = (5/3)*100 + (5/7)*100 = 5000/21.

So the function of x and t is:

(17) 5000/21 - 10t = (5/3)*100^2/5*x^3/5 + (5/7)*(1/100)^2/5*x^7/5.

The original question was how long does it take the bird to reach the nest, in other words what is t when x=0. By substituting x=0 in the above equation t = Const/10 = 500/21.

Tristan Simbulan sent in the following solution which you may find better:

The initial position of the bird is ( 100, 0 ) and the nest is ( 0, 0 ) I am using the velocity vector component parallel to y and x.

Let a be the angle between x axis and bird's line of sight towards ( 0, 0 ).

(1)The wind velocity minus y component of the bird's velocity = dy/dt = 2 - 5 sin a = 2 - 5 * y/sqrt(x^2 + y^2)

(2) Velocity x component = dx/dt = - 5 cos a = - 5 * x / sqrt( x^2 + y^2 )

(3) (dy/dt)/(dx/dt) = dy/dx = [ 5y -2 * sqrt(x^2 + y^2) ] / 5x 5xdy - 5ydx + 2 * sqrt(x^2 + y^2) dx = 0 Let y = vx, dy = vdx + x dv , the result by substitution is dv/sqrt(1 + v^2) + 2/5 * dx/x = 0.

ln[ v + sqrt(1 + v^2) ] + 2/5 ln x = C since v = y/x, by substitution ln [ y/x + sqrt(1 + y^2/x^2) ] + 2/5*ln x = C At y = 0 and x = 100 , then C = ln 100^(2/5).

The equation of the curve is sqrt(x^2 + y^2) = x^(3/5) *100^(2/5) - y.

Square both sides then y = [ 100^(4/5) * x^(3/5) - x^(7/5) ] / [ 2 * 100^(2/5)].

Solve (2) Velocity x component = dx/dt = - 5 cos a = - 5 * x / sqrt( x^2 + y^2 ) in terms of y.

By algebraic manipulation using fractional exponent and factoring the result is 100^(4/5) * x^(-2/5) dx + x^2/5 dx = - 10^(9/5) dt 5/3 *100^(4/5) * x^(3/5) + 5/7 * x^(7/5) = - 10^(9/5) * t + C At x =100 and t = 0 , C = 100^(7/5) * 50/21.

At x = 0 the result is 0 = - 10^(9/5) * t + C , substitute C = 100^(7/5) * 50/21 Solve: 10^(9/5)*t = 100^(7/5) * 50/21.

Therefore t = [100^(7/5)] / [10^(9/5)] * 50/21

t = [ 10^(14/5) ] / 10^(9/5) * 50/21 = 10 *50/21 = 500/21.

Go back to Problem 55
Go back to Shack's Math Problems

Problem 56 Answer

Problem 56 Solution

Let events A₁,...,A_k form a partition of the space S such that Pr(A_j)>0 for j=1,...,k, and let B be any event such that Pr(B)>0. Then for i=1,...,k,

Pr(A_i|B) = Pr(A_i)*Pr(B|A_i) / Sum for j=1 to k of Pr(A_j)*Pr(B|A_j).

The source of this problem is the May 1997 issue of the Actuarial Review. Applying this theorm:

B=boy selected
A₁=boy added
A₂=girl added
g=number of girls before baby is added.

Pr(A₂|B) = (1/2)*(2/(3+g)) / [(1/2)*(2/(3+g)) + (1/2)*(3/(3+g))] = 0.4

Problem 57 Answer

For n=3 (8 by 8) use four sets of the 4 by 4 configuration. Use three of them in such a way that the three extra squares form one extra triominoe, the fourth one should have the extra square in the corner.:

For each additional n repeat the n-1 by n-1 configuration four times, rotating in such as way as to create one extra triominoe.

Here is a nice graphic from Chris Johnson that explains the solution more clearly.

Thanks to Guy de Kindler for this one.

Problem 58 Answer

Problem 60 Answer

The second player should switch on 8 or less.

The expected return of the first player is ~ 0.081930 .

Problem 60 Solution

Following is an e-mail sent by Steve Schaefer, arguing that a random strategy on the part of player 1 is better than a fixed on.

Mike,

I agree with your answer to #60, but I'm unconvinced by your solution.  I'm
pretty sure that you didn't sufficiently verify the answer.

I agree that you found the best answer with integer solutions using a
spreadsheet, but I don't think you ruled out non-integer solutions.

I found the integer solution P1 trades when holding 7 or less and P2 draws
when holding 8 or less, but I wasn't convinced that P1's optimum might not
involve trading _sometimes_ when holding an 8.

As it turns out, the integer solution is the best for that choice of rules. 
If P1 trades the 8 with probablility p, while P2 continues to draw for 8 or
less, then P1's expected outcome is (180-2p)/2197.  If P1 holds the 7 with
probability q, while P2 continues to draw for 8 or less, then P1's expected
outcome is (180-29q)/2197.  In this case, minor variations from the optimum
integer solution result in a lower expected outcome for P1.

If, however, we change the rules for ties, we can get a different situation.
Consider rules where P2 wins all ties, but P2 is required to draw to break a
tie if P1 and P2 just traded equal cards.  (Incidentally, this seems to be
the best game on several different levels.)

The best integer solution for P1 is to trade 7 and below.  If P1 doesn't
trade, then P2 will draw to 8 and below.  The expected outcome is in P2's
favor, but it is only 3/2197.  If P1 trades eights as well, then P2 will
change strategies and draw to a nine; then P2's expected outcome improves to
5/2197.  But, it is wrong to conclude that P1 should never trade eights!

If P1 occasionally trades eights, then P2 can't afford to shift strategy. 
In fact if P1 trades eights five out of seven times (at random), then P2's
expected outcome is reduced to 1.571/2197.  (That's 11/15379).

If P1 executes this strategy, it doesn't matter whether P2 draws to a nine
or not.  Except that if P2 always draws to a nine, then P1 might want to try
cheating a little to trade eights less than five out of seven times.  If P1
could cheat completely and stop trading eights altogether, then P1 could
have an advantage of 7/2197!  Similarly, if P2 never draws to a nine, then
P1 might want to try cheating the other way.  If P1 could cheat completely
and always trade eights, then P1's advantage would only be 1/2197.

So, P2 needs a randomized strategy, too!  In fact, the equation for P2's
expected outcome is:

2197*y = x2*(-7+12*x1)+(1-x2)*(3-2*x1),

where x1 is the probability that P1 will trade an eight and x2 is the
probability that P2 will draw to a nine.  If each player tries to maximize
their own outcome against an opponent doing the same, then the two partial
derivatives will be zero:

0 = 2197* dy/dx1 = 14*x2 - 2
0 = 2197* dy/dx2 = 14*x1 - 10

So, we get x1 = 5/7 and x2 = 1/7.  That is, player 1 will always trade a
seven or less and will never trade a nine or more, but will trade an eight
five times out of seven at random.  When player 1 elects not to trade, then
player 2 will always draw to an eight or less and will always keep a ten or
more, but will draw to a nine one time out of seven at random.

As it turns out, this variation on the rules gives the smallest margin for
either player that I could find.

In short, your solution sounded like you only considered integer strategies,
whereas non-integer strategies are definitely possible.  If we pick more
interesting rules, then we get a more interesting solution.

-Steve

Problem 61 Answer

Problem 61 Solution

It may not seem reasonable to take the last step on faith but millions of computer trials have verified the answer.

Problem 62 Answer - Probabilities in Poker

 
                                           Approximate            
    Hand             Exact Probability     Probability
---------------   ----------------------   ------------
Royal Flush:              4 in 2,598,960   1 in 649,740 
Straight Flush:          36 in 2,598,960   1 in 72,193 
Four of a Kind:         624 in 2,598,960   1 in 4,165
Full House:           3,744 in 2,598,960   1 in 694
Flush:                5,108 in 2,598,960   1 in 509
Straight:            10,200 in 2,598,960   1 in 255
Three of a Kind:     54,912 in 2,598,960   1 in 47
Two Pair:           123,552 in 2,598,960   1 in 21
One Pair:         1,098,240 in 2,598,960   1 in 2.4
Nothing:          1,302,540 in 2,598,960   1 in 2.0

Problem 62 Solution - Probabilities in Poker

Probabilities of drawing

 
                                           Approximate            
    Hand             Exact Probability     Probability
---------------   ----------------------   ------------
Royal Flush:              4 in 2,598,960   1 in 649,740 
Straight Flush:          36 in 2,598,960   1 in 72,193 
Four of a Kind:         624 in 2,598,960   1 in 4,165
Full House:           3,744 in 2,598,960   1 in 694
Flush:                5,108 in 2,598,960   1 in 509
Straight:            10,200 in 2,598,960   1 in 255
Three of a Kind:     54,912 in 2,598,960   1 in 47
Two Pair:           123,552 in 2,598,960   1 in 21
One Pair:         1,098,240 in 2,598,960   1 in 2.4
Nothing:          1,302,540 in 2,598,960   1 in 2.0

Derivations

If you had x cards and chose y of them, the number of unique sets you could create would be x!/(y!*(x-y)!). For the purposes of this document I shall notate this is (x:y).

The number of ways to arrange 5 cards out of 52 is (52:5) = 2,598,960. The odds of drawing any given hand are the number of ways it can be arranged divided by the total number of ways to arrange five cards above. Below are the number of ways to arrange each hand.

Royal Flush

The number of different royal flushes are four (one for each suit).

Straight Flush

The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen, or King. Thus there are 9 possible high cards, and 4 possible suits, creating 9 * 4 = 36 different possible straight flushes.

Four of a Kind

There are 13 different possible ranks of the 4 of a kind. The fifth card could be anything of the remaining 48. Thus there are 13 * 48 = 624 different four of a kinds.

Full House

There are 13 different possible ranks for the three of a kind, and 12 left for the two of a kind. There are 4 ways to arrange three cards of one rank (4 different cards to leave out), and (4:2) = 6 ways to arrange two cards of one rank. Thus there are 13 * 12 * 4 * 6 = 3,744 ways to create a full house.

Flush

There are 4 suits to choose from and (13:5) = 1,287 ways to arrange five cards in the same suit. Then subtract the royal and straight flushes to avoid double counting. The total number of flushes is 4 * 1,287 - 4 - 36 = 5,108.

Straight

The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen,King, or Ace. Thus there are 10 possible high cards. Each card may be of four different suits. Then subtract the royal and straight flushes to avoid double counting. Thus the number of ways to arrange a straight is 10 * 4⁵ - 4 - 36 = 10,200.

Three of a Kind

There are 13 ranks to choose from for the three of a kind and 4 ways to arrange 3 cards among the four to choose from. There are (12:2) = 66 ways to arrange the other two ranks to choose from for the other two cards. In each of the two ranks there are four cards to choose from. Thus the number of ways to arrange a three of a kind is 13 * 4 * 66 * 4² = 54,912.

Two Pair

There are combin(13,2)=78 ways to choose 2 ranks out of 13 for the two pair. Then there are 11 ranks left for the singleton. For each pair there are combin(4,2)=6 ways to choose 2 suits out of 4. For the singleton there are 4 possible suits to chose from. So the total two pair combinations are 78*11*6*6*4 = 123,552.

There are 13 ranks to choose from for the pair and (4:2) = 6 ways to arrange the two cards in the pair. There are (12:3) = 220 ways to arrange the other three ranks of the singletons, and four cards to choose from in each rank. Thus there are 13 * 6 * 220 * 4³ = 1,098,240 ways to arrange a pair.

Nothing

There must be five different ranks represented, of which there are (13:5) = 1,287 possible combinations. Each rank has four cards to choose from. Finally subtract the number of straights, flushes, straight flushes, and royal flushes, to avoid double counting. Thus the number of ways to arrange nothing is 1,287 * 4⁵ - 4 - 36 - 5,108 - 10,200 = 1,302,540.

Problem 63 Answer

Problem 64 Answer

Problem 64 Solution

Let a be the distance from the point (0,0) to a point on the figure (known as a cardoid). Next consider the angle t formed from the points (-1,0), (0,0) and a at (0,0).

After working through the trigonomtry (this step is left to you) the length of a in terms of t is 2*(1-cos(t)).

The formula for arc length is the integral of ( a² + (da/dt)² )^1/2.

In this case it is the integral from 0 to 2*pi of:

( 4*(1-cos(t))² + 4*sin²(t) )^1/2. =

8^1/2 * (1 - cos(t))^1/2.

After consulting a table of integrals we find that the integral of (1 - cos(t))^1/2 is -2*2^1/2*cos(t/2).

Taking the integral the answer works out to 16.

I'd like to thank Andrea for correcting an earlier mistake in my answer.

Problem 65 Answer

If the y/x is an integer then the answer is y/x, otherwise it is y/z.

Problem 66 Answer

Problem 66 Solution

Next order the two points such that p₁ <= p₂.

Next put the two points in cartesial coordinates (x,y,z).

Through a little geometry it can be found that:
(x₁,y₁,z₁) = (cos(p₁)*sin(q₁), sin(p₁)*sin(q₁), cos(q₁)) and
(x₂,y₂,z₂) = (cos(p₂)*sin(q₂), sin(p₂)*sin(q₂), cos(q₂)).

The angle between two vectors is defined as cos(t)=(v dot w)/(|v| * |w|).

Let v donate the vector from (0,0,0) to (x₁,y₁,z₁) and
let w donate the vector from (0,0,0) to (x₂,y₂,z₂).

The dot product of v and w is:
cos(p₁)*sin(q₁)*cos(p₂)*sin(q₂) + sin(p₁)*sin(q₁)*sin(p₂)*sin(q₂) + cos(q₁)*cos(q₂).

If we rotate the longitude of both points by p₁ then p₁ becomes 0 and p₂ becomes p₂-p₁, in which case the dot product is:
cos(0)*sin(q₁)*cos(p₂-p₁)*sin(q₂) + sin(0)*sin(q₁)*sin(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂). =
sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂).

Now, both |u| and |v| must be 1 since they lie on the surface of a unit sphere. Thus the angle between the two points must be cos^-1[sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂) / (1 * 1). = sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂)].

This angle is the ratio of the distance along the surface between the two points to the circumference of the earth. Thus the distance is the angle times the circumference of the earth which equals:
cos^-1[(sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂))] * r.

Problem 69 answer

Problem 69 Solution

Let x stand for the point at which the first player is indiffent between standing on one turn or taking another. Let z be the score on the first spin. The probability of winning by taking another turn is the integral from z to 1 of t² which equals (1-z³)/3. By equating this with the probability of winning by not taking another turn, z², you find the indifference point: (1-z³)/3= z². So the indifference point is x where x³+x²-1=0, x =~ 0.53208889 .

The probability of winning is the integral from 0 to x of (1-z³)/3 plus the integral from x to 1 of z². After integration this answer is (x - x⁴/4)/3 + (1 -x ³)/3 = (-x⁴/4 -x³ + x + 1)/3 =~ .45380187 .

Problem 70 Answer

Note: In all fairness to the state I should mention that you can buy three tickets for the price of two, increasing the expected return to 56.97 cents (but still a waste of money).

Problem 70 Solution

Lets assume the chosen numbers are 1-2-3-4-5. Of the 19,575,738 ways to arrange the six drawn balls how many will fit the drawn numbers to make a winner? Take the ratio of that number of 19,575,738 and you have the probability.

There are 34 ways to draw the 6 numbers to match 1-2-3-4-5 (you must match all of the first 5, the bonus ball can be any of the remaining 34).

The number of ways to match four plus the bonus ball are 170 (the bonus ball can be any of the 5 you got right, and the last ball can be any of the remaining 34).

The number of ways to match four w/o the bonus ball are 5610 (5 different ways to arrange the 4 correct numbers * the 5th number can be any of the remaining 34 * the bonus ball can be any of the remaining 33).

Keep following the above logic for all possible winning combinations:

                 Ways     Probability           Expected
Type of win      to win   of winning   Payoff   return
--------------   ------   -----------  ------   --------
Pick 5               34   .0000017368  $50000   .0868
Pick 4 + BB         170   .0000086842    $600   .0052
Pick 4 w/o BB      5610   .0002865792    $400   .1146
Pick 3 + BB       11220   .0005731585     $30   .0172
Pick 3 w/o BB    179520   .0091705355     $15   .1376
Pick 2 + BB      179520   .0091705355      $2   .0183

Total            376074                         .3798

Note: In all fairness to the state I should mention that you can buy three tickets for the price of two, increasing the expected return to 56.97 cents (but still a waste of money).

Problem 71 Answer

Problem 73 Answer

Problem 73 Solution

Number   Probability
  of         of
Trials     Success
------   -----------
  1           p
  2           pq
  3           pq2
  4           pq3
  .           .
  .           .
  .           .

The mean number of trials (m) is: m = p + 2pq + 3pq² + 4pq³ + ...

mq = pq + 2pq² + 3pq³ + ...

m-qm= p + pq + pq² + pq³ + ...

m(1-q) = 1.

m = 1/(1-q) = 1/p.

Second, the answer to the problem can be express as the sum of the following:

• Number of trials to get a first toy
• Number of trials to get a second toy once you have one toy
• Number of trials to get a third toy once you have two toys
• Number of trials to get the final toy once you have three toys

Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.

Summing these yields 1 + 4/3 + 2 + 4 = 25/3.

I'd like to thank Michael Brasher for providing this solution. Below is my old solution which is much less elegant, I recommend you ignore it.

The probability that in j boxes there will be three or less kinds of toys is 4*(3/4)^j where j>3.

The probability that in j boxes there will be exactly three kinds of toys is 4*(3/4)^j - 12*(1/2)^j + 12*(1/4)^j. This is rather hard to explain why but basically there are 4 different sets of 3 toys, 6 different sets of 2 toys, and 4 different sets of 1 toy, use this information to not double count combinations of 2 prizes or 1 prize.

Thus the answer is 1/4*sum for j=3 to infinity of j*(4*(3/4)^j - 12*(1/2)^j + 12*(1/4)^j).

Theorem: The sum for j=1 to infinity of i*(1/n)ⁱ = n/(n-1)².

From there is is just simple math to find that the answer is 25/3.

Problem 74 Answer

I got this answer from a Usenet discussion on probability, I can't prove it myself at this time.

Problem 75 Answer

Here are some approxmiate probabilities for number of people n:

n       Prob(no matches)
--      ----------
2	50.000000%
3	33.333333%
4	37.500000%
5	36.666667%
6	36.805556%
7	36.785714%
8	36.788194%
9	36.787919%
10	36.787946%
11	36.787944%
12	36.787944%
13	36.787944%
14	36.787944%
15	36.787944%

Problem 75 Solution

There are 12! ways of drawing names.

Next determine the number of these 12! ways that have no matches.

Subtract from 12! the number of ways that one person matches, which would be 12*11! (12 people times 11! ways to arrange the other 11 names).

However this would be double counting those combinations in which two people match, which is (12:2)*10! = 12!/2!. Note that (x:y) means x choose y or x!/(y!*(x-y)!). So add this group back in.

Yet we must take out from those we put in those combinations in which three people match, which is (12:3)*9! = 12!/3!.

So keep repeating this process and the numerator or number of combinations in which there are no matches is 12! - 12! + 12!/2! - 12!/3! + 12!/4! - 12/5! ... + 12!/12!.

Divide this by the total number of combinatios, 12!:

1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! ... +1/12!.

This same method works for any n.

Note that it approaches 1/e.

Problem 76 Answer

Problem 76 Solution

In this case A is drawing a black marble and B is having already drawn a black marble.

Pr(A and B) = (1/2) * [(3/4)² + (1/4)²] = 5/16.

Pr(B) = 1/2.

Pr(A|B) = Pr(A and B)/Pr(B) = (5/16)/(1/2) = 10/16 = 5/8.

Problem 77 Answer

Problem 77 Solution

In this case Pr(A and B) = 5*(The integral from 0.4 to 0.6 of p²) =

5*(p³/3 from 0.4 to 0.6) =

(5/3)*(.216 - .064) =

19/75 =~ 0.2533333

Pr(B) = 5 * The integral from 0.4 to 0.6 of p = 0.5 .

Thus the answer is 38/75 =~ .5066667

Problem 78 Answer

Thanks to Marilyn Vos Savant for this one.

Problem 78 Solution

George can paint at a rate of 1/30th of a house per hour.

Combine their efforts and they can paint 1/20 + 1/30 = 1/12th of a house in an hour.

So if their rate is 1/12th of a house per hour it will take 12 hours to paint an entire house.

Thanks to Marilyn Vos Savant for this one.

Problem 80 Answer

Problem 81 Answer

Problem 81 Solution

Both X and Y should choose their strategy randomly. Let x be the probability that country X attacks by sea. Let y be the probability that country Y defends by sea.

The probability of a successful invasion is :
f(x,y)=.8xy + x(1-y) + (1-x)y + .6(1-x)(1-y) =
.8xy + x - xy + y - xy + .6 - .6x - .6y + .6xy =
-.6xy + .4x + .4y + .6 .

Y is obviously going to try to minimize the probability of a successful attack. Taking the derivative of f(x,y) with respect to y yields:

-.6x + .4 = 0.
x=2/3.

Thus X should attack by sea with probability 2/3 and by land with probability 1/3. Y should also defend by sea with probability 2/3 and by land with probability 1/3. The probability of a successful invation is -.6*(2/3)*(2/3) + .4*(2/3) + .4*(2/3) + .6 = 13/15.

Problem 82 Answer

Problem 82 Solution

First review the probability of throwing any given number on any given throw:

• 2: 1/36
• 3: 2/36
• 4: 3/36
• 5: 4/36
• 6: 5/36
• 7: 6/36
• 8: 5/36
• 9: 4/36
• 10: 3/36
• 11: 2/36
• 12: 1/36

From the above the probalility of rolling 7 is 6/36 = 1/6 and the probability of rolling an 11 is 2/36 = 1/18. These are the numbers that win on the first throw.

Next lets assume the point thrown on the first roll is a 4, what is the probability of throwing it again before a 7?

Let pr(x) stand for the probability of event x happening on any given roll. The answer is:

pr(4) +
pr(anything other than 4 and 7) * pr(4) +
pr(anything other than 4 and 7)² * pr(4) +
pr(anything other than 4 and 7)³ * pr(4) +
pr(anything other than 4 and 7)⁴ * pr(4) +
+ ...

Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 1-3/36-6/36 = 27/36 = 3/4.

pr(rolling a 4 before a 7)
= 1/12 + (3/4 * 1/12) + ((3/4)² * 1/12) + ((3/4)³ * 1/12) + ...
= 1/12 * sum for i = 0 to infinity of (3/4)ⁱ
= 1/12 * (1/(1-3/4))
= 1/12 * 4 = 1/3.

The probability of rolling a 10 is the same as the probability of rolling a 4 so pr(rolling a 10 before a 7) also equals 1/3.

Next assume the point thrown is a 5.

Pr(5) = 4/36 = 1/9, pr(anything other than 5 and 7) = 1-4/36-6/36 = 26/36 = 13/18.

pr(rolling a 5 before a 7)
= 1/9 + (13/18 * 1/9) + ((13/18)² * 1/9) + ((13/18)³ * 1/9) + ...
= 1/9 * sum for i = 0 to infinity of (13/18)ⁱ
= 1/9 * (1/(1-13/18))
= 1/9 * 18/5 = 2/5.

The probability of rolling a 9 is the same as the probability of rolling a 5 so pr(rolling a 9 before a 7) also equals 2/5.

Next assume the point thrown is a 6.

Pr(6) = 5/36, pr(anything other than 6 and 7) = 1-5/36-6/36 = 25/36.

pr(rolling a 6 before a 7)
= 5/36 + (25/36 * 5/36) + ((25/36)² * 5/36) + ((25/36)³ * 5/36) + ...
= 5/36 * sum for i = 0 to infinity of (25/36)ⁱ
= 5/36 * (1/(1-25/36))
= 5/36 * 36/11 = 5/11.

The probability of rolling an 8 is the same as the probability of rolling a 6 so pr(rolling a 8 before a 7) also equals 5/11.

Now we're ready to add all this together...

The probability of winning the pass line bet is:
pr(7) + pr(11) + pr(4)*pr(4 before a 7) + pr(5)*pr(5 before a 7) + pr(6)*pr(6 before a 7) + pr(8)*pr(8 before a 7) + pr(9)*pr(9 before a 7) + pr(10)*pr(10 before a 7)
= 6/36 + 1/18 + (1/12 * 1/3) + (1/9 * 2/5) + (5/36 * 5/11) + (5/36 * 5/11) + (1/9 * 2/5) + (1/12 * 1/3)
= 6/36 + 1/18 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36
= 330/1980 + 110/1980 + 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980
= 976/1980 = 244/495.

The expected return on this wager is the product of the probability of winning and the ratio of what you keep (including your original wager) to the original bet if you do win (in this case 2). Thus the expected return is 244/495 * 2 = 488/495 =~ 98.59%.

The house advantage is what the casino gets to keep, on average, which is 1 minus the expected return, which equals 1-(488/495) = 7/495 =~ 1.41%

In the appendix to this problem I show the house advantage on the don't pass bet.

Problem 83 Answer

Problem 83 Solution

The table below shows the total suffering given the four possibilities:

                      Your partner
                 -----------------------
                 confess     not confess
  
Y  confess:     10*2+10=30    0+20=20
o
u  not confess: 2*20+0=40     2*3+3=9

10p+20 = 31p+9
11 = 21p
p=11/21 =~ 52.4%.

Thus if you think your partner's probability of confessing is greater than 11/21 you should confess, otherwise you should keep your mouth shut.

Problem 84 Answer

First weigh any 4 pearls against any other 4.

If one side weighs heavier than the other then:

Label the pearls on the heavy side H, the pearls on the light side L, and the 4 other pearls G. Next, weigh two H pearls and two L pearls against 2 G pearls, 1 H pearl, and 1 L pearl (HHLL vs GGHL).

If the HHLL side goes down then either one of the two H pearls on the left side from step 2 is heavy or the L pearl on the right side of step 2 is light. Finally weigh ( HL vs GG ). If the HL side goes down the H pearl is heavy, if the HL side goes up the L pearl is light, if they stay the same the other H pearl is heavy.

If the HHLL side (from step 2) goes up then either one of the two L pearls on the left side from step 2 is light or the H pearl on the right side of step 2 is heavy. Finally weigh ( HL vs GG ). If the HL side goes down the H pearl is heavy, if the HL side goes up the L pearl is light, if they stay the same the other L pearl is light.

If the two sides from step 2 stay the same you have left one H pearl and one L pearl. Weigh ( HL vs GG ). If the left side goes down the H pearl is heavy, if it goes up the L pearl is light.

If the two sides from step one stay the same then label the 8 pearls used in step 1 as G and the 4 others pearls as B. For the second weighing weigh 3 B pearls against 3 G pearls ( BBB vs GGG).

If the left side goes down one of the B pearls is heavy. Weigh them 2 of them against each other the heavy side will have the heavy pearl, if the sides stay the same the other pearl is heavy.

If the left side goes up one of the B pearls is light. Weigh them 2 of them against each other the light side will have the light pearl, if the sides stay the same the other pearl is light.

If the two sides from the second weighing stay the same then the last B pearl is heavy or light. Use the third weighing against a good pearl to determine if it is heavy or light.

Problem 85 Answer

Problem 85 Solution

If only pirate 5 were left he would get everything.

If only pirate 4 and 5 were left then pirate 4 has no hope, regardless of what he suggests pirate 5 will vote to throw him overboard, and overboard he will go. Thus it doesn't make any difference what pirate 4 suggests.

If only pirates 3,4, and 5 were left then pirate 3 could count on pirate 4's support as long as pirate 4 got at least 1 coin. This is because if pirate 3 goes overboard pirate 4 knows he will get nothing. So pirate 3 would suggest (0,1,999) to which he and pirate 4 would vote 'yes.' Note that according to the problem pirate 4 would vote 'no' to (0,0,1000), even though a 'yes' vote would save his own life his pleasure of throwing pirate 3 over would be worth his own life.

If only pirates 2,3,4, and 5 were left then pirate 2 could count on pirate 3 voting 'no' unless pirate 3 got more than 999 coins, which would leave everybody else with 0, which obviously wouldn't pass. So pirate 2 should not count on pirate 3's support so he shouldn't waste any coins on him. If pirate 2 goes overboard he knows the money will be distributed (0,1,999). So if he offers 2 coins to pirate 4 and 1 coin to pirate 5 then he will get their support since they will come out better voting 'yes.' Thus pirate 2's suggestion would be (1,2,0,997) to which pirates 2, 4 and 5 would vote 'yes.'

Thus if all five pirates were left pirate 1 could count on pirate 2 voting 'no' since pirate 2 would get 997 if pirate 1 went overboard. So pirate 1 should not waste any coins on pirate 2 since he will vote 'no' anyway and concentrate on winning 2 votes from the other 3. If pirate 1 goes overboard the money will be distributed (1,2,0,997). So it would cost 2 coins to get pirate 5's support, 3 coins to get priate 4's support, and 1 coin to get pirate 3's support. Since he only needs to buy two votes the cheapest two are pirates 5 and 3. Thus pirate 1 would suggest (2,0,1,0,997) to which pirate 1, 3, and 5 would vote 'yes'.

Thanks to The Ultimate Puzzle Site which has also posted this problem, click here to see their solution if this one was too confusing.

Problem 86 Answer

Problem 86 Solution

f'(t)=t*f(t).

Let y=f(t).

dy/dt=ty.

dy=ty dt.

dy/y = t dt.

Integrating both sides:

ln(y) = t²/2 + C.

y = exp(t²/2 + C).

y = exp(t²/2). Since y=1 when t=0

Problem 87 Answer

Problem 87 Solution

The plane will run out of fuel after 10/3 hours.

The distance the plane can cover is the integral from 0 to 10/3 of (6,000,000/(20,000-3,000t)) dt.

= 6,000,000 * -1/3000 * ln(20,000-3,000t) from 10/3 to 0.

= -2,000 * ln(1/2)

= 2000 * ln(2) =~ 1386.3 miles .

Problem 88 Answer

First, surgeon 1 operates with glove A inside glove B.

Second, take glove A out of glove B.

Third, surgeon 2 operates using glove B.

Fourth, turn glove A inside out.

Fifth, surgeon 3 operates with glove A inside glove B.

Note: Another way to ask this is how do you have safe sex with three men, one woman, and two condoms?

Problem 89 answer

Problem 89 Solution

Solution 1

The house advantage in just the pass line bet is 7/495, see the solution to problem 82 for an explanation of this step.

There is a 1/3 chance that the come out roll will be a 2,3,7,11, or 12, click here if this step needs explanation.

There is a 2/3 chance that double odds will be taken. Given that $1 is bet on the pass line the expected money bet per round is 1/3*$1 + 2/3*$3 = $7/3. Since $1 is always bet on the pass line 1/(7/3)=3/7 of money bet, overall, is on the pass line, and the other 4/7 is on the odds.

Thus the house advantage is 3/7*(7/495) + 4/7*0 = 3/495 = 1/165. The players advantage is just the opposite of this, -1/165.

Solution 2

The player's advantage is the sum over all possible outcomes of the product of the probability of the outcome and the net profit or loss, divided by the sum over all possible outcomes of the expected amount bet.

Lets say the original wager on the pass line is 5. By summing the following possible outcomes you will have the expected gain or loss.

• Roll 2 on come out roll: 1/36 * -5 = -5/36 = -4950/35640
• Roll 3 on come out roll: 2/36 * -5 = -10/36 = -9900/35640
• Roll 4 on come out roll, then win: 3/36 * 3/9 * (5+20) = 225/324 = 24750/35640
• Roll 4 on come out roll, then lose: 3/36 * 6/9 * (-5-10) = 270/324 = -29700/35640
• Roll 5 on come out roll, then win: 4/36 * 4/10 * (5+15) = 320/360 = 31680/35640
• Roll 5 on come out roll, then lose: 4/36 * 6/10 * (-5-10) = -360/360 = -35640/35640
• Roll 6 on come out roll, then win: 5/36 * 5/11 * (5+12) = 425/396 = 38250/35640
• Roll 6 on come out roll, then lose: 5/36 * 6/11 * (-5-10) = -450/396 = -40500/35640
• Roll 7 on come out roll: 6/36 * -5 = -30/36 = 29700/35640
• Roll 8 on come out roll, then win: 5/36 * 5/11 * (5+12) = 425/396 = 38250/35640
• Roll 8 on come out roll, then lose: 5/36 * 6/11 * (-5-10) = -450/396 = -40500/35640
• Roll 9 on come out roll, then win: 4/36 * 4/10 * (5+15) = 320/360 = 31680/35640
• Roll 9 on come out roll, then lose: 4/36 * 6/10 * (-5-10) = -360/360 = -35640/35640
• Roll 10 on come out roll, then win: 3/36 * 3/9 * (5+20) = 225/324 = 24750/35640
• Roll 10 on come out roll, then lose: 3/36 * 6/9 * (-5-10) = 270/324 = -29700/35640
• Roll 11 on come out roll: 2/36 * 5 = 10/36 = 5/18 = 9900/35640
• Roll 12 on come out roll: 1/36 * -5 = -5/36 = -4950/35640

Now lets sum the expected amount wagered:

If the first roll is a 2,3,7,11, or 12 the wager is 5.
If the first roll is anything else the wager is 15.
Thus the expected wager is (1+2+6+2+1)/36 * 5 + (3+4+5+5+4+3)/36 * 15 = 60/36 + 360/36 = 420/36 = 35/3.

Thus the player's advantage is (-2520/35640) / (35/3) = -7560/1,247,400 = -378/62370 = -1/165.

In other words the house has an advantage of approximately 0.61%.

Problem 90 Answer

Problem 90 Solution

p * [ 1/(1+i) + 1/(1+i)² + 1/(1+i)³ + ... + 1/(1+i)³⁶⁰ ] =

p * [ (1 - 1/(1+i)³⁶⁰) * ( 1/(1+i) + 1/(1+i)² + 1/(1+i)³ + ... ) ].

Now recall that that the sum for n=1 to infinity of xⁿ = x/1-x, where x is less than 1, so we can further simplify:

p * [ (1 - 1/(1+i)³⁶⁰) * (1/(1+i))/(1-(1/(1+i))) =

p * [ (1 - 1/(1+i)³⁶⁰) / i = $100,000.

Solving for p (remember that i=.075/12):

p = $100,000 * i / (1 - 1/(1+i)³⁶⁰) =~ $699.21

For your own information here are what the monthly payments would be under various other interest rates:

0%	 $277.78
1%	 $321.64
2%	 $369.62
3%	 $421.60
4%	 $477.42
5%	 $536.82
6%	 $599.55
7%	 $665.30
8%	 $733.76
9%	 $804.62
10%	 $877.57
11%	 $952.32
12%	$1028.61
13%	$1106.20
14%	$1184.87
15%	$1264.44
16%	$1344.76
17%	$1425.68
18%	$1507.09
19%	$1588.89
20%	$1671.02
25%	$2084.58
30%	$2500.34
35%	$2916.76
40%	$3333.36
45%	$3750.01
50%	$4166.67
55%	$4583.33
60%	$5000.00
65%	$5416.67
70%	$5833.33
75%	$6250.00
80%	$6666.67
85%	$7083.33
90%	$7500.00
95%	$7916.67

Problem 91 Answer

Click here for specific answers given x and y.

Problem 92 Solution

Problem 92 Solution

cos(x)=AD/3, sin(x)=BD/3, cos(y)=CD/7, sin(y)=BD/7.

cos²(x)+sin²(x) = AD²/9 + BD²/9 = 1.

cos²(y)+sin²(y) = CD²/49 + BD²/49 = 1.

BD² = 9 - AD² = 49 - CD².

Remember that AD + CD = 9, or CD = 9 - AD.

9 - AD² = 49 - (9-AD)².

9 = 49 - 81 + 18AD.

AD = 41/18.

(41/18)² + BD² = 9.

BD = (9-(41/18)²)^1/2 = 1.95236

Others have pointed out an alternate solution using Heron's Formula, which states the area of a triange is sqr(s*(s-a)*(s-b)*(s-c)), where s=(a+b+c)/2, and a, b, and c are the lengths of the three sides.

In this case a, b, and c are 3, 7, and 9.

s=(3+7+9)/2=9.5

area=sqr(9.5*(9.5-3)*(9.5-7)*(9.5-9)) = sqr(1235/16) = 8.785641695

The area must also equal (1/2)*9*h, where h is the height.

So sqr(1235/16) = (1/2)*9*h

h=sqr(1235/16)*(2/9) = 1.95236

Problem 93 Answer

The answer is (7,000,000/30) × ln(8/1) seconds = 485,203 seconds = 5.6158 days.

Problem 93 Solution

Solution

Problem 94 Answer

Problem 94 Solution

The probability of 0 potholes is e^-6*6⁰/0! = e^-6.

The probability of 1 pothole is e^-6*6¹/1! = 6*e^-6.

The probability of 2 potholes is e^-6*6²/2! = 18*e^-6.

The probability of 3 potholes is e^-6*6³/3! = 36*e^-6.

Take the sum the answer is 61*e^-6 =~ 0.151204 .

This problem was taken from the May 1982 Actuary Exam 110, problem 20.

Problem 95 Answer

Second, A and B cross together and D returns (11 minutes consumed).

Third, C and D cross together (2 minutes consumed).

Thanks to Antonio Cordoba for this problem. He says that Microsoft applicants are tested with this one.

Problem 95 Solution

Problem 96 Answer

Problem 97 Answer

Problem 97 Solution

Since the radius of the lighthouse is 1 the length of the straight part of the chain is the same as x. Through a little trigonomtry we can get the coordinates of the dog given x as (cos(x)+x*sin(x),sin(x)-x*cos(x)). The distance from (0,0) to the dog, given x, conveniently works out to (1+x²)^1/2, using the length of a line segment formula.

Lets call y the angle fromed by (1,0), (0,0), and the location of the dog. Through some more trigonomty (again left up to you) we can solve for y in terms of x as y=x-tan^-1. Now we are ready to integrate.

The area in terms of polar coordinates is 1/2 * the integral over the range of the angle of the radius squared. We can not simply take the integral from 0 to Π of 1+x² because this is not the radius formed by the angle x, but the radius formed by y. We could take the integral over the range of the angle y but finding the coordinates of the dog in terms of y is very hard (I couldn't do it). However we can take the ingregal as x goes from 0 to Π of 1+x² multiplied by the change in y given a change a change in x. Since y=x-tan^-1(x). dy/dx = 1-1/(1+x²)!

So now the area becomes the integral as x goes from 0 to Π of 1/2 the product of 1+x² and 1-1/(1+x²) which equals 1/2 the integral from 0 to Π of x² which equals Π³/6.

This integral, however, does not cover the area bounded by the triangle (0,0), (-1,0) and (-1,Π). So we must add this area which is Π/2. Yet we must subtract the area inside the lighthouse which the dog may not go in, which is also Π/2. So these two modifications conventiently cancel each other out.

The area of the quarter circle to the left of x=-1 and above y=0 is Π³/4, since the length of the chain is Π and the dog covers a semicircle to the left of the lighthouse. So the total area above y=0 is 5/12 * Π³. Finally double this for the area under y=0 and the total area is 5/6 * Π³

Here is another solution I received from Patrick.

I would like to thank Steven Lutz for sending me this problem.

Here are the four possible answers, the last one appears in the picture below.

1. 1x9, 2x8, 3x6, 4x7, 5x10 = 11x11 square
2. 1x6, 2x10, 3x9, 4x7, 5x8 = 11x11 square
3. 1x2, 3x7, 4x6, 5x10, 8x9 = 13x13 square
4. 1x2, 3x8, 4x5, 7x9, 6x10 = 13x13 square

Go back to Problem 98

Go back to Shack's Math Problems

This solution is hard to explain so if you don't understand my explanation please forgive me.

First consider the minimum and maximum possible areas the rectangles can cover, not neccessarily forming a square. The maximum area would be (10*9)+(8*7)+(6*5)+(4*3)+(2*1)=190. The minimum area would be (10*1)+(9*2)+(8*3)+(7*4)+(6*5)=110. The three possible squares between these two bounds are 121 (11 by 11), 144 (12 by 12), and 169 (13 by 13).

Try to tile a 12 by 12 square. It makes common sense that one rectangle should fit in each corner of the square with the fifth one in the middle. There is no other possibility since the sides of the rectangles are all different lengths. Each side of the square shall consist of the sides of two different rectangles. Thus the sides will consist of these sums: 10+2, 9+3, 8+4, 7+5. The middle rectangle will be the leftover two numbers, 6 by 1.

It also stands to reason that each side of the square will consist of the long side of one rectangle and the short side of another. To form a middle rectangle of side 6 we must subtract the side of one rectangle from the side of another, the possibilities being 10-4=6 and 8-2=6. 9-3=6 is not possible since these two dimensions must form a side of the square. In the 10-4 case we would have the 10+2 side of the square opposite of the 8+4 side. The other two opposite sides being 9+3 and 7+5, all odd numbers. Note that it is impossible to subtract one odd number from another and obtain 1. Thus the 10-4 case is impossible. Under the 8-2 case we still have the 10+2 and the 8+4 on opposite sides of the square. Again it will be impossible to form a 1 remainder with odd numbered sides only the other way. Thus the 12 by 12 square is impossible to tile.

The 13 by 13 square will have sides of lengths 10+3, 9+4, 8+5, and 7+6, with a 2 by 1 rectangle in the middle. Through a process of elimination as in the 12 by 12 case we can't help but stumple upon the solutions.

The 11 by 11 square can have sides of lengths 10+1, 9+2, 8+3, 7+4, and 6+5. Four of these will go along the sides and the fifth shall form the middle rectangle. Again it is a process of trial and error to find the solutions.

This is a picture of just one solution:

Go back to Problem 98

Go back to Shack's Math Problems

Problem 99 Answer

Problem 99 Solution

From the beginning you have enough grain to make 19 one-way trips to some cache point along the way. The first question is, far out should you transport that grain? The answer is far enough so that you leave a cache somewhere of 9000 pounds of grain. Why 9000? Every cache point should have some amount of grain evenly divisible by 1000. This way the camel can start out every trip with 1000 pounds, and not have a remainder on the last trip. So you have 1000 pounds of grain to let the camel eat, divided by 19 one-way trips. 1000 pounds divided by 19 trips is 1000/19=~52.63 pounds, or 52.63 miles per trip. Each trip you will eat 52.63 miles along the way there, deposit 894.73 pounds of grain, and take back 52.63 pounds of grain to eat on the way home, except the last trip in which you don't need to return. The total deposit works out to (10*894.73)+52.63 = 9000 pounds, as arranged. So you will be able to cache 9000 pounds of grain 52.63 miles.

Next repeat the above but only allowing 17 one-way trips. 1000 divided by 17 is 1000/17 =~ 58.82 . So you will be able to cache 8000 pounds of grain 111.46 miles from city A.

Below is a summary of the entire journey:

1. Cache 9000 pounds over 19 one-way trips, moving 52.63 miles, or 52.63 miles from city A
2. Cache 8000 pounds over 17 one-way trips, moving 58.82 miles, or 111.46 miles from city A
3. Cache 7000 pounds over 15 one-way trips, moving 66.67 miles, or 178.12 miles from city A
4. Cache 6000 pounds over 13 one-way trips, moving 76.92 miles, or 255.04 miles from city A
5. Cache 5000 pounds over 11 one-way trips, moving 90.91 miles, or 345.95 miles from city A
6. Cache 4000 pounds over 9 one-way trips, moving 111.11 miles, or 457.07 miles from city A
7. Cache 3000 pounds over 7 one-way trips, moving 142.86 miles, or 599.92 miles from city A
8. Cache 2000 pounds over 5 one-way trips, moving 200.00 miles, or 799.92 miles from city A
9. Take 1000 pounds directly to city B, comsuming 200.08 pounds on the way, leaving 599.84 pounds and take back 200.08 pounds to eat on the way back.
10. Take the last 1000 pounds directly to city B, comsuming 200.08 pounds on the way, and depositing an additional 799.92 pounds of grain, for a total deposit of 1399.77 pounds.

I would like to thank Scott Morris for this problem.

Problem 100 Answer

Given any right triangle you can create a figure like the one above using four of them. This is because the sum of the three angles always is 180 degrees. The area of the outside square is c². We can also find the area by summing the area of the individual parts: 4*(a*b/2)+(a-b)² = 2ab + a² -2ab + b² = a² + b². Thus a² + b² = c².

I didn't assign a diffulty level because I saw Carl Sagan explain this proof on television so I never had a chance to solve it myself.

Problem 101 Answer

         Truth  Lying  Random
         Guy    Guy    Guy
         -----  -----  ------
I          A      B      C
II         A      C      B
III        B      A      C
IV         B      C      A
V          C      A      B
VI         C      B      A

Label the three men A, B, and C and tell each man who is what letter.

• Step 1: Ask A, "Is B more likely to tell the truth than C?" If yes go to step 2, if no go to step 5.
• Step 2: Ask C, "Are you the random guy?" If yes go to step 3, if no go to step 4.
• Step 3: Ask C, "Is A the truth guy?" If yes then scenario IV, if no then scenario II.
• Step 4: Ask C, "Is A the lying guy?" If yes then scenario V, if no then scenario VI.
• Step 5: Ask B, "Are you the random guy?" If yes then step 6, if no then step 7.
• Step 6: Ask B, "Is A the truth guy?" If yes then scenario VI, if no then scenario I.
• Step 7: Ask B, "Is A the lying guy?" If yes then scenario III, if no then scenario IV.

Problem 102 Answer

Problem 102 Solution

f(x)=2*pi*x/pi = 2x.

The mean distance to the center is the integral from 0 to 1 of f(x)*x:

Integral from 0 to 1 of 2x*x = 2x³/3 from 0 to 1 = 2/3.

Problem 103 Answer

Thanks go to Stephanie Mullen for this one.

Problem 102 Solution

f(x)=2*pi*x/pi = 2x.

The mean distance to the center is the integral from 0 to 1 of f(x)*x:

Integral from 0 to 1 of 2x*x = 2x³/3 from 0 to 1 = 2/3.

Problem 104 Answer

If any of the following conditions are true the second player should spin again.

1. His score is less than the first player's score.
2. His score is 50 cents or less.
3. His score is 65 cents or less and he has tied the first player.

If there is only one person to beat, then the third player should spin again if his score is less than the current highest score. If his first spin ties the highest score, then he should spin again if the tie is at 45 cents or less. If the tie is at 50 cents he is indifferent to spinning to staying pat.

If there are two tied players to beat, then the third player should spin again if his score is less than the current high scores. If his first spin ties the high scores, then he should spin again if the tie is at 65 cents or less.

The following table shows the probability of each player winning, according to player 1's initial spin. The bottom row shows the overal probabilities for each player winning before the spin-off begins.

Here are the winning number of combinations out of 6*20⁶ possible.

Problem 105 Answer

Problem 105 Solution

Let I(a,b) denote the integral from a to b. It is obvious that f(x)=0 for x<1.

It is obvious that f(x)=1 for 1<=x<2.

If 2<=x<=3 then f(x)=1+2*(x-2)/(x-1). This can be found by using simple geometry which is left for you to do.

Now lets solve x=4. The first car will divide the rest of the street into a small and a large portion, lets call them s and l. s is equally likely to range from 0 to 3/2. l is equally likely to range from 3/2 to 3. The probability that the smaller portion is less than 1 is 2/3. Likewise the probability that smaller portion is greater than 1 is 1/3. If the smaller portion is greater than 1 then exactly 3 cars will be able to park. If the smaller portion is less than one the number than will be able to park is 1+f(l).

f(x) = 1/3*3 + 2/3*(1+f(l), where l goes from 2 to 3)

= 1 + 2/3*(1 + I(2,3) 1+2*(x-2)/(x-1) dx)

= 1 + 2/3 + 2/3 + 4/3*I(2,3) (x-2)/(x-1) dx, next let y=x-1

= 1 + 4/3 * 4/3 * I(1,2) (y-1)/y dy

= 7/3 + 4/3 * I(1,2) 1 - 1/y dy

= 7/3 + 4/3 * (y-ln(y) for y from 1 to 2)

= 7/3 + 4/3 * (2-ln(2)-1+ln(1))

= 7/3 + 4/3 - 4/3*ln(2)

= 11/3 - 4/3*ln(2)

=~ 2.7425

Problem 106 Answer

Problem 106 Solution

uwzx forms a trapazoid with sides of length a,a+c,c,2*sqr(ac) per the pythagorean formula. Likewise the bottom of trapazoid vyzw is 2*sqr(bc) and the bottom of trapazoid uvyx is 2*sqr(ab).

The sum of the bottoms of trapazoids uwzx and vyzw is equal to the bottom of trapazoid uvyx:

2*sqr(ac)+2*sqr(bc)=2*sqr(ab)

sqr(ac)+sqr(bc)=sqr(ab)

sqr(c)*(sqr(a)+sqr(b))=sqr(ab)

sqr(c)=sqr(ab)/(sqr(a)+sqr(b))

c=ab/(a+2*sqr(ab)+b)

Note: Steve Schaefer points out the following general relationship c^-1/2 = a^-1/2 + b^-1/2

Problem 107 Answer

Problem 108 Answer

Problem 108 Solution

After the first person is finished the remaining pile will have 4, 8, 12, 16, 20, ...

The remaining pile after the first person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 16, 36, 56, 76, 96, ...

After the second person is finished the remaining pile will have 12, 28, 44, 60, 76, ...

The remaining pile after the second person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 76, 156, 236, 316, 396, ...

After the third person is finished the remaining pile will have 60, 124, 188, 252, 316 ...

The remaining pile after the third person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 316, 636, 956, 1276, 1596, ...

After the fourth person is finished the remaining pile will have 252, 508, 764, 1020, 1276 ...

The remaining pile after the fourth person must be a number such that you can subtract one then multiply by 4/5 and get an integer. The smallest possibilities is 1276

After the fifth person is finished the remaining pile will have 1020 ...

So the fifth person will get 1276-1020-1 = 255.

The fifth person will leave behind a pile of 1020.

The fourth person will have left behind a pile of 1020+255+1 = 1276.

The fourth person will get 1276/4 = 319.

The third person will have left behind a pile of 1276+319+1 = 1596.

The third person will get 1596/4 = 399.

The second person will have left behind a pile of 399+1596+1 = 1996.

The second person will get 1996/4 = 499.

The first person will have left behind a pile of 499+1996+1 = 2496.

The first person will get 2496/4 = 624.

The original pile must have had 624+2496+1 = 3121.

Problem 109 Answer

Problem 109 Solution

Pr(AB) = Probability that bag with 99 reds is drawn and 1 red is drawn from that bag = 1/2 * 99/100 = 99/200.

Pr(B) = Probability that a red is drawn = ((1/2)*(99/100) + (1/2)*(1/100)) = 100/200 = 1/2.

Pr(AB)/Pr(B) = (99/200)/(100/200) = 99/100.

Problem 110 Answer

Let the corners of the square be at (0,0),(1,0),(1,1), and (0,1).

The solution is to dig the following ditches:

• (0,0) to (x,x)
• (1,0) to (x,x)
• (0,1) to (x,x)
• (1/2,1/2) to (1,1)

Where x = (3-sqr(3))/6 =~ 0.2113248654

The total length of all ditches is aproximately 2.6389584338

Problem 111 Answer

The radius of the exterior solution is 6.

Problem 111 Solution

The triangle marked by the centers of the three outer circles conveniently forms a right triangle with sides 3, 4, and 5 (3² + 4² = 5²) as seen by the diagram below.

Next draw segments from the center of the interior circle and perpendicular to the outer triangle sides of length 3 and 4.

There are four right triangles in this picture, two of which are the same. Using the pythagorean formula we know:

(1): x² + y² = (1+r)²
(2): x² + (3-y)² = (r+2)²
(3): y² + (4-x)² = (r+3)²

Combining equations (1) and (2) we get y=(3-r)/3
Combining equations (1) and (3) we get x=(2-r)/2

Substituting these values for x and y back into equation 1 we get:

((2-r)/2)² + ((3-r)/3)² = r² + 2r + 1

This leads to...

23*r² + 132r - 36 = 0

Putting this through the quadratic formula we get r=6/23, -6

So the interior circle has radius 6/23 and the outer circle has radius 6.

Here is my old solution, as well as some comments from another reader.

The solution to this problem is rather hard to explain so I'm only going to go over it briefly.

First arrange two of the circles so that their centers lay on the y axis. Then use your trigonomtry to determine the coordinates of the center of the third circle (2.4,-.2) if the large circle is on top, the middle circle on bottom, and the third circle is to the right. See problem 92 for help in how to find such coordinates.

Second, invert all three circles such that the third circle is mapped onto itself. Note that the top and bottom circles are mapped to a horizontal lines that are tanjent to the third circle.

Note: To invert any figure take all points (t,r) and map them onto the point (t,c/r) where t is the angle in polor coordinates and r is the distance from the center. In this case choose the constant c carefully such that the third circle is mapped onto itself (okay, c=4.8 in this case).

The image of the center circle will be tanjent to both horizontal lines and the third circle. Going to the right of the third circle will lead us to the interior circle.

Consider the line that goes through (0,0) and the center of the circle in the above step. The distance from the center of the points where this line intersect the circle are sqr(19.4)-1 and sqr(19.4)+1. Invert these two points and their distance from the center are 4.8/(sqr(19.4)-1) and 4.8/(sqr(19.4)+1). Both these points lay on a diameter of the interior circle so take their difference and divide by 2 to get the radius. The answer simplifies to 6/23.

To find the exterior solution shift the third circle to the left rather than the right. The radius of the exterior solution is exactly 6!

Note: The general case if the third circle has coordinates (x,y) and radius r then the radii of the solution circles are (x² + y² - r²) / (4xr +/- (x²+y²+3*r²)).

The + will determine the interior solution, the - the exterior solution.

I would like to propose a solution, also using some trigonometry. First of
all, some definitions:

   The centres of the four circles:
     'A' for radius 3.
     'B' for radius 2
     'C' for radius 1
     'Z' for the middle circle.

   Two angles:
     'a' between CA and CZ
     'b' between CB and CZ

   Our unknown:
     'x' is the radius of the middle circle

Then, we make some obvious statements:
   CA = 1 + 3 = 4
   CB = 1 + 2 = 3
   AB = 2 + 3 = 5
   AZ = 3 + x
   BZ = 2 + x
   CZ = 1 + x

Followed by a trivial consequence:   
(1)   a + b = PI/2  (because CA^2 + CB^2 = AB^2)

Now we establish two equations using trygonometry:
(2)   BZ^2 = CZ^2 + CB^2 - 2*CZ*CB*cos(b)
(3)   AZ^2 = CZ^2 + CA^2 - 2*CZ*CA*cos(a)

By replacing the values we know and introducing 'x', (2) and (3) become:
(2)   (2 + x)^2 = (1 + x)^2 + 9 - 2*3*(1 + x)*cos(b)
(3)   (3 + x)^2 = (1 + x)^2 + 16 - 2*4*(1 + x)*cos(a)

By expanding the squares and simplifying (I also omit some '*' to make the
formulae more readable) we obtain:

(2)   -6 + 2x + 6(1 + x)cos(b) = 0
(3)   -8 + 4x + 8(1 + x)cos(a) = 0

By adding '+2 -2' to the left side of (2) and '+4 -4' to the left side of
(3), and then grouping '(1 + x)' terms, we obtain:

(2)   (1 + x)(1 + 3cos(b)) = 4
(3)   (1 + x)(1 + 2cos(a)) = 3

>From (1), we can express cos(a) in terms of cos(b):
(1)   cos(a) = cos(PI/2 - b) = sin(b) = sqrt(1 - (cos(b)^2))

By substituting cos(a) in (3):
(3)   (1 + x)(1 + 2sqrt(1 - (cos(b)^2))) = 3

Now we can calculate cos(b) from (2) and substitute it into (3):
(2)   cos(b) = (4/(1 + x) - 1)/3    ==>
(3)   (1 + x)(1 + 2sqrt(1 - ((4/(1 + x) - 1)/3)^2)) = 3

For better reading, we can define y = 1/(1 + x), with which (3) becomes (we
can multiply both sides by (1 + x) because x > 0; therefore, (1 + x) > 0 too):

   1 + (2/3)sqrt(9 - (4y - 1)^2) = 3y

A bit of simple algebra brings us to:

   4sqrt(2 + y - 2y^2) = 9y - 3

We will verify later that '(2 + y - 2y^2) >= 0'. For the time being, we
square both sides and we get (after some simplification):

   145y^2 - 86y -23 = 0

This resolves into: (43 +- 72)/145.  By taking the positive solution, we
obtain:

   y = 23/29   ==>   x = 29/23 - 1 = 6/23.

To check that '2 + y - 2y^2 >= 0', we calculate:

   2 + y - 2y^2 = 2 + 23/29 - 2(23/29)^2 = 1.5+ > 0

I haven't worked on the "external" circle (yet), and my solution relies on
the fact that the centres of the original circles form a right triangle.
You do show a more general solution, but I confess (please, do not take
offense) that I am a bit skeptical that such a simple equation would apply.
In fact, I did not manage to understand it. Perhaps (time permitting), you
could add an explanatory figure.

You say that the two centres A and B are along the Y-axis. What you don't
say is that the X-axis is the tangent between the two bigger circles. This
implies the two following equations (C must be below the X-axis, because
the biggest circle is above it. Therefore, we can write the formulae in
such a way that y calculates to a negative number):
(1)   BC^2 - (2 + y)^2 = x^2
(2)   AC^2 - (3 - y)^2 = x^2

This system provides x = 2.4 and y = -0.2, which is what you say in your
solution.

So far so good, but then I really get lost...


Ciao,    Giulio

Problem 112 Answer

Problem 112 Solution

The standard deviation of the number of reds is sqr(1,000,000 * 0.4 * (1-.4)) =~ 489.9

Let l denote the lower bound of the range, let u denote the upper bound, and let x denote the number of reds drawn:

Pr (l <= x <= u) = .95

Pr (l-400,000 <= x-400,000 <= u-400,000) = .95

Pr ((l-.5-400,000)/489.9 <= Z <= (u+.5-400,000)/489.9) = .95 where Z denotes a random variable distributed according to the standard normal distribution.

Next we want the probability that the number of reds will fall on either side of this range to be .025 per side.

Pr( Z <= (u+.5-400,000)/489.9) = .975

(u+.5-400,000)/489.9 = 1.96

u-399,999.5 = 960.2

u =~ 400,960

Pr( Z <= (l-400,000.5)/489.9) = .025

(l-400,000.5)/489.9 = -1.96

l = 399,040

So the range is 399,040 to 400,960

Problem 113 Answer

Problem 113 Solution

Q = The sum for i=1 to 10 of ( (N_i - np_i)²/np_i )

= The sum for i=1 to 10 of ( (N_i - 1000)²/1000 )

= (676+121+16+64+324+64+6241+64+16+1024)/1000 = 8.61

According the the chi-squared table with 9 degrees of freedom a Q statistic of greater than 16.92 shold fail the test. 8.61 is well under 16.92 so it should pass the test. In other words the digits do appear to be random.

At a level of significance such that 10,000 random digits passed 60% of the time the 10,000 digits of e would still pass, but at a 50% level they would not.

Problem 114 Answer

Problem 114 Solution

Let S_A² = The sum for i=1 to 8 of (s_i-1.5125)² = .10875, where S_i is the ith sample from location A.

Let S_B² = The sum for i=1 to 6 of (s_i-1.5125)² = .167673, where S_i is the ith sample from location B.

Let U be the statistic for testing the the hypothesis that the mean of location A does not equal the mean of location B.

U = (sqr(8+6-2)*(1.5125-1.6683)) / (sqr(1/8 + 1/6)+sqr(.10875+.167673)) = -1.693354 .

The t value, with 12 degrees of freedom, at the .95 level of significance is 1.782 .

Thus if -1.782 <= U <= 1.782 the test will pass, which it does.

Problem 115 Answer

Problem 115 Solution

If the number of glasses is a power of 2 (32, 64, 128, 256, etc.) then the correct procedure on the first step would be to test exactly half the glasses. Then test half of the glasses in the positive half next. Keep repeating this procedure. Each step will narrow down the field of poisoned glasses by a factor of 2.

If the number number of glasses is not a power of 2 then there are lots of ways you can test them and still achieve the same maximum number of tests. One way would be to find the closest power of 2 less than n, set this many glasses aside, and then test the others. The only n for which this remainder is 1, between 100 and 200, is 129. In the case of 129 the first test would likely be negative. Then proceed with the procedure for testing 128. Between the first test and the log₂128 the maximum number of tests is 8.

Credit for this problem belongs to Stephen Barr (Mathematical brain benders; ISBN 0-486-24260- 9 published by mamillan NY 1969), previously published in 'Scientific American.'

Problem 116 Answer

Problem 116 Solution

a₀ = 0

a₁ = 9/19*a₂
a₂ = 9/19*a₃ + 10/19*a₁
a₃ = 9/19*a₄ + 10/19*a₂
.
.
.
a₂₇ = 9/19*a₂₈ + 10/19*a₂₆
a₂₈ = 9/19*a₂₉ + 10/19*a₂₇
a₂₉ = 9/19*a₃₀ + 10/19*a₂₈
a₃₀ = 1

Divide the left side into two parts:

9/19*a₁ + 10/19*a₁ = 9/19*a₂
9/19*a₂ + 10/19*a₂ = 9/19*a₃ + 10/19*a₁
9/19*a₃ + 10/19*a₃ = 9/19*a₄ + 10/19*a₂
.
.
.
9/19*a₂₇ + 10/19*a₂₇ = 9/19*a₂₈ + 10/19*a₂₆
9/19*a₂₈ + 10/19*a₂₈ = 9/19*a₂₉ + 10/19*a₂₇
9/19*a₂₉ + 10/19*a₂₉ = 9/19*a₃₀ + 10/19*a₂₈

Rearange with 10/19 terms on the left side and 9/19 terms on the right:

10/19*(a₁) = 9/19*(a₂ - a₁)
10/19*(a₂ - a₁) = 9/19*(a₃ - a₂)
10/19*(a₃ - a₂) = 9/19*(a₄ - a₃)
.
.
.
10/19*(a₂₇ - a₂₆) = 9/19*(a₂₈ - a₂₇)
10/19*(a₂₈ - a₂₇) = 9/19*(a₂₉ - a₂₈)

Next multiply both sides by 19/9:

10/9*(a₁) = (a₂ - a₁)
10/9*(a₂ - a₁) = (a₃ - a₂)
10/9*(a₃ - a₂) = (a₄ - a₃)
.
.
.
10/9*(a₂₇ - a₂₆) = (a₂₈ - a₂₇)
10/9*(a₂₈ - a₂₇) = (a₂₉ - a₂₈)

Next telescope sums:

(a₂ - a₁) = 10/9*(a₁)
(a₃ - a₂) = (10/9)²*(a₁)
(a₄ - a₃) = (10/9)³*(a₁)
.
.
.
(a₂₉ - a₂₈) = (10/9)²⁸*(a₁)
(a₃₀ - a₂₉) = (10/9)²⁹*(a₁)

Next add the above equations:

(a₃₀ - a₁) = a₁ * ((10/9) + (10/9)² + (10/9)³ + ... + (10/9)²⁹)

1 = a₁ * (1 + (10/9) + (10/9)² + (10/9)³ + ... + (10/9)²⁹)

a₁ = 1 / (1 + (10/9) + (10/9)² + (10/9)³ + ... + (10/9)²⁹)

a₁ = (10/9 - 1) / ((10/9)³⁰ - 1)

Now that we know a₁ we can find a₂₀:

(a₂ - a₁) = 10/9*(a₁)
(a₃ - a₂) = (10/9)²*(a₁)
(a₄ - a₃) = (10/9)³*(a₁)
.
.
.
(a₁₉ - a₁₈) = (10/9)¹⁸*(a₁)
(a₂₀ - a₁₉) = (10/9)¹⁹*(a₁)

Add the above equations together:

(a₂₀ - a₁) = a₁ * ((10/9) + (10/9)² + (10/9)³ + ... + (10/9)¹⁹)
a₂₀ = a₁ * ((10/9)²⁰ - 1)) / (10/9 - 1))
a₂₀ = [ (10/9 - 1) / ((10/9)³⁰ - 1) ] * [ ((10/9)²⁰ - 1) / (10/9 - 1) ]
a₂₀ = ((10/9)²⁰ - 1) / ((10/9)³⁰ - 1) =~ .3198

Problem 118 Answer

Problem 118 Solution

Next put 1716 locks on the safe, one for each way to group 7 pirates. For each lock give 7 keys to a unique group of 7 pirates. This way any given lock will have a keyholder in any group of 7 or more. For any group of 6 there will be exactly one lock in which the other 7 pirates have the key. Obviously any group of less than 6 would also be missing at least one key to at least one lock.

Here are the number of keys required for other numbers of pirates:

Number 
  of      
Pirates   Number of Locks
------- -------------------
3	                  3
5	                 10
7	                 35
9	                126
11	                462
13	              1,716
15	              6,435
17	             24,310
19	             92,378
21	            352,716
23	          1,352,078
25	          5,200,300
27	         20,058,300
29	         77,558,760
31	        300,540,195
33	      1,166,803,110
35	      4,537,567,650
37	     17,672,631,900
39	     68,923,264,410
41	    269,128,937,220
43	  1,052,049,481,860
45	  4,116,715,363,800
47	 16,123,801,841,550
49	 63,205,303,218,876
51	247,959,266,474,050
53	973,469,712,824,060

Problem 120 Answer

Problem 120 Solution

The problem is to find E₁.

E₅=0 since the player already has a yahtzee and needs no further rolls.

E₄=1+(5/6)*E₄+(1/6)*E₅

6*E₄=6+5*E₄+E₅

E₄=6+E₅

E₄=6

Next solve for E₃

E₃ = 1 + (25/36)*E₃ + (10/36)*E₄ + (1/36)*E₅

36*E₃ = 36 + 25*E₃ + 10*E₄ + E₅

11*E₃ = 36 + 10*E₄ + E₅

11*E₃ - 10*E₄ + E₅ = 36

Next solve for E₂

At this point it starts to get less obvious what the probabilities are to advance closer to a yahtzee. Lets assume the two of a kind is of the number a. Below are the number of ways to advance closer and the number of corresponding combinations.

abc: 3!*combin(5,2)=60
abb: 3*5=15
bbb: 5

So there are 60+15+5=80 ways to progress to a three of a kind.

aab: 3*5=15 So there are 15 ways to progress to a four of a kind.

There is obviously only 1 way to progress to a five of a kind.

cde: 3!*combin(5,3)=60
bbc: 5*4*3=60
So there are 60+60=120 ways to stay at a pair.

There are 6³=216 ways to roll 3 dice.

E₂ = 1 + (120/216)*E₂ + (80/216)*E₃ + (15/216)*E₄ + (1/216)*E₅

216*E₂ = 216 + 120*E₂ + 80*E₃ + 15*E₄ + 1*E₅

96*E₂ = 216 + 80*E₃ + 15*E₄ + 1*E₅

96*E₂ = 216 + 80*E₃ + 15*E₄

96*E₂ - 80*E₃ - 15*E₄ = 216

Next solve for E₁. Lets assume that the one of a kind is already an a.

There is obviously 1 way to get to a yahtzee.

Going to a four of a kind:

aaab: 5*4=20
bbbb: 5

20+5=25

Going to a three of a kind:

aabc: combin(5,2)*4!/2! = 10*24/2 = 120
aabb: 5*combin(4,2) = 5*6 = 30
abbb: 5*4 = 20
bccc: 5*4*4 = 80
120+30+20+80 = 250

Going to a pair:

abcd: combin(5,3)*4! = 240
abbc: 5*4*4!/2! = 240
bbcc: combin(5,2)*combin(4,2) = 60
bbcd: 5*combin(4,2)*4!/2! = 360
240+240+60+360 = 900

Staying at one of a kind:

bcde: combin(5,4)*4! = 120

E₁ = 1 + (120/1296)*E₁ + (900/1296)*E₂ + (250/1296)*E₃ + (25/1296)*E₄ + (1/1296)*E₅

1296*E₁ = 1296 + 120*E₁ + 900*E₂ + 250*E₃ + 25*E₄ + 1*E₅

1176*E₁ = 1296 + 900*E₂ + 250*E₃ + 25*E₄ + 1*E₅

1176*E₁ = 1296 + 900*E₂ + 250*E₃ + 25*E₄

1176*E₁ - 900*E₂ - 250*E₃ - 25*E₄ = 1296

So now we have four equations and four unknowns and can solve with matrix algebra. To summarize here is what we know thus far:

1176*E₁ - 900*E₂ - 250*E₃ - 25*E₄ = 1296

96*E₂ - 80*E₃ - 15*E₄ = 216

11*E₃ - 10*E₄ + E₅ = 36

E₄=6

Here is the matrix that needs to be solved.

I'll leave the solving to the matrix to you. Here are the final answers.

E₁ = 151496136/13660416 =~ 11.09015538

E₂ = 11046/1056 =~ 10.46022727

E₃ = 96/11 =~ 8.727272727

E₄ = 6

Thanks to Nick Hobson for suggesting a similar problem.

Problem 123 Answer

Problem 123 Solution

Solution #1

Solution #2

Let x be the height of the pentagon and r be the distance from the center to any corner. The height of the pentagon was solved in problem 47 as (2*5^1/2+5)^1/2/2. The area of the pentagon equals 10 times the area of the triangle shown in the following diagram.

We already know r so let's solve for x in terms of r using the pythagorean formula on the triangle in the diagram above.

(1/2)² + (x-r)² = r²

1/4 + x² -2xr + r² = r²

1/4 + x² -2xr = 0

r = 1/(8x) + x/2

So x-r = x/2 - 1/(8x)

The area of each triangle is 1/2*base*height = x/8 - 1/(32x)

The area of the pentagon is 10 of these triangles = 10x/8 - 10/(32x) = 5x/4 - 5/(16x) = (4x² - 1)/(32x).

We know from problem 47 that x = (2*5^1/2+5)^1/2/2. Substituting this value for x in the above equation results in an area of (25+10*5^1/2)^1/2/4 =~ 1.7205.

Solution #3

The next step is hard to explain without a picture, however I will try. Consider three consecutive corners of the pentagon and label them a, b, and c. Next draw a line from a to c. Call d the point in the middle of segment ac. Now consider the triangle abd. This triangle has angles 36, 54, and 90. We also know the hypotenuse is 1 and longer base is half the distance between any two non-adjacent corners, or (1+sqr(5))/4. From this we can deduce that cos(36)=(1+sqr(5))/4.

Next extend a segment from all five corners of the pentagon and all five midpoints of each side. This will divide the pentagon into 10 equal triangles. Each of these triangles conveniently also has angles of 36, 54, and 90. Next lets define the following:

r=distance from any corner of pentagon to center of pentagon
y=distance from the midpoint of any side to the center of pentagon

r+y is the height of the pentagon, which from problem 47 is sqr(5+2*sqr(5))/2.

The ratio y/r = cos(36) = (1+sqr(5))/4.

Now we have two equations and two unknowns. At this point we grind out the algebra to find that:

r=2*sqr(5+2*sqr(5))/(5+sqr(5))=~0.850651

y=sqr(25+10*sqr(5))/10=~0.688191

The pentagon is composed of 10 right triangles, each of area 1/2*1/2*y. The area of the pentagon is thus (10/4)*y = sqr(25+10*sqr(5))/4 =~ 1.720477.

Thanks to Mark Hoyle for some help with this problem.

Problem 125 Solution

Björn from Germany suggested another solution. Let both hands be at 12. Over the next 12 hours they will cross 11 times. Thus they every 12/11 hours.

Problem 126 Answer

Strategy of y Call x's bid X.

If x bids more than $1750 then y should bid X - X*(sqr((2000-X)/1000)-(2000-x)/2000)+ $.01 .

If x bids less than $1750 then y should bid 2X-$1999.99 .

Strategy of x x should bid $1750.01 .

Problem 128 Solution

E(x)=1 + (E(x)+E(y))/2
E(y)=1 + (E(x))/2

Solving for E(x) and E(y) yields E(x)=6 and E(y)=4. Since the first state is x the answer is 6.

Approach the second problem in the same way, however this time from state y you either end or stay in state y, thus the equations are:

E(x)=1 + (E(x)+E(y))/2
E(y)=1 + (E(y))/2

In this case E(x)=4 and E(y)=2, the answer being E(x) or 4.

Thanks to Alon Amit for this problem.

Problem 129 Answer

Problem 129 Solution

Note that the manner of systematically taking the sums of each set of faces is the same as adding exponents when taking the product of (x⁶+x⁵+x⁴+x³+x²+x¹)² = x¹²+ 2x¹¹+ 3x¹⁰+ 4x⁹+ 5x⁸+ 6x⁷+ 5x⁶+ 4x⁵+ 3x⁴+ 2x³+ x². The coefficient for xⁿ corresponds to the number of ways a total of n can be thrown.

The challenge is to find two other polynomials whose product is also x¹²+ 2x¹¹+ 3x¹⁰+ 4x⁹+ 5x⁸+ 6x⁷+ 5x⁶+ 4x⁵+ 3x⁴+ 2x³+ x². Let's start by factoring each standard die.

x⁶+x⁵+x⁴+x³+x²+x¹ =
x*(x⁵+x⁴+x³+x²+x¹+1) =
x*(x⁶-1)/(x-1) =
x*(x³-1)*(x³+1)/(x-1) =
x*(x-1)*(x²+x+1)*(x+1)*(x²-x+1)/(x-1) =
x*(x²+x+1)*(x+1)*(x²-x+1)

So the product of two dice is
(x⁶+x⁵+x⁴+x³+x²+x¹)² =
(x*(x²+x+1)*(x+1)*(x²-x+1))² =
x²*(x²+x+1)²*(x+1)²*(x²-x+1)²