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 (0,1081/2). 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=(-1081/2/6)*x + 1081/2.
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.
Michael Shackleford, A.S.A.