Solution to Problem36

Congratulations to this week's winner

Mike Fitzpatrick

Correct solutions were also received from Ray Kremer, William Webb, Greg Falcon, and Jeff Dowin. Two partial and one incorrect solution were also received.
Greg Falcon and William Webb provided very different solutions. First I'll sketch Greg Falcon's solution. In the diagram on the left F is the distance along the wall from the top of the ladder to the top of the box and G is the horizontal distance from the bottom of the ladder to the right hand edge of the box. The point P is halfway down the ladder, that is, 3.5 feet from both top and bottom, and D is the distance from P to the upper right hand corner of the box, marked C. The three triangles are all similar, which gives the relationship F = 1/G. Pythagoras's Theorem gives the relationships
1 + F2 = (3.5 + D)2 and 1 + G2 = (3.5 - D)2
F2 = (4.5 + D)(2.5 + D) and G2 = (4.5 - D)(2.5 - D)
Multiply these last equations together and use F = 1/G to get
1 = (20.25 - D2)(6.25 - D2)
which yields a quadratic equation in D2. The rest is straightforward. The height of the ladder is

William Webb's solutions is trigonometric in nature. Start by labeling the lower right hand angle A. The distance from C to B is csc(A) while that from C to T is sec(A). He now uses various trig identities to show that sin(2A) is a root of the quadratic equation 49x2 - 4x - 4 = 0. The rest follows easily.

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Page last updated 23 April 1998.