A 3' cube sits on level ground against a vertical wall. A 12' ladder on the same ground leans against the wall such that it touches the top edge of the box.

How far from the wall must the foot of the ladder be, if it is to reach maximum height whilst meeting the foregoing conditions?

There are two possible ways that the situation described could occur, but they are reflections of each other.

Let the distance from the edge of the box to the foot of the ladder be x, and the distance from the top of the box to the top of the ladder be y.

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|y \


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     x

We also know that each edge of this square is three feet long.

Also, because all the angles are the same, the two triangles formed are similar.
Thus,
y/3 = 3/x
y = 9/x

We also know by the large triangle and the Pythagorean theorem that
(x+3)² + (y+3)² = 12²
Substituting 9/x for y, we get
(x+3)² + (9/x)² = 144

I'm not sure how to solve this order of equation, but putting it into Maple gives a value of 1.0865999806791402. Thus, the length from the wall to the foot of the ladder (x+3) is 4.0865999806791402', or about 4'1".

In this scenario, y+3 is √(12² - 4.0865..²), about 11.828499482999559, or close to 11'10". So, by moving the bottom of the ladder four feet, the top of the ladder only goes down two inches! I found that pretty interesting.

Posted by DJ on 2003-12-09 11:59:36