Hi,

This is a very challenging problem. I've been working on this for a while. (it was a word document at first).

Thanks. Any help is greatly appreciated

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- Feb 17th 2008, 05:16 PMJonanoviChallenging Ladder Problem
Hi,

This is a very challenging problem. I've been working on this for a while. (it was a word document at first).

Thanks. Any help is greatly appreciated - Feb 17th 2008, 06:03 PMDavePercy
The first thing to do is give names to some of the unlabeled distances in the picture. Lets call the height of the short ladder $\displaystyle h_1$ and the height of the long one $\displaystyle h_2$. We can also divide $\displaystyle d$ into two parts where it intersects the dotted red line; call the one on the left $\displaystyle d_1$ and the one on the right $\displaystyle d_2$.

Now we have two pairs of similar right triangles; each ladder forms a triangle with the building and with the dotted red line. We can set up some proportions for these triangles:

$\displaystyle \frac {d} {h_1} = \frac {d_2} {c} $

$\displaystyle \frac {d} {h_2} = \frac {d_1} {c} $

By multiplying both sides of each equation by $\displaystyle c$ and seeing that $\displaystyle d_1 + d_2 = d$ we have

$\displaystyle \frac {d c} h_1 + \frac {d c} h_2 = d$

which after dividing both sides by $\displaystyle d c$ becomes

$\displaystyle \frac 1 h_1 + \frac 1 h_2 = \frac 1 c$

Now all we have to do is use the Pythagorean Theorum to get $\displaystyle h_1 = \sqrt {20^2 - d^2}$ and a similar expression for $\displaystyle h_2$ and substitute.

The easiest way to do part b is to plug in 12 for $\displaystyle d$, then simplify the left side before taking the reciprocal of each side.

Part c is basically guess and check; use a calculator or computer program to plug lots of two-decimal-place values into that expression until you can change the sign by changing the plugged in value by only .01.