# Inscribed Square Problem

Here’s another fun little geometry problem from MathChallenges.net. I found a different solution then what they gave. Let me know how you solve it.

## 3 thoughts on “Inscribed Square Problem”

1. This is a very intresting problem. We used to do some of these problems in Geometry with Mrs. Polishcuck.
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What I did was…

So let’s name the sides of the square “s”.

and we know that a^2 and b^2 gets you the hypotunse.

We also know that the height of the mini top triangle is (a-s). and the width of the bottom right triangle is (b-s).

By finding and adding the hypotunse of these two mini triangles, you get the hypotunese of the big triangle.

In math terms…..
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*Note: Sq( ) stands for square root.

sq(a^2 + b^2) = sq((a-s)^2 + s^2) + sq((b-s)^2 + s^2)

If you symplify it, you should get the answer….

I know that it gets really messy, and that this is probally not the most practical approach…but it should be correct.

Sorry that I didn’t have time to find out the final answer…since I had a english essay to finish. ^_^

but theoratically this approach should work despite the superlong squareroots and FOIL-ing that you have to do. 😄

2. Spoiler Alert! Don’t read ahead if you don’t want to see the way I solved it.

I also solved it in a different way.
The way I solved it was by finding the area of the overall triangle, and setting it equal to the area of the sum of the small triangles and the square.
Labeling the sides of the square g I said:
(Attempting Latex, I hope it parses)

$\\ \frac{1}{2}ab = g^2 + \frac{1}{2}(b-g)(g) + \frac{1}{2}g(a-g) \\ ab = 2g^2 + (b-g)g + (a-g)g \\ ab = 2g^2 + (b + a - 2g)g \\ ab = 2g^2 + gb + ga - 2g^2 \\ ab = gb + ga \\ \frac{ab}{a+b} = g$

• Great! That’s how I did it too! I’m not sure what made me think to use area. And you’ll notice, if you look at their solution, they didn’t do it that way. They use a similar-triangles approach. Good work! 🙂

(And great work with the $LaTeX$ !)