# A grade school question from China

My wife showed me a homework problem of a friend’s son in China that he could not solve, and his mother also could not solve. I thought it was quite an interesting problem, so I thought I would share.

I do not have an acceptable diagram (which admittedly would make the problem much easier to digest), so I will be as precise as possible in describing the problem. Suppose that you have a triangle $ABC$ with a right angle at vertex $B$. The length of the side $AC$ is equal to 14. Construct squares $AGFB$ (read counter-clockwise) on side $AB$ and $BEDC$ on side $BC$. Draw the line from vertex $A$ to the vertex $D$ of square $BCDE$, and let $H$ be the intersection of the line segment $AD$ and $BC$. Suppose further that the line segment $GH$  is parallel to $AC$. What is the area of the quadrilateral $BHDE$?

We let $x$ denote the side length of $AB$, $y$ denote the side length of $BC$, so that $x^2 + y^2 = 14^2 = 196$. Put $z$ for the length of $BH$. Since $AC$ and $GH$ are parallel, it follows that $z = y - x$. Therefore the desired area is given by $[AED] - [ABH]$ (the square brackets denote the area of the polygon with the given vertices), or

$\frac{y(x+y)}{2} - \frac{x(y-x)}{2} = \frac{x^2 + y^2}{2} = 98.$

This problem is nice because it requires adroit use of many different geometric facts, and once the proper principles are applied, the solution is beautiful and elegant. Hard to imagine an 11 year old being able to do this regularly though!

Edit: in the original post there was an error where I forgot to divide by 2 in the penultimate step. Silly mistake!