# 1997 AJHSME Problems/Problem 15

## Problem

Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is $[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy]$ $\text{(A)}\ \dfrac{\sqrt{3}}{3} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{2}{3} \qquad \text{(D)}\ \dfrac{\sqrt{5}}{3} \qquad \text{(E)}\ \dfrac{7}{9}$

## Solution 1

Since we are dealing with ratios, let the big square have sides of $3$ and thus an area of $3^2 = 9$. Chosing a multiple of $3$ will avoid fractions in the rest of the answer.

To find the area of the inscribed square, subtract off the areas of the four triangles. Each triangle has an area of $\frac{1}{2}\cdot 2\cdot 1 = 1$.

Thus, the area of the inscribed square is $9 - 4\cdot 1 = 5$, and the ratio of areas is $\frac{5}{9}$, giving option $\boxed{B}$.

## Solution 2

Let the side of the big square be $3x$. The area of this square is $3x\cdot 3x = 9x^2$

You can use the Pythagorean Theorem on the any of the right triangles to find the length of the side of the inscribed square. One leg is $x$, while the other leg is $2x$. Thus, the length of the hypotenuse $c$, which is also the side of the inscrubed square, is: $a^2 + b^2 = c^2$ $x^2 + (2x)^2 = c^2$ $x^2 + 4x^2 = c^2$ $5x^2 = c^2$

Once you get $c^2$, you can stop, because $c^2$ is the area of the inscribed square. The ratio of the areas of the small square to the big square is $\frac{5x^2}{9x^2} = \frac{5}{9}$, giving option $\boxed{B}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 