# 2009 UNCO Math Contest II Problems/Problem 9

## Problem

A square is divided into three pieces of equal area by two parallel lines as shown. If the distance between the two parallel lines is $8$ what is the area of the square?

$[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); draw((1,0)--(0,2/3),black); draw((1,1/3)--(0,1),black); [/asy]$

## Solution

Let x be the length of a side. Then the square has area $x^2$ and each portion has area $x^2 \times\frac{1}{3}$ If x is the base of one of the triangles, then the height will be $\frac{2x}{3}$. By the pythaogrean theorem, longer side of the parallelogram has length $\sqrt(x^2+(\frac{2x}{3})^2)$ Thus sqrt(13)*x/3*8 = x^2/3. Solving this gives x = 8*sqrt(13). Thus, the area of the square is 64*13 = 832.

## Solution 2 (Better Explained)

$[asy] unitsize(135); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A, B, C, D, E, F; A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1); E=(0,-1/3); F=(1,-2/3); draw(A--B--C--D--cycle); label("A",A,N); label("B",B,N); label("C",C,E); label("D",D,S); label("E",E,W); label("F",F,SE); draw(A--F); draw(C--E); draw(A--C); draw(E--F); [/asy]$

In $\triangle$ADC, $\frac {AE}{ED}$ = $\frac {1}{2}$ because $\frac {[ACE]}{[CED]}$ = $\frac {1}{2}$ (According to question)

$[ECD]$ = $2[EFC]$ (AECF is a parallelogram and $2[EFC]$ = $[AECF]$)

$2[EFC]$ = $[EDC]$ (same reason as before)

$[EFC]$ = $\frac {1}{2}$ * 8 * EC

$[EDC]$ = $\frac{1}{2}$ * EC * (Perpendicular from D to EC)

2 * $\frac {1}{2}$ * 8 * EC = $\frac{1}{2}$ * EC * (Perpendicular from D to EC)

(Perpendicular from D to EC) = 16

Let AE = $x$, then ED = $2x$. So the side length of square is $3x$.

(Perpendicular from D to EC) = $\frac {ED * DC}{EC}$

EC = $\sqrt{13}x$ (By Pythagoras Theorem)

16 = $\frac {2x * 3x}{\sqrt{13}x}$

16 = $\frac {6x^{2}}{\sqrt{13}x}$

$x$ = $\frac {8\sqrt{13}}{3}$

$3x$ = $8\sqrt{13}$

Area of square ABCD = $9x^{2}$ = $(8\sqrt{13})^{2}$ = 832