# 2010 AMC 12A Problems/Problem 3

## Problem

Rectangle $ABCD$, pictured below, shares $50\%$ of its area with square $EFGH$. Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$. What is $\frac{AB}{AD}$?

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray); draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); draw((25,15)--(25,20)--(50,20)--(50,15)--cycle); label("A",(0,20),W); label("B",(50,20),E); label("C",(50,15),E); label("D",(0,15),W); label("E",(0,25),NW); label("F",(25,25),NE); label("G",(25,0),SE); label("H",(0,0),SW); [/asy]$

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

## Solution 1

If we shift $A$ to coincide with $E$, and add new horizontal lines to divide $EFGH$ into five equal parts:

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray); draw((25,20)--(25,25)--(50,25)--(50,20)--cycle); draw((0,5)--(25,5)); draw((0,10)--(25,10)); draw((0,15)--(25,15)); label("A=E",(0,25),W); label("B",(50,25),E); label("C",(50,20),E); label("D",(0,20),W); label("F",(25,25),NE); label("G",(25,0),SE); label("H",(0,0),SW); [/asy]$

This helps us to see that $AD=a/5$ and $AB=2a$, where $a=EF$. Hence $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$.

## Solution 2

From the problem statement, we know that $$\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}$$

If we let $a = EF$ and $b = AD$, we see $$[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}$$. Hence, $\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{\bold{10}}$

## Video Solution 1 (Logic and Word Analysis)

~Education, the Study of Everything