# Difference between revisions of "2014 AMC 10A Problems/Problem 16"

## Problem

In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?

$[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("E",E,E); label("F",F,S); label("G",G,W); label("H",H,N); label("\frac12",(0.25,0),S); label("\frac12",(0.75,0),S); label("1",(1,0.5),E); label("1",(1,1.5),E); [/asy]$

$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$

## Solution

### Solution 1

Note that the region is a kite; hence its diagonals are perpendicular and it has area $\dfrac{ab}{2}$ for diagonals of length $a$ and $b$. Since $HF=1$ as both $H$ and $F$ are midpoints of parallel sides of rectangle $GECD$ and $CE=1$, we let $b=HF=1$. Now all we need to do is to find $a$.

Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. This gives us $a=IJ$. Now let $D=(0,0)$. We thus find that the equation of $\overleftrightarrow{AF}$ is $4x+y=2$ and that of $\overleftrightarrow{DH}$ is $2x-y=0$. Solving this system gives us $x=\dfrac{1}{3}$, so the $x$-coordinate of $I$ is $\dfrac{1}{3}$; in other words, $I$ is $\dfrac{1}{3}$ from $\overline{AD}$. By symmetry, $J$ is also the same distance from $\overline{BC}$, so as $CD=1$ we have $a=IJ=1-\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{1}{3}$. Hence the area of the kite is $\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(E)}\ \dfrac{1}{6}}$.

### Solution 2

Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\dfrac{1}{2}$. We will solve for the areas of $ADI$ and $BCJ$ in terms of x by noting that the area of each triangle is the length of the perpendicular from $I$ to $AD$ and $J$ to $BC$ respectively. Because the area of $x$ = $\dfrac{1}{2}* IJ$ based on the area of a kite formula, $\dfrac{ab}{2}$ for diagonals of length $a$ and $b$, $IJ = 2x$. So each perpendicular is length $\dfrac{1-2x}{2}$. So taking our numbers and plugging them into $[ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]$ gives us $2 = \dfrac{5}{2} - 3x$ Solving this equation for $x$ gives us $x = {\textbf{(C)}\ \frac{1}{6}}$