# Difference between revisions of "2016 AMC 8 Problems/Problem 22"

## Problem

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$. The area of the "bat wings" (shaded area) is

$[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("A",(3.05,4.2)); label("B",(2,4.2)); label("C",(1,4.2)); label("D",(0,4.2)); label("E", (0,-0.2)); label("F", (3,-0.2)); label("1", (0.5, 4), N); label("1", (1.5, 4), N); label("1", (2.5, 4), N); label("4", (3.2, 2), E); [/asy]$

$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$

## Solution 1

Draw G in between B and C Draw H, J, K beneath C, G, B respectively.

$[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, grey); fill((3,0)--(2,4)--(1.5,3)--cycle, grey); draw((1,0)--(1,4)); draw((1.5,0)--(1.5,4)); draw((2,0)--(2,4)); label("A",(3.05,4.2)); label("B",(2,4.2)); label("C",(1,4.2)); label("D",(0,4.2)); label("E", (0,-0.2)); label("F", (3,-0.2)); label("G", (1.5, 4.2)); label("H", (1, -0.2)); label("J", (1.5, -0.2)); label("K", (2, -0.2)); label("1", (0.5, 4), N); label("1", (2.5, 4), N); label("4", (3.2, 2), E); [/asy]$

Let us take a look at rectangle CDEH. I have labeled E' for convenience.

$[asy] fill((0,0)--(1,4)--(1,2)--cycle, grey); draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)); draw((0,0)--(1,4)--(1,2)--(0,0)); label("C",(1,4.2)); label("D",(0,4.2)); label("E", (0,-0.2)); label("H", (1, -0.2)); label("E'", (1.2, 2)); [/asy]$

We can clearly see that CEE' has $\frac{1}{4}$ the area of CDEH because it has half the base and the same height. Similarly, the shaded regions in CGJH, BGJK, and ABKF all have $\frac{1}{4}$ the area of their rectangle. So, the total shaded region is just $\frac{1}{4}$ the area of the total region, or $\frac{1}{4} \times 3 \times 4$, or $\boxed{\textbf{(C) }3}$

## Solution 2

The area of trapezoid $CBFE$ is $\frac{1+3}2\cdot 4=8$. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\frac12$ and $\frac92$. Subtracting these areas from the trapezoid, we get $8-\frac12-\frac92 =\boxed3$. Therefore, the answer to this problem is $\boxed{\textbf{(C) }3}$

## Solution 3 (Coordinate Geometry)

Set coordinates to the points:

Let $E=(0,0)$, $F=(3,0)$

$[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label(scale(0.7)*"A(3,4)",(3.25,4.2)); label(scale(0.7)*"B(2,4)",(2.1,4.2)); label(scale(0.7)*"C(1,4)",(0.9,4.2)); label(scale(0.7)*"D(0,4)",(-0.3,4.2)); label(scale(0.7)*"E(0,0)", (0,-0.2)); label(scale(0.7)*"Z(\frac{3}{2},3)", (1.5,1.8)); label(scale(0.7)*"F(3,0)", (3,-0.2)); label(scale(0.7)*"1", (0.3, 4), N); label(scale(0.7)*"1", (1.5, 4), N); label(scale(0.7)*"1", (2.7, 4), N); label(scale(0.7)*"4", (3.2, 2), E); [/asy]$

Now, we easily discover that line $CF$ has lattice coordinates at $(1,4)$ and $(3,0)$. Hence, the slope of line $CF=-2$

Plugging in the rest of the coordinate points, we find that line $CF=-2x+6$

Doing the same process to line $BE$, we find that line $BE=2x$.

Hence, setting them equal to find the intersection point...

$y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3$.

Hence, we find that the intersection point is $(\frac{3}{2},3)$. Call it Z.

Now, we can see that

$E=(0,0)$

$Z=(\dfrac{3}{2},3)$

$C=(1,4)$.

Now use the Shoelace Theorem.

$\frac{(0\cdot 3 + \dfrac{3}{2}\cdot 4 + 1\cdot 0)-(\dfrac{3}{2}\cdot 0 + 1\cdot 3 + 4\cdot 0)}{2} = \frac{6-3}{2} = \frac{3}{2}$

Using the Shoelace Theorem, we find that the area of one of those small shaded triangles is $\frac{3}{2}$.

Now because there are two of them, we multiple that area by $2$ to get $\boxed{\textbf{(C) }3}$