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

Let G be the midpoint 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. First of all, we can see that EE'H and CE'B are similar triangles because all their three angles are the same. Furthermore, since EH=CB, we can confirm that EE'H and CE'B are identical triangles. Thus, CE'=E'H, which is half of CH.

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

Then we can 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 two non-colored isosceles triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger 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 5 (Pick's Theorem)

Solution 2-4 are easily better, but if you really wanted to you could use Pick's Theorem for each half of the "bat wings". Unfortunately it isn't immediately applicable since the point common to each bat wing does not lie on a lattice point. We can remedy this by pretending the figure is twice as big and at the end divide the area by 4 (since the area of similar shapes scales quadratically with the scaling factor).

$[asy] // Original drawing code draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$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("$2$", (0.5, 4), N); label("$2$", (1.5, 4), N); label("$2$", (2.5, 4), N); label("$8$", (3.2, 2), E); // Draw the grid lines for (real i=0.5; i<3; i+=0.5) { draw((i,0)--(i,4), gray+linewidth(0.5)); // Vertical grid lines } for (real j=0.5; j<4; j+=0.5) { draw((0,j)--(3,j), gray+linewidth(0.5)); // Horizontal grid lines } // Boundary points with green dots and black border filldraw(circle((0,0), 0.05), green, black+linewidth(0.5)); filldraw(circle((.5,1), 0.05), green, black+linewidth(0.5)); filldraw(circle((1,2), 0.05), green, black+linewidth(0.5)); filldraw(circle((1.5,3), 0.05), green, black+linewidth(0.5)); filldraw(circle((1,4), 0.05), green, black+linewidth(0.5)); filldraw(circle((.5,2), 0.05), green, black+linewidth(0.5)); // Interior points with red dots and black border filldraw(circle((.5,1.5), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,2.5), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,3), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,3.5), 0.05), red, black+linewidth(0.5)); [/asy]$

Now we can safely use Pick's Theorem on the scaled-up wings:

$A'=2\left(\frac{\textcolor{green}{b}}{2}+\textcolor{red}{i}-1\right)=2\left(\frac{6}{2}+4-1\right)=12$

And finally we scale this down to get the original area:

$A=\frac14A'=\frac14 12=\boxed{\textbf{(C) }3}$

~proloto

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/oBzkBYeHFa8

~Education, the Study of Everything

Video Solutions

https://youtu.be/Tvm1YeD-Sfg - Happytwin
https://youtu.be/q3MAXwNBkcg ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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