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

Latest revision as of 15:22, 31 August 2024

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

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 $\boxed{\textbf{(C) }3}$

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/oBzkBYeHFa8

~Education, the Study of Everything

Video Solutions

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 24: / Line 24: @@
 ==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 <math>\frac{1}{4}</math> 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 <math>\frac{1}{4}</math> the area of their rectangle. So, the total shaded region is just <math>\frac{1}{4}</math> the area of the total region, or <math>\frac{1}{4} \times 3 \times 4</math>, or <math>\boxed{\textbf{(C) }3}</math>
-==Solution 2==
 The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math>
-== 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:
-<cmath>A'=2\left(\frac{\textcolor{green}{b}}{2}+\textcolor{red}{i}-1\right)=2\left(\frac{6}{2}+4-1\right)=12</cmath>
-And finally we scale this down to get the original area:
-<cmath>A=\frac14A'=\frac14 12=\boxed{\textbf{(C) }3}</cmath>
-~proloto
 ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
@@ Line 132: / Line 34: @@
 ==Video Solutions==
-*https://youtu.be/Tvm1YeD-Sfg - Happytwin
 *https://youtu.be/q3MAXwNBkcg ~savannahsolver

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