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

(→‎Solution 2)
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Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math>
 
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math>
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 +
==Solution 4 (Fastest)==
 +
<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);
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fill((3,0)--(2,4)--(1.5,3)--cycle, black);
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label("$A$",(3.05,4.2));
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label("$B$",(2,4.2));
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label("$C$",(1,4.2));
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label("$D$",(0,4.2));
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label("$E$", (0,-0.2));
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label("$F$", (3,-0.2));
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label("$G$", (1.5, 3.2), N);
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label("$1$", (0.5, 4), N);
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label("$1$", (1.5, 4), N);
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label("$1$", (2.5, 4), N);
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label("$4$", (3.2, 2), E);
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</asy>
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 +
First, it is easy to see that <math> \triangle CGB \sim \triangle EGF </math>. Therefore, the ratio of the height of <math> \triangle CBG </math> to the height of <math> \triangle EFG </math> is <math> \frac{1}{3} </math>. Thus, the area of <math> \triangle CBG </math> is <math> \frac{1\cdot1}{2} = \frac{1}{2} </math>, and the area of <math> \triangle CBE </math> is <math> \frac{1\cdot4}{2} = 2 </math>. So, the area of <math> \triangle CGE </math> is <math> 2-\frac{1}{2} </math>. Besides, since trapezoid <math> CBEF </math> is isosceles, <math> \triangle CGE \cong \triangle BGF </math>. Hence, the area of the "bat wings" is <math> 2\cdot(2-\frac{1}{2})= \boxed{\textbf{(C) }3} </math>.
 +
 +
~[[User:Bloggish|Bloggish]]
  
 
==Video Solutions==
 
==Video Solutions==

Revision as of 08:04, 28 January 2023

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 triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the largere 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 $\boxed{\textbf{(C) }3}$

Solution 4 (Fastest)

[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("$G$", (1.5, 3.2), N); 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]

First, it is easy to see that $\triangle CGB \sim \triangle EGF$. Therefore, the ratio of the height of $\triangle CBG$ to the height of $\triangle EFG$ is $\frac{1}{3}$. Thus, the area of $\triangle CBG$ is $\frac{1\cdot1}{2} = \frac{1}{2}$, and the area of $\triangle CBE$ is $\frac{1\cdot4}{2} = 2$. So, the area of $\triangle CGE$ is $2-\frac{1}{2}$. Besides, since trapezoid $CBEF$ is isosceles, $\triangle CGE \cong \triangle BGF$. Hence, the area of the "bat wings" is $2\cdot(2-\frac{1}{2})= \boxed{\textbf{(C) }3}$.

~Bloggish

Video Solutions

Video Solution by OmegaLearn

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

See Also

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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