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

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Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?
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Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA=1</math>. The area of the "bat wings" (shaded area) is
 
<asy>
 
<asy>
 
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));
 
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));
Line 5: Line 5:
 
fill((0,0)--(1,4)--(1.5,3)--cycle, black);
 
fill((0,0)--(1,4)--(1.5,3)--cycle, black);
 
fill((3,0)--(2,4)--(1.5,3)--cycle, black);
 
fill((3,0)--(2,4)--(1.5,3)--cycle, black);
label("A",(3.05,4.2));
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label("$A$",(3.05,4.2));
label("B",(2,4.2));
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label("$B$",(2,4.2));
label("C",(1,4.2));
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label("$C$",(1,4.2));
label("D",(0,4.2));
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label("$D$",(0,4.2));
label("E", (0,-0.2));
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label("$E$", (0,-0.2));
label("F", (3,-0.2));
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label("$F$", (3,-0.2));
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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);
 
</asy>
 
</asy>
  
 
<math>\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</math>
 
<math>\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</math>
  
==Solution==
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==Solution 1==
The area of the trapezoid containing the shaded region and the two isosceles triangles is <math>\frac{1+3}2\cdot 4=8</math>. Next we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio, so the height of the bigger one is 3, while the height of the smaller one is 1. 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>. The answer is (C).
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Draw G in between B and C
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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));
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draw((3,0)--(1,4)--(0,0));
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fill((0,0)--(1,4)--(1.5,3)--cycle, grey);
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fill((3,0)--(2,4)--(1.5,3)--cycle, grey);
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draw((1,0)--(1,4));
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draw((1.5,0)--(1.5,4));
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draw((2,0)--(2,4));
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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, 4.2));
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label("$H$", (1, -0.2));
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label("$J$", (1.5, -0.2));
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label("$K$", (2, -0.2));
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label("$1$", (0.5, 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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Let us take a look at rectangle CDEH. I have labeled E' for convenience.
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<asy>
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fill((0,0)--(1,4)--(1,2)--cycle, grey);
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draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0));
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draw((0,0)--(1,4)--(1,2)--(0,0));
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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("$H$", (1, -0.2));
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label("$E'$", (1.2, 2));
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</asy>
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We can clearly 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 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 one 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 3 (Coordinate Geometry)==
 +
 
 +
Set coordinates to the points:
 +
 
 +
Let <math>E=(0,0)</math>, <math>F=(3,0)</math>
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 +
<asy>
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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(scale(0.7)*"$A(3,4)$",(3.25,4.2));
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label(scale(0.7)*"$B(2,4)$",(2.1,4.2));
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label(scale(0.7)*"$C(1,4)$",(0.9,4.2));
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label(scale(0.7)*"$D(0,4)$",(-0.3,4.2));
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label(scale(0.7)*"$E(0,0)$", (0,-0.2));
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label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8));
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label(scale(0.7)*"$F(3,0)$", (3,-0.2));
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label(scale(0.7)*"$1$", (0.3, 4), N);
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label(scale(0.7)*"$1$", (1.5, 4), N);
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label(scale(0.7)*"$1$", (2.7, 4), N);
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label(scale(0.7)*"$4$", (3.2, 2), E);
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</asy>
 +
 
 +
Now, we easily discover that line <math>CF</math> has lattice coordinates at <math>(1,4)</math> and <math>(3,0)</math>. Hence, the slope of line <math>CF=-2</math>
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Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math>
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 +
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.
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 +
Hence, setting them equal to find the intersection point...
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<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.
 +
 
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Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
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Now, we can see that
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<math>E=(0,0)</math>
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<math>Z=(\dfrac{3}{2},3)</math>
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<math>C=(1,4)</math>.
 +
 
 +
Now use the [[Shoelace Theorem|shoelace theorem]].
 +
 
 +
<math>\frac{(0*3 + \dfrac{3}{2}*4 + 1*0)-(\dfrac{3}{2}*0 + 1*3 + 4*0)}{2} = \frac{6-3}{2} = \frac{3}{2}</math>
 +
 
 +
Using the well known [[Shoelace Theorem|shoelace theorem]], we find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>.
  
 +
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math>
 +
==Video Solution==
 +
https://youtu.be/Tvm1YeD-Sfg - Happytwin
  
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:04, 22 September 2020

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.

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

Using the well known 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 $\boxed{\textbf{(C) }3}$

Video Solution

https://youtu.be/Tvm1YeD-Sfg - Happytwin

2016 AMC 8 (ProblemsAnswer KeyResources)
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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