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

(→‎Solution 3 (Coordinate Geometry))
(→‎Solution 2)
 
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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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== Problem ==
 +
 
 +
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 18: Line 21:
  
 
<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 1==
 
==Solution 1==
−
Draw G in between B and C
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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>.
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Draw H, J, K beneath C, G, B respectively .
 
−
<asy>
 
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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>
 
−
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)==
+
~23orimy412uc3478
  
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Set coordinates to the points:
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==Solution 2==
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Plot the point <math>G</math> where the two "wings" intersect. Now notice how <math>\triangle CBG\sim\triangle EFG</math>. Since the length of <math>\overline {CB}</math> is one third that of <math>\overline {EF}</math>, then that means <math>\triangle EFG</math>'s height is <math>3</math> times bigger than  <math>\triangle CBG</math>. Since both of their heights (<math>h</math>) add up to four, then we have the equation <math>3h+h=4 \implies h=1</math>. Now that we now the height and length of both triangles, we can use complementary counting, <math>\text{Area}-\text{Unshaded Region}</math>.
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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));
 
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draw((3,0)--(1,4)--(0,0));
 
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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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<math>\text {Total Area}=12</math>
  
−
Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math>
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<math>[\triangle CDE]=2</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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<math>[\triangle ABF]=2</math>
  
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Hence, setting them equal to find the intersection point...
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<math>[\triangle CBG]=\frac1{2}</math>
  
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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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<math>[\triangle EFG]=\frac{9}{2}</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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<math>\text {Unshaded Region}=9\implies\text{"Bat Wings"}=\boxed{\textbf{(C) }3}</math>
  
−
Now, we can see that
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[https://aops.com/wiki/index.php/User:Am24 ~AM24]
  
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<math>E=(0,0)</math>
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/oBzkBYeHFa8
  
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<math>Z=(\dfrac{3}{2},3)</math>
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~Education, the Study of Everything
  
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<math>C=(1,4)</math>.
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==Video Solutions==
  
−
Now use the [[Shoelace Theorem|shoelace theorem]].
+
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
  
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<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>
+
==Video Solution by OmegaLearn==
 +
https://youtu.be/FDgcLW4frg8?t=4448
  
−
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>.
+
~ pi_is_3.14
  
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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>
+
== Video Solution only problem 22's by SpreadTheMathLove==
−
==Video Solution==
+
https://www.youtube.com/watch?v=sOF1Okc0jMc
−
https://youtu.be/Tvm1YeD-Sfg - Happytwin
 
  
 +
==See Also==
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 14:54, 24 August 2025

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

~23orimy412uc3478

Solution 2

Plot the point $G$ where the two "wings" intersect. Now notice how $\triangle CBG\sim\triangle EFG$. Since the length of $\overline {CB}$ is one third that of $\overline {EF}$, then that means $\triangle EFG$'s height is $3$ times bigger than $\triangle CBG$. Since both of their heights ($h$) add up to four, then we have the equation $3h+h=4 \implies h=1$. Now that we now the height and length of both triangles, we can use complementary counting, $\text{Area}-\text{Unshaded Region}$.

$\text {Total Area}=12$

$[\triangle CDE]=2$

$[\triangle ABF]=2$

$[\triangle CBG]=\frac1{2}$

$[\triangle EFG]=\frac{9}{2}$

$\text {Unshaded Region}=9\implies\text{"Bat Wings"}=\boxed{\textbf{(C) }3}$

~AM24

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/oBzkBYeHFa8

~Education, the Study of Everything

Video Solutions

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=4448

~ pi_is_3.14

Video Solution only problem 22's by SpreadTheMathLove

https://www.youtube.com/watch?v=sOF1Okc0jMc

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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