Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 5 (Pick's Theorem))
(→‎Solution 2)
 
(14 intermediate revisions by 10 users not shown)
Line 24: Line 24:
  
 
==Solution 1==
 
==Solution 1==
−
Let G be the midpoint B and C
+
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>.
−
Draw H, J, K beneath C, G, B, respectively.
 
  
−
<asy>
+
~23orimy412uc3478
−
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.  
+
==Solution 2==
 +
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>.
 +
 
 +
<math>\text {Total Area}=12</math>
 +
 
 +
<math>[\triangle CDE]=2</math>
 +
 
 +
<math>[\triangle ABF]=2</math>
  
−
<asy>
+
<math>[\triangle CBG]=\frac1{2}</math>
−
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>
+
<math>[\triangle EFG]=\frac{9}{2}</math>
  
−
==Solution 2==
+
<math>\text {Unshaded Region}=9\implies\text{"Bat Wings"}=\boxed{\textbf{(C) }3}</math>
−
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>
+
 
 +
[https://aops.com/wiki/index.php/User:Am24 ~AM24]
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
Line 73: Line 50:
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
 +
==Video Solutions==
  
−
 
−
==Video Solutions==
 
−
*https://youtu.be/Tvm1YeD-Sfg - Happytwin
 
 
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
 
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
−
https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14
+
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==
 
==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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&oldid=258781"