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"

(→‎Video Solution)
(→‎Solution 2)
 
(28 intermediate revisions by 14 users not shown)
Line 22: Line 22:
 
<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>
  
−
==Solutions==
 
−
===Solution 1===
 
−
Draw G in between B and C
 
−
Draw H, J, K beneath C, G, B respectively.
 
  
−
<asy>
+
==Solution 1==
−
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));
+
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((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.  
+
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/oBzkBYeHFa8
  
−
<asy>
+
~Education, the Study of Everything
−
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 <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)===
+
==Video Solutions==
  
−
Set coordinates to the points:
+
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
−
 
 
−
Let <math>E=(0,0)</math>, <math>F=(3,0)</math>
 
−
 
 
−
<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 <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>
 
−
 
 
−
Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math>
 
−
 
 
−
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.
 
−
 
 
−
Hence, setting them equal to find the intersection point...
 
−
 
 
−
<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.
 
−
 
 
−
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
 
−
 
 
−
Now, we can see that
 
−
 
 
−
<math>E=(0,0)</math>
 
−
 
 
−
<math>Z=(\dfrac{3}{2},3)</math>
 
−
 
 
−
<math>C=(1,4)</math>.
 
−
 
 
−
Now use the [[Shoelace Theorem|Shoelace Theorem]].
 
−
 
 
−
<math>\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}</math>
 
−
 
 
−
Using the [[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
 
−
 
 
−
https://youtu.be/q3MAXwNBkcg
 
−
 
 
−
~savannahsolver
 
−
 
 
−
== Video Solution ==
 
−
https://youtu.be/FDgcLW4frg8?t=4448
 
−
 
 
−
~ pi_is_3.14
 
  
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14
  
 +
==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}}

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&oldid=226113"