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

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==Solution 1==
 
==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 <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 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>
 
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>
 
==Solution 3 (Coordinate Geometry)==
 
 
Set coordinates to the points:
 
 
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>
 
 
==Solution 4==
 
<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 <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]]
 
 
== Solution 5 (Pick's Theorem) ==
 
 
Solution 2/4 are easily better, but if you really wanted to you could use [[Pick's Theorem]] for each half of the "bat wings".  Unfortunately it isn't immediately applicable since the point common to each bat wing does not lie on a lattice point.  We can remedy this by pretending the figure is twice as big and at the end divide the area by 4 (since the area of similar shapes scales quadratically with the scaling factor).
 
 
<asy>
 
// Original drawing code
 
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("$2$", (0.5, 4), N);
 
label("$2$", (1.5, 4), N);
 
label("$2$", (2.5, 4), N);
 
label("$8$", (3.2, 2), E);
 
 
// Draw the grid lines
 
for (real i=0.5; i<3; i+=0.5) {
 
    draw((i,0)--(i,4), gray+linewidth(0.5));  // Vertical grid lines
 
}
 
for (real j=0.5; j<4; j+=0.5) {
 
    draw((0,j)--(3,j), gray+linewidth(0.5));  // Horizontal grid lines
 
}
 
 
// Boundary points with green dots and black border
 
filldraw(circle((0,0), 0.05), green, black+linewidth(0.5));
 
filldraw(circle((.5,1), 0.05), green, black+linewidth(0.5));
 
filldraw(circle((1,2), 0.05), green, black+linewidth(0.5));
 
filldraw(circle((1.5,3), 0.05), green, black+linewidth(0.5));
 
filldraw(circle((1,4), 0.05), green, black+linewidth(0.5));
 
filldraw(circle((.5,2), 0.05), green, black+linewidth(0.5));
 
 
// Interior points with red dots and black border
 
filldraw(circle((.5,1.5), 0.05), red, black+linewidth(0.5));
 
filldraw(circle((1,2.5), 0.05), red, black+linewidth(0.5));
 
filldraw(circle((1,3), 0.05), red, black+linewidth(0.5));
 
filldraw(circle((1,3.5), 0.05), red, black+linewidth(0.5));
 
 
 
</asy>
 
 
Now we can safely use Pick's Theorem on the scaled-up wings:
 
 
<cmath>A'=2\left(\frac{\textcolor{green}{b}}{2}+\textcolor{red}{i}-1\right)=2\left(\frac{6}{2}+4-1\right)=12</cmath>
 
 
And finally we scale this down to get the original area:
 
 
<cmath>A=\frac14A'=\frac14 12=\boxed{\textbf{(C) }3}</cmath>
 
 
~proloto
 
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
Line 207: Line 34:
  
 
==Video Solutions==
 
==Video Solutions==
*https://youtu.be/Tvm1YeD-Sfg - Happytwin
+
 
 
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
 
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
  

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 (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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