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

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<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===
+
==Solution 1==
Draw G in between B and C
+
Let G be the midpoint B and C
Draw H, J, K beneath C, G, B respectively.
+
Draw H, J, K beneath C, G, B, respectively.
  
 
<asy>
 
<asy>
Line 50: Line 50:
 
</asy>
 
</asy>
  
Let us take a look at rectangle CDEH. I have labeled E' for convenience.  
+
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>
 
<asy>
Line 63: Line 63:
 
</asy>
 
</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>
+
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===
+
==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>
+
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)===
+
==Solution 3 (Coordinate Geometry)==
  
 
Set coordinates to the points:
 
Set coordinates to the points:
Line 120: Line 120:
 
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</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===
+
==Solution 4==
https://youtu.be/Tvm1YeD-Sfg - Happytwin
+
<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!!!)==
 +
https://youtu.be/oBzkBYeHFa8
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
  
https://youtu.be/q3MAXwNBkcg
+
==Video Solutions==
 +
*https://youtu.be/Tvm1YeD-Sfg - Happytwin
 +
*https://youtu.be/q3MAXwNBkcg ~savannahsolver
  
~savannahsolver
+
==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 16:12, 18 January 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

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 $\frac{1}{4}$ 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 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 $\boxed{\textbf{(C) }3}$

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.

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

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

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 $\triangle CGB \sim \triangle EGF$. Therefore, the ratio of the height of $\triangle CBG$ to the height of $\triangle EFG$ is $\frac{1}{3}$. Thus, the area of $\triangle CBG$ is $\frac{1\cdot1}{2} = \frac{1}{2}$, and the area of $\triangle CBE$ is $\frac{1\cdot4}{2} = 2$. So, the area of $\triangle CGE$ is $2-\frac{1}{2}$. Besides, since trapezoid $CBEF$ is isosceles, $\triangle CGE \cong \triangle BGF$. Hence, the area of the "bat wings" is $2\cdot(2-\frac{1}{2})= \boxed{\textbf{(C) }3}$.

~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:

\[A'=2\left(\frac{\textcolor{green}{b}}{2}+\textcolor{red}{i}-1\right)=2\left(\frac{6}{2}+4-1\right)=12\]

And finally we scale this down to get the original area:

\[A=\frac14A'=\frac14 12=\boxed{\textbf{(C) }3}\]

~proloto

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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