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| ==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.
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| <asy>
| | ~23orimy412uc3478 |
| 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. 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.
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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>
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| 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>
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| ==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 largere 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>
| | 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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| ==Solution 3 (Coordinate Geometry)==
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| Set coordinates to the points:
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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> | |
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| 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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| Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</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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| Hence, setting them equal to find the intersection point...
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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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| Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
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| Now, we can see that
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| <math>E=(0,0)</math>
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| <math>Z=(\dfrac{3}{2},3)</math> | | <math>\text {Total Area}=12</math> |
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| <math>C=(1,4)</math>. | | <math>[\triangle CDE]=2</math> |
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| Now use the [[Shoelace Theorem|Shoelace Theorem]].
| | <math>[\triangle ABF]=2</math> |
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| <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> | | <math>[\triangle CBG]=\frac1{2}</math> |
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| Using the [[Shoelace Theorem|Shoelace Theorem]], we find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>.
| | <math>[\triangle EFG]=\frac{9}{2}</math> |
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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>
| | <math>\text {Unshaded Region}=9\implies\text{"Bat Wings"}=\boxed{\textbf{(C) }3}</math> |
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| ==Solution 4 (Fastest)==
| | [https://aops.com/wiki/index.php/User:Am24 ~AM24] |
| <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("$A$",(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, 3.2), N);
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| label("$1$", (0.5, 4), N);
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| label("$1$", (1.5, 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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| 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>.
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| ~[[User:Bloggish|Bloggish]]
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| ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
| Line 148: |
Line 50: |
| ~Education, the Study of Everything | | ~Education, the Study of Everything |
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| | ==Video Solutions== |
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| ==Video Solutions==
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| *https://youtu.be/Tvm1YeD-Sfg - Happytwin
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| *https://youtu.be/q3MAXwNBkcg ~savannahsolver | | *https://youtu.be/q3MAXwNBkcg ~savannahsolver |
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| ==Video Solution by OmegaLearn== | | ==Video Solution by OmegaLearn== |
| https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14 | | https://youtu.be/FDgcLW4frg8?t=4448 |
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| | ~ pi_is_3.14 |
| | |
| | == Video Solution only problem 22's by SpreadTheMathLove== |
| | https://www.youtube.com/watch?v=sOF1Okc0jMc |
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| ==Solution 5== | | ==See Also== |
| | {{AMC8 box|year=2016|num-b=21|num-a=23}} |
| | {{MAA Notice}} |
| | [[Category:Introductory Geometry Problems]] |
Problem
Rectangle 
below is a 
rectangle with 
. 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]](//latex.artofproblemsolving.com/a/0/7/a07a9e12af7228a9f783f34687384df0042b721e.png)
Solution 1
The area of trapezoid 
is 
. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a 
ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is 
while the height of the smaller one is 
Thus, their areas are 
and 
. Subtracting these areas from the trapezoid, we get 
. Therefore, the answer to this problem is 
.
~23orimy412uc3478
Solution 2
Plot the point 
where the two "wings" intersect. Now notice how 
. Since the length of 
is one third that of 
, then that means 
's height is 
times bigger than 
. Since both of their heights (
) add up to four, then we have the equation 
. Now that we now the height and length of both triangles, we can use complementary counting, 
.
![$[\triangle CDE]=2$](//latex.artofproblemsolving.com/c/0/7/c0705fdf411556a177878bacc20389edb52d6878.png)
![$[\triangle ABF]=2$](//latex.artofproblemsolving.com/e/0/6/e0668b6377c904b859a61e0f810a89e07fd44987.png)
![$[\triangle CBG]=\frac1{2}$](//latex.artofproblemsolving.com/e/b/8/eb8347fa2a37c7499e73d07fbfb2caa9f2aea0f5.png)
![$[\triangle EFG]=\frac{9}{2}$](//latex.artofproblemsolving.com/b/3/f/b3fc8df9c086f613690383eab6558bd8f7989d58.png)
~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
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
below is a 
rectangle with 
. 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]](http://latex.artofproblemsolving.com/a/0/7/a07a9e12af7228a9f783f34687384df0042b721e.png)
is 
. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a 
ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is 
while the height of the smaller one is 
Thus, their areas are 
and 
. Subtracting these areas from the trapezoid, we get 
. Therefore, the answer to this problem is 
.
where the two "wings" intersect. Now notice how 
. Since the length of 
is one third that of 
, then that means 
's height is 
times bigger than 
. Since both of their heights (
) add up to four, then we have the equation 
. Now that we now the height and length of both triangles, we can use complementary counting, 
.
![$[\triangle CDE]=2$](http://latex.artofproblemsolving.com/c/0/7/c0705fdf411556a177878bacc20389edb52d6878.png)
![$[\triangle ABF]=2$](http://latex.artofproblemsolving.com/e/0/6/e0668b6377c904b859a61e0f810a89e07fd44987.png)
![$[\triangle CBG]=\frac1{2}$](http://latex.artofproblemsolving.com/e/b/8/eb8347fa2a37c7499e73d07fbfb2caa9f2aea0f5.png)
![$[\triangle EFG]=\frac{9}{2}$](http://latex.artofproblemsolving.com/b/3/f/b3fc8df9c086f613690383eab6558bd8f7989d58.png)
