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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>
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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, 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==
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| − | 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> | |
| − | | |
| − | ==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>
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| − | | |
| − | <math>C=(1,4)</math>.
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| − | | |
| − | Now use the [[Shoelace Theorem|Shoelace Theorem]].
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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>
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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>.
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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>
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| − | | |
| − | ==Solution 4 (Fastest)==
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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("$A$",(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!!!)== |
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| | ==Video Solutions== | | ==Video Solutions== |
| − | *https://youtu.be/Tvm1YeD-Sfg - Happytwin
| + | |
| | *https://youtu.be/q3MAXwNBkcg ~savannahsolver | | *https://youtu.be/q3MAXwNBkcg ~savannahsolver |
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| | https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14 | | https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14 |
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| − | ==Bonus Simple Solution== | + | ==See Also== |
| − | To solve the problem, one way is to find the total area of the rectangle, and subtract the areas that aren't shaded from it to get the area of the "bat wings". The total area of the rectangle is 3 * 4 which is 12. The areas of triangles ABF and DCE are both 4*1/2 which is 2. There are two triangles so the area will be 4. Subtract 4 from 12, which is 8. Let us name the point inside the rectangle that is connected to the "bat wings" G. It is clear that the triangle CBG is equilateral, so the area is 1*1/2, which is 0.5. Subtract 0.5 from 8, which is 7.5. The area of triangle EFG is 3*3/2 is 4.5. Subtract 4.5 from 7.5, which is 3. So, the answer is \boxed{\textbf{(C) }3}.
| + | {{AMC8 box|year=2016|num-b=21|num-a=23}} |
| | + | {{MAA Notice}} |
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 
