Difference between revisions of "2016 AMC 8 Problems/Problem 22"
Reddragon644 (talk | contribs) (→Solution) |
Hashtagmath (talk | contribs) |
||
(21 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
− | Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. | + | == Problem == |
+ | |||
+ | Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA=1</math>. The area of the "bat wings" (shaded area) is | ||
+ | |||
<asy> | <asy> | ||
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); | draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); | ||
Line 19: | 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> | ||
− | ==Solution== | + | |
+ | ==Solution 1== | ||
+ | Draw G in between 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. | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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> | 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 | + | ==Solution 3 (Coordinate Geometry)== |
− | + | Set coordinates to the points: | |
Let <math>E=(0,0)</math>, <math>F=(3,0)</math> | Let <math>E=(0,0)</math>, <math>F=(3,0)</math> | ||
<asy> | <asy> | ||
− | draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); | + | draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); |
− | draw((3,0)--(1,4)--(0,0)); | + | draw((3,0)--(1,4)--(0,0)); |
− | fill((0,0)--(1,4)--(1.5,3)--cycle, black); | + | fill((0,0)--(1,4)--(1.5,3)--cycle, black); |
− | fill((3,0)--(2,4)--(1.5,3)--cycle, black); | + | fill((3,0)--(2,4)--(1.5,3)--cycle, black); |
− | label("$A(3,4)$",(3.25,4.2)); | + | label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); |
− | label("$B(2,4)$",(2.1,4.2)); | + | label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); |
− | label("$C(1,4)$",(0.9,4.2)); | + | label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); |
− | label("$D(0,4)$",(-0.3,4.2)); | + | label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); |
− | label("$E(0,0)$", (0,-0.2)); | + | label(scale(0.7)*"$E(0,0)$", (0,-0.2)); |
− | label("$Z(\frac{3}{2},3)$", (1.5,1.8)); | + | label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); |
− | label("$F(3,0)$", (3,-0.2)); | + | label(scale(0.7)*"$F(3,0)$", (3,-0.2)); |
− | label("$1$", (0.3, 4), N); | + | label(scale(0.7)*"$1$", (0.3, 4), N); |
− | label("$1$", (1.5, 4), N); | + | label(scale(0.7)*"$1$", (1.5, 4), N); |
− | label("$1$", (2.7, 4), N); | + | label(scale(0.7)*"$1$", (2.7, 4), N); |
− | label("$4$", (3.2, 2), E); | + | label(scale(0.7)*"$4$", (3.2, 2), E); |
</asy> | </asy> | ||
Line 62: | Line 108: | ||
<math>E=(0,0)</math> | <math>E=(0,0)</math> | ||
− | <math>Z=(\ | + | <math>Z=(\dfrac{3}{2},3)</math> |
<math>C=(1,4)</math>. | <math>C=(1,4)</math>. | ||
− | Shoelace | + | 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 | + | 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> | Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math> | ||
+ | |||
+ | ==Video Solutions== | ||
+ | *https://youtu.be/Tvm1YeD-Sfg - Happytwin | ||
+ | *https://youtu.be/q3MAXwNBkcg ~savannahsolver | ||
+ | * 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 17:34, 21 April 2021
Contents
Problem
Rectangle below is a rectangle with . The area of the "bat wings" (shaded area) is
Solution 1
Draw G in between B and C Draw H, J, K beneath C, G, B respectively.
Let us take a look at rectangle CDEH. I have labeled E' for convenience.
We can clearly see that CEE' has 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 the area of their rectangle. So, the total shaded region is just the area of the total region, or , or
Solution 2
The area of trapezoid is . Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one 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
Solution 3 (Coordinate Geometry)
Set coordinates to the points:
Let ,
Now, we easily discover that line has lattice coordinates at and . Hence, the slope of line
Plugging in the rest of the coordinate points, we find that line
Doing the same process to line , we find that line .
Hence, setting them equal to find the intersection point...
.
Hence, we find that the intersection point is . Call it Z.
Now, we can see that
.
Now use the Shoelace Theorem.
Using the Shoelace Theorem, we find that the area of one of those small shaded triangles is .
Now because there are two of them, we multiple that area by to get
Video Solutions
- https://youtu.be/Tvm1YeD-Sfg - Happytwin
- https://youtu.be/q3MAXwNBkcg ~savannahsolver
- 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.