Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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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. | + | label("$A(3,4)$",(3.25,4.2)); |
| − | label("$B(2,4)$",(2,4.2)); | + | label("$B(2,4)$",(2.1,4.2)); |
| − | label("$C(1,4)$",( | + | label("$C(1,4)$",(0.9,4.2)); |
| − | label("$D(0,4)$",(-0. | + | label("$D(0,4)$",(-0.3,4.2)); |
label("$E(0,0)$", (0,-0.2)); | label("$E(0,0)$", (0,-0.2)); | ||
label("$Z(\frac{3}{2},3)$", (1.5,1.8)); | label("$Z(\frac{3}{2},3)$", (1.5,1.8)); | ||
label("$F(3,0)$", (3,-0.2)); | label("$F(3,0)$", (3,-0.2)); | ||
| − | label("$1$", (0. | + | label("$1$", (0.3, 4), N); |
label("$1$", (1.5, 4), N); | label("$1$", (1.5, 4), N); | ||
| − | label("$1$", (2. | + | label("$1$", (2.7, 4), N); |
label("$4$", (3.2, 2), E); | label("$4$", (3.2, 2), E); | ||
</asy> | </asy> | ||
Revision as of 21:34, 8 January 2018
Rectangle 
below is a 
rectangle with 
. What is the area of the "bat wings" (shaded area)?
![[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)
Solution
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[SOMEBODY PROVE THIS], 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 2
Setting coordinates!
Let 
, 
![[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("$A(3,4)$",(3.25,4.2)); label("$B(2,4)$",(2.1,4.2)); label("$C(1,4)$",(0.9,4.2)); label("$D(0,4)$",(-0.3,4.2)); label("$E(0,0)$", (0,-0.2)); label("$Z(\frac{3}{2},3)$", (1.5,1.8)); label("$F(3,0)$", (3,-0.2)); label("$1$", (0.3, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.7, 4), N); label("$4$", (3.2, 2), E); [/asy]](http://latex.artofproblemsolving.com/b/6/0/b600aa543093c904bd8674a0c641655693361988.png)
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
.
Shoelace!
Using the well known Shoelace Formula(https://en.m.wikipedia.org/wiki/Shoelace_formula), 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
| 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 | ||
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