2016 AMC 8 Problems/Problem 22
Contents
Problem
Rectangle below is a rectangle with . The area of the "bat wings" (shaded area) is
Solution 1
Let G be the midpoint 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. 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.
Then we can 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 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
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
Solution 4
First, it is easy to see that . Therefore, the ratio of the height of to the height of is . Thus, the area of is , and the area of is . So, the area of is . Besides, since trapezoid is isosceles, . Hence, the area of the "bat wings" is .
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).
Now we can safely use Pick's Theorem on the scaled-up wings:
And finally we scale this down to get the original area:
~proloto
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solutions
- https://youtu.be/Tvm1YeD-Sfg - Happytwin
- https://youtu.be/q3MAXwNBkcg ~savannahsolver
Video Solution by OmegaLearn
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.