2016 AMC 10B Problems/Problem 23
Contents
Problem
In regular hexagon , points , , , and are chosen on sides , , , and respectively, so lines , , , and are parallel and equally spaced. What is the ratio of the area of hexagon to the area of hexagon ?
Solution 1
We draw a diagram to make our work easier:
Assume that is of length . Therefore, the area of is . To find the area of , we draw , and find the area of the trapezoids and .
From this, we know that . We also know that the combined heights of the trapezoids is , since and are equally spaced, and the height of each of the trapezoids is . From this, we know and are each of the way from to and , respectively. We know that these are both equal to .
We find the area of each of the trapezoids, which both happen to be , and the combined area is .
We find that is equal to .
Solution 2
First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are small triangles in hexagon , and small triangles in the whole hexagon.
Thus, the answer is .
Solution 3 (Similar Triangles)
Extend and to meet at point , as shown in the diagram. Then . Then and . Subtracting , we find that and . Subtracting again, we find that Finally,
Solution 4 (Extending Lines)
Refer to the diagram from Solution 1.
Let us start by connecting points to to create a new line segment . We drop a perpendicular line segment from to side at point . Since , and , we know that
Therefore, is a 30-60-90 triangle. Assume that . Therefore, we know that , .
Let us draw such that is perpendicular to . Since the distance between the parallel lines are equal, we know that is half of the distance between the parallel lines. Hence, will be one-third the length of .
From this, we also know that .
Hence, we know that Since we know that is also a 30-60-90 triangle. From this, we know that .
Let us drop another perpendicular line segment from to point such that is perpendicular to .
Since and since is the side length of the regular hexagon, we know that
By symmetry, we also know that .
Therefore, we can find the length of . Hence, we know that
Now, we can find the area of trapezoid
By symmetry, we know that the area of trapezoid
Using the area of a hexagon formula, we get that the area of regular hexagon
Hence, the area of hexagon
Hence, the ratio of the area of hexagon to the area of hexagon is
~yk2007 :D
Solution 5
We will do this by area subtraction. Drawing in the required lines, we drop an altitude from A to WZ and an altitude from E to XY. You get 2 30-60-90 triangles, and given that the side length is , AZ is , which means that WZ is , adding in AB. Using the 30-60-90 triangle again, you get that the area of trapezoid is . Repeat on the other side to get that the area of both of these trapezoids combined are . Finding the area of the hexagon, dividing, and subtracting, gets you C.
-dragoon (LATEX help)
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=638
~ pi_is_3.14
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 AMC 10 Problems and Solutions |
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