2011 AMC 10B Problems/Problem 20
- The following problem is from both the 2011 AMC 12B #16 and 2011 AMC 10B #20, so both problems redirect to this page.
Contents
Problem
Rhombus has side length and °. Region consists of all points inside the rhombus that are closer to vertex than any of the other three vertices. What is the area of ?
Solution
Suppose that is a point in the rhombus and let be the perpendicular bisector of . Then if and only if is on the same side of as . The line divides the plane into two half-planes; let be the half-plane containing . Let us define similarly and . Then is equal to . The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:
Since and are equilateral, contains , contains and , and contains . Then with and so . Multiply this by 4 and it turns out that the pentagon has area .
Solution 2
We follow the steps shown above until we draw pentagon . We know that rhombus can be divided into equilateral triangles and . Using the special right triangle rules, we find the height of the equilateral triangles (and the height of the rhombus) to be . Therefore, the area of is . We now have to take off the areas , , and to get the desired shape. is just half of and and are each , for a total area of .
Solution 3
We split rhombus into two equilateral triangles, and . In triangle , the probability that a randomly selected point is closer to than either other point is (why?). Similarly, in triangle , the same principle applies. Thus, the area of the region closer to than , , or is . Since and are congruent, we have , and we are done.
Solution 4
Since and are halfway between and , respectively, we know that . By symmetry, is equilateral, so and therefore and are 30-60-90 right triangles. Thus, . We know that , so therefore . Summing these three regions, we get . ~ Technodoggo, Asymptote diagram modified from Solution 1
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |
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