Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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− | It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]], we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>. | + | It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on triangle ABC, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>. |
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore | If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore |
Revision as of 18:05, 16 January 2017
Problem
Equiangular hexagon has side lengths and . The area of is of the area of the hexagon. What is the sum of all possible values of ?
Solution
Solution 1
It is clear that is an equilateral triangle. From the Law of Cosines on triangle ABC, we get that . Therefore, the area of is .
If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore
.
Based on the initial conditions,
Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .
Solution 2
As above, we find that the area of is .
We also find by the sine triangle area formula that , and thus This simplifies to .
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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