Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
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]], 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>. | ||
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Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>. | Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>. | ||
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+ | ===Solution 2=== | ||
+ | As above, we find that the area of <math>\triangle ACE</math> is <math>\frac{\sqrt3}4(r^2+r+1)</math>. | ||
+ | |||
+ | We also find by the sine [[triangle]] area formula that <math>ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4</math>, and thus | ||
+ | <cmath>\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4}\right)=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}</cmath> | ||
+ | This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>. | ||
== See also == | == See also == |
Revision as of 20:52, 7 February 2011
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, 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
\[\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4}\right)=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}\] (Error compiling LaTeX. ! Extra }, or forgotten \right.)
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 |