Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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<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> | <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>. | This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>. | ||
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+ | ===Solution 3 (no trig)=== | ||
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+ | Extend <math>AB</math> so that it creates right triangle <math>\triangle AEC</math> where <math>\angle E = 90^\circ</math>. It is given that the hexagon is equiangular, therefore <math>\angle ABC = 120^\circ</math>. <math>\angle ABC</math> and <math>\angle EBC</math> are supplementary so <math>\angle EBC = 60^\circ</math>. | ||
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+ | We can use either Pythagorean theorem or the properties of a <math>30-60-90</math> triangle to find the length of <math>BE={r \over 2}</math> and <math>CE = {\sqrt 3 \over 2 }r</math>. The legs of <math>\triangle AEC</math> are <math>1 + {r \over 2}</math> and <math>{\sqrt 3 \over 2 }r</math>. | ||
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+ | Using Pythagorean theorem, we get <math>AC = (r^2+r+1)</math>. We can then follow <math>\textbf {Solution 1}</math> to solve for <math>r</math>. <math>\boxed{\textbf{(E)}\ 6}</math>. | ||
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+ | Alternatively, we can find the area of <math>\triangle ABC</math>. We know that the three smaller triangles: <math>\triangle ABC</math>, <math>\triangle CDE</math>, and <math>\triangle EFA</math> are congruent because of <math>S-A-S</math>. Therefore one of the smaller triangles accounts for <math>10\%</math> of the total area. The height of the smaller triangle <math>\triangle ABC</math> is just <math>CE</math> so the area is <math>{1 \cdot {\sqrt 3 \over 2 }r \over 2}</math>. We can then find the area of the hexagon using <math>\textbf {Solution 1}</math>. | ||
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+ | We can even find the area of <math>\triangle ACE</math> and <math>\triangle ABC</math> and solve for <math>r</math> because the ratio of the areas is <math>7</math> to <math>1</math>. | ||
== See also == | == See also == |
Revision as of 10:09, 2 July 2018
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 (or Girard identities, or Newton-Girard identities) 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 .
Solution 3 (no trig)
Extend so that it creates right triangle where . It is given that the hexagon is equiangular, therefore . and are supplementary so .
We can use either Pythagorean theorem or the properties of a triangle to find the length of and . The legs of are and .
Using Pythagorean theorem, we get . We can then follow to solve for . .
Alternatively, we can find the area of . We know that the three smaller triangles: , , and are congruent because of . Therefore one of the smaller triangles accounts for of the total area. The height of the smaller triangle is just so the area is . We can then find the area of the hexagon using .
We can even find the area of and and solve for because the ratio of the areas is 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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