Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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== Solution == | == Solution == | ||
+ | this diagram only shows one possible value of r | ||
+ | <asy> | ||
+ | unitsize(5cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | |||
+ | draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle); | ||
+ | draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle); | ||
+ | |||
+ | label("$E$",(0,0),SW); | ||
+ | label("$F$",(-0.5,0.866),W); | ||
+ | label("$A$",(-0.414, 1.015),NW); | ||
+ | label("$B$",(0.586,1.015),NE); | ||
+ | label("$C$",(0.6716,0.866),E); | ||
+ | label("$D$",(0.1716, 0),SE); | ||
+ | label("1",(-0.25,0.433),SW); | ||
+ | label("r",(-0.457,0.9405),NW); | ||
+ | label("1",(0.086,1.015),N); | ||
+ | label("r",(0.6288,0.9405),NE); | ||
+ | label("1",(0.4216,0.433),SE); | ||
+ | label("r",(0.0858,0),S); | ||
+ | |||
+ | </asy> | ||
+ | ~Brian | ||
===Solution 1=== | ===Solution 1=== | ||
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>. | 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>. | ||
Line 27: | Line 50: | ||
===Solution 3 (no trig)=== | ===Solution 3 (no trig)=== | ||
+ | <asy> | ||
+ | unitsize(5cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | |||
+ | draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle); | ||
+ | draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle); | ||
+ | draw((0.586,1.015)--(0.6716,1.015)--(0.6716,0.866)); | ||
+ | |||
+ | label("$E$",(0,0),SW); | ||
+ | label("$F$",(-0.5,0.866),W); | ||
+ | label("$A$",(-0.414, 1.015),NW); | ||
+ | label("$B$",(0.586,1.015),N); | ||
+ | label("$C$",(0.6716,0.866),E); | ||
+ | label("$D$",(0.1716, 0),SE); | ||
+ | label("$M$",(0.6716,1.015),NE); | ||
+ | label("1",(-0.25,0.433),SW); | ||
+ | label("r",(-0.457,0.9405),NW); | ||
+ | label("1",(0.086,1.015),N); | ||
+ | label("r",(0.6288,0.9405),W); | ||
+ | label("1",(0.4216,0.433),SE); | ||
+ | label("r",(0.0858,0),S); | ||
+ | label("$\frac{r}{2}$",(0.6289,1.015),N); | ||
+ | label("$\frac{\sqrt{3}}{2}r$",(0.6716,0.9405),E); | ||
− | Extend <math>AB</math> so that it creates right triangle <math>\triangle | + | </asy> |
+ | Extend <math>AB</math> to point M so that it creates right triangle <math>\triangle AMC</math> where <math>\angle M = 90^\circ</math>. It is given that the hexagon is equiangular, therefore <math>\angle MBC = \frac{360}{6} = 60^\circ</math>. (exterior angles of a polygon add up to 360 <math>^\circ</math>) | ||
− | We can use either Pythagorean theorem or the properties of a <math>30-60-90</math> triangle to find the length of <math> | + | We can use either Pythagorean theorem or the properties of a <math>30-60-90</math> triangle to find the length of <math>BM={r \over 2}</math> and <math>CM = {\sqrt 3 \over 2 }r</math>. The legs of <math>\triangle AMC</math> are <math>1 + {r \over 2}</math> and <math>{\sqrt 3 \over 2 }r</math>. |
− | 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>. | + | Using Pythagorean theorem, we get <math>AC^{2} = (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>. |
− | 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> | + | 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>CM</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>. |
Latest revision as of 04:12, 16 January 2020
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
this diagram only shows one possible value of r ~Brian
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 to point M so that it creates right triangle where . It is given that the hexagon is equiangular, therefore . (exterior angles of a polygon add up to 360 )
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 .
~Zeric Hang
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.