Difference between revisions of "2007 AIME I Problems/Problem 15"
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Denote the length of a side of the triangle <math>x</math>, and of <math>\overline{AE}</math> as <math>y</math>. The area of the entire equilateral triangle is <math>\frac{x^2\sqrt{3}}{4}</math>. Add up the areas of the triangles using the <math>\frac{1}{2}ab\sin C</math> formula (notice that for the three outside triangles, <math>\sin 60 = \frac{\sqrt{3}}{2}</math>): <math>\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5 \cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\sqrt{3}</math>. This simplifies to <math>\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)</math>. Some terms will cancel out, leaving <math>y = \frac{5}{3}x - 22</math>. | Denote the length of a side of the triangle <math>x</math>, and of <math>\overline{AE}</math> as <math>y</math>. The area of the entire equilateral triangle is <math>\frac{x^2\sqrt{3}}{4}</math>. Add up the areas of the triangles using the <math>\frac{1}{2}ab\sin C</math> formula (notice that for the three outside triangles, <math>\sin 60 = \frac{\sqrt{3}}{2}</math>): <math>\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5 \cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\sqrt{3}</math>. This simplifies to <math>\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)</math>. Some terms will cancel out, leaving <math>y = \frac{5}{3}x - 22</math>. | ||
− | <math>\angle FEC</math> is an [[external angle]] to <math>\triangle AEF</math>, from which we find that <math> | + | <math>\angle FEC</math> is an [[external angle]] to <math>\triangle AEF</math>, from which we find that <math>60 + \angle CED = 60 + \angle AFE</math>, so <math>\angle CED = \angle AFE</math>. Similarly, we find that <math>\angle EDC = \angle AEF</math>. Thus, <math>\triangle AEF \sim \triangle CDE</math>. Setting up a [[ratio]] of sides, we get that <math>\frac{5}{x-y} = \frac{y}{2}</math>. Using the previous relationship between <math>x</math> and <math>y</math>, we can solve for <math>x</math>. |
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
− | <math> | + | <math>xy - y^2 = 10</math> |
<math>\frac{5}{3}x^2 - 22x - \left(\frac{5}{3}x - 22\right)^2 - 10 = 0</math> | <math>\frac{5}{3}x^2 - 22x - \left(\frac{5}{3}x - 22\right)^2 - 10 = 0</math> | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:13, 4 July 2013
Problem
Let be an equilateral triangle, and let and be points on sides and , respectively, with and . Point lies on side such that angle . The area of triangle is . The two possible values of the length of side are , where and are rational, and is an integer not divisible by the square of a prime. Find .
Solution
Denote the length of a side of the triangle , and of as . The area of the entire equilateral triangle is . Add up the areas of the triangles using the formula (notice that for the three outside triangles, ): . This simplifies to . Some terms will cancel out, leaving .
is an external angle to , from which we find that , so . Similarly, we find that . Thus, . Setting up a ratio of sides, we get that . Using the previous relationship between and , we can solve for .
Use the quadratic formula, though we only need the root of the discriminant. This is . The answer is .
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.