Difference between revisions of "2007 AIME I Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
First of all, assume <math>EC=x,BD=m, ED=a, EF=b</math>, then we can find <math>BF=m-3, AE=2+m-x</math> | First of all, assume <math>EC=x,BD=m, ED=a, EF=b</math>, then we can find <math>BF=m-3, AE=2+m-x</math> | ||
− | It is not hard to find <math>ab*sin60^{\circ}*\frac{1}{2}=14\sqrt{3}, ab=56</math>, we apply LOC on <math>\triangle{DEF}, \triangle{ | + | It is not hard to find <math>ab*sin60^{\circ}*\frac{1}{2}=14\sqrt{3}, ab=56</math>, we apply LOC on <math>\triangle{DEF}, \triangle{BFD}</math>, getting that <math>(m-3)^2+m^2-m(m-3)=a^2+b^2-ab</math>, leads to <math>a^2+b^2=m^2-3m+65</math> |
Apply LOC on <math>\triangle{CED}, \triangle{AEF}</math> separately, getting <math>4+x^2-2x=a^2; 25+(2+m-x)^2-5(2+m-x)=b^2.</math> Add those terms together and use the equality <math>a^2+b^2=m^2-3m+65</math>, we can find: | Apply LOC on <math>\triangle{CED}, \triangle{AEF}</math> separately, getting <math>4+x^2-2x=a^2; 25+(2+m-x)^2-5(2+m-x)=b^2.</math> Add those terms together and use the equality <math>a^2+b^2=m^2-3m+65</math>, we can find: | ||
<math>2x^2-(2m+1)x+2m-42=0</math> | <math>2x^2-(2m+1)x+2m-42=0</math> |
Latest revision as of 21:14, 6 August 2022
Contents
[hide]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 exterior 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 .
Solution 2
First of all, assume , then we can find It is not hard to find , we apply LOC on , getting that , leads to Apply LOC on separately, getting Add those terms together and use the equality , we can find:
According to basic angle chasing, , so , the ratio makes , getting that Now we have two equations with , and values for both equations must be the same, so we can solve for in two equations. , then we can just use positive sign to solve, simplifies to , getting , since the triangle is equilateral, , and the desired answer is
~bluesoul
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.