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 [[ | + | <math>\angle FEC</math> is an [[exterior 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>. |
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− | Use the [[quadratic formula]], though we only need the root of the [[discriminant]]. This is <math>\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}</math><math> = \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}</math>. The answer is <math>989</math>. | + | Use the [[quadratic formula]], though we only need the root of the [[discriminant]]. This is <math>\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}</math><math> = \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}</math>. The answer is <math>\boxed{989}</math>. |
+ | |||
+ | ==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> | ||
+ | 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: | ||
+ | <math>2x^2-(2m+1)x+2m-42=0</math> | ||
+ | |||
+ | According to basic angle chasing, <math>\angle{A}=\angle{C}; \angle{AFE}=\angle{CED}</math>, so <math>\triangle{AFE}\sim \triangle{CED}</math>, the ratio makes <math>\frac{5}{x}=\frac{2+m-x}{2}</math>, getting that <math>x^2-(2+m)x+10=0</math> | ||
+ | Now we have two equations with <math>m</math>, and <math>x</math> values for both equations must be the same, so we can solve for <math>x</math> in two equations. | ||
+ | <math>x=\frac{2m+1 \pm \sqrt{4m^2+4m+1-16m+336}}{4}; x=\frac{4+2m \pm \sqrt{4m^2+16m-144}}{4}</math>, then we can just use positive sign to solve, simplifies to <math>3+\sqrt{4m^2+16m-144}=\sqrt{4m^2-12m+337}</math>, getting <math>m=\frac{211-3\sqrt{989}}{10}</math>, since the triangle is equilateral, <math>AB=BC=2+m=\frac{231-3\sqrt{989}}{10}</math>, and the desired answer is <math>\boxed{989}</math> | ||
+ | |||
+ | ~bluesoul | ||
== See also == | == See also == |
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 |
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