Difference between revisions of "2007 AIME I Problems/Problem 15"
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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. | 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{251-3\sqrt{989}}{10}</math>, and the desired answer is <math>\boxed{989}</math> | <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{251-3\sqrt{989}}{10}</math>, and the desired answer is <math>\boxed{989}</math> | ||
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+ | ~bluesoul | ||
== See also == | == See also == |
Revision as of 01:03, 1 February 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 inequality , 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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