Difference between revisions of "2005 AIME I Problems/Problem 15"
m (→Solution) |
(→Solution) |
||
Line 16: | Line 16: | ||
Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline{AD}</math>, we have | Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline{AD}</math>, we have | ||
− | <cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = | + | <cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.</cmath> |
Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get | Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get | ||
− | <cmath>(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath> | + | <cmath>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath> |
Thus <math>c = 2</math> or <math> = 10</math>. We discard the value <math>c = 10</math> as extraneous (it gives us an [[equilateral triangle]]) and are left with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>. | Thus <math>c = 2</math> or <math> = 10</math>. We discard the value <math>c = 10</math> as extraneous (it gives us an [[equilateral triangle]]) and are left with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>. |
Revision as of 19:08, 27 May 2017
Problem
Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find
Solution
Let , and be the points of tangency of the incircle with , and , respectively. Without loss of generality, let , so that is between and . Let the length of the median be . Then by two applications of the Power of a Point Theorem, , so . Now, and are two tangents to a circle from the same point, so and thus . Then so and thus .
Now, by Stewart's Theorem in triangle with cevian , we have
Our earlier result from Power of a Point was that , so we combine these two results to solve for and we get
Thus or . We discard the value as extraneous (it gives us an equilateral triangle) and are left with , so our triangle has area and so the answer is .
See also
2005 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.