Difference between revisions of "2008 AIME II Problems/Problem 11"
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Let <math>X</math> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle. | Let <math>X</math> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle. | ||
− | To find <math> | + | To find <math>XC</math>, consider the right triangle <math>PCX</math>. Since <math>\odot P</math> is tangent to <math>\overline{AC},\overline{BC}</math>, then <math>PC</math> [[bisect]]s <math>\angle ACB</math>. Let <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we recognize the <math>7 - 24 - 25</math> [[right triangle]], except scaled by <math>4</math>. |
So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3}</math>. | So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3}</math>. |
Revision as of 13:45, 13 November 2016
Problem
In triangle , , and . Circle has radius and is tangent to and . Circle is externally tangent to and is tangent to and . No point of circle lies outside of . The radius of circle can be expressed in the form , where , , and are positive integers and is the product of distinct primes. Find .
Solution
Let and be the feet of the perpendiculars from and to , respectively. Let the radius of be . We know that . From draw segment such that is on . Clearly, and . Also, we know is a right triangle.
To find , consider the right triangle . Since is tangent to , then bisects . Let ; then . Dropping the altitude from to , we recognize the right triangle, except scaled by .
So we get that . From the half-angle identity, we find that . Therefore, . By similar reasoning in triangle , we see that .
We conclude that .
So our right triangle has sides , , and .
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get , for a final answer of .
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.