Difference between revisions of "2008 AIME II Problems/Problem 11"
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By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>. | By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>. | ||
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+ | == Solution 2(pure synthetic) == | ||
+ | Refer to the above diagram. Let the larger circle have center <math>O_1</math>, the smaller have center <math>O_2</math>, and the incenter be <math>I</math>. We can easily calculate that the area of <math>\triangle ABC = 2688</math>, and <math>s = 128</math> and <math>R = 21</math>, where <math>R</math> is the inradius. | ||
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+ | Now, Line <math>\overline{AI}</math> is the perpendicular bisector of <math>\overline{BC}</math>, as <math>\triangle ABC</math> is isosceles. Letting the point of intersection be <math>X</math>, we get that <math>BX = 28</math> and <math>IX = 21</math>, and <math>B, O_2, I</math> are collinear as <math>O_2</math> is equidistant from <math>\overline{AB}</math> and <math>\overline{BC}</math>. By Pythagoras, <math>BI = 35</math>, and we notice that <math>\triangle BIX</math> is a 3-4-5 right triangle. | ||
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+ | Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam | ||
== See also == | == See also == |
Revision as of 19:32, 30 August 2019
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 .
Solution 2(pure synthetic)
Refer to the above diagram. Let the larger circle have center , the smaller have center , and the incenter be . We can easily calculate that the area of , and and , where is the inradius.
Now, Line is the perpendicular bisector of , as is isosceles. Letting the point of intersection be , we get that and , and are collinear as is equidistant from and . By Pythagoras, , and we notice that is a 3-4-5 right triangle.
Letting be the desired radius and letting be the projection of onto , we find that . Similarly, we find that the distance between the projection from onto , , and , is . From there, we let the projection of onto be , and we have , , and . We finish with Pythagoras on , whence we get the desired answer of . - Spacesam
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.