Difference between revisions of "2007 AIME II Problems/Problem 15"
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[[Image:2007 AIME II-15.png]] | [[Image:2007 AIME II-15.png]] | ||
=== Solution 1 === | === Solution 1 === | ||
− | First, apply [[Heron's formula]] to find that | + | First, apply [[Heron's formula]] to find that <math>[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84</math>. The semiperimeter is <math>21</math>, so the [[inradius]] is <math>\frac{A}{s} = \frac{84}{21} = 4</math>. |
− | Now consider the [[incenter]] <math>I</math> of <math>\triangle ABC</math>. Let the [[radius]] of one of the small circles be <math>r</math>. Let the centers of the three little circles tangent to the sides of <math>\triangle ABC</math> be <math>O_A</math>, <math>O_B</math>, and <math>O_C</math>. Let the center of the circle tangent to those three circles be <math>O</math>. The [[homothety]] <math>\mathcal{H}\left(I, \frac{4-r}{4}\right)</math> maps <math>\triangle ABC</math> to <math>\triangle XYZ</math>; since <math>OO_A = OO_B = OO_C = 2r</math>, <math>O</math> is the circumcenter of <math>\triangle XYZ</math> and <math>\mathcal{H}</math> therefore maps the circumcenter of <math>\triangle ABC</math> to <math>O</math>. Thus, <math>2r = R \cdot \frac{4 - r}{4}</math>, where <math>R</math> is the [[circumradius]] of <math>\triangle ABC</math>. Substituting <math>R = \frac{abc}{[ABC]} = \frac{65}{8}</math>, <math>r = \frac{260}{129}</math> and the answer is <math>\boxed{389}</math>. | + | Now consider the [[incenter]] <math>I</math> of <math>\triangle ABC</math>. Let the [[radius]] of one of the small circles be <math>r</math>. Let the centers of the three little circles tangent to the sides of <math>\triangle ABC</math> be <math>O_A</math>, <math>O_B</math>, and <math>O_C</math>. Let the center of the circle tangent to those three circles be <math>O</math>. The [[homothety]] <math>\mathcal{H}\left(I, \frac{4-r}{4}\right)</math> maps <math>\triangle ABC</math> to <math>\triangle XYZ</math>; since <math>OO_A = OO_B = OO_C = 2r</math>, <math>O</math> is the circumcenter of <math>\triangle XYZ</math> and <math>\mathcal{H}</math> therefore maps the circumcenter of <math>\triangle ABC</math> to <math>O</math>. Thus, <math>2r = R \cdot \frac{4 - r}{4}</math>, where <math>R</math> is the [[circumradius]] of <math>\triangle ABC</math>. Substituting <math>R = \frac{abc}{4[ABC]} = \frac{65}{8}</math>, <math>r = \frac{260}{129}</math> and the answer is <math>\boxed{389}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 17:17, 27 November 2020
Problem
Four circles and with the same radius are drawn in the interior of triangle such that is tangent to sides and , to and , to and , and is externally tangent to and . If the sides of triangle are and the radius of can be represented in the form , where and are relatively prime positive integers. Find
Contents
[hide]Solution
Solution 1
First, apply Heron's formula to find that . The semiperimeter is , so the inradius is .
Now consider the incenter of . Let the radius of one of the small circles be . Let the centers of the three little circles tangent to the sides of be , , and . Let the center of the circle tangent to those three circles be . The homothety maps to ; since , is the circumcenter of and therefore maps the circumcenter of to . Thus, , where is the circumradius of . Substituting , and the answer is .
Solution 2
Consider a 13-14-15 triangle. [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is , where is the semiperimeter. Scale the triangle with the inradius by a linear scale factor,
The circumradius is where and are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, .
Cut and combine the triangles, as shown. Then solve for :
The solution is .
See also
2007 AIME II (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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