Difference between revisions of "2007 AIME I Problems/Problem 9"

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== Solution ==
 
== Solution ==
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[[Image:AIME_2007_-9.PNG]]
 
Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly <math>EB=DB</math>. Let <math>EB=x</math>. Draw the two [[perpendicular]] radii to D and E. Now we have a [[cyclic quadrilateral]]. Let the radius be length <math>r</math>. We see that since the cosine of angle ABC is <math>\frac{15}{17}</math> the cosine of angle EBD is <math>-\frac{15}{17}</math>. Since the measure of the angle opposite to EBD is the [[complement]] of this one, its cosine is <math>\frac{15}{17}</math>. Using the law of cosines, we see that <math>x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}</math>
 
Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly <math>EB=DB</math>. Let <math>EB=x</math>. Draw the two [[perpendicular]] radii to D and E. Now we have a [[cyclic quadrilateral]]. Let the radius be length <math>r</math>. We see that since the cosine of angle ABC is <math>\frac{15}{17}</math> the cosine of angle EBD is <math>-\frac{15}{17}</math>. Since the measure of the angle opposite to EBD is the [[complement]] of this one, its cosine is <math>\frac{15}{17}</math>. Using the law of cosines, we see that <math>x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}</math>
 
This tells us that <math>r=4x</math>.
 
This tells us that <math>r=4x</math>.

Revision as of 00:41, 16 March 2007

Problem

In right triangle $ABC$ with right angle $C$, $CA = 30$ and $CB = 16$. Its legs $CA$ and $CB$ are extended beyond $A$ and $B$. Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$, the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$, and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

AIME 2007 -9.PNG Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly $EB=DB$. Let $EB=x$. Draw the two perpendicular radii to D and E. Now we have a cyclic quadrilateral. Let the radius be length $r$. We see that since the cosine of angle ABC is $\frac{15}{17}$ the cosine of angle EBD is $-\frac{15}{17}$. Since the measure of the angle opposite to EBD is the complement of this one, its cosine is $\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}$ This tells us that $r=4x$.

Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle G. Draw the radii to F and G and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, 8x+2.4x+x=34. Solving we find that $4x=\frac{680}{57}$ so our answer is 737.

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions