Difference between revisions of "2007 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | + | === Solution 1 === | |
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| − | == Solution | ||
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>. | ||
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. | 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 <math>2.4x</math>. 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 <math>4x=\frac{680}{57}</math> so our answer is 737. | + | Using the law of cosines again, we find that the length of our tangents is <math>2.4x</math>. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, <math>\displaystyle 8x+2.4x+x=34</math>. Solving we find that <math>4x=\frac{680}{57}</math> so our answer is 737. |
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| + | === Solution 2 === | ||
| + | [[Image:AIME_2007_-9.PNG]] | ||
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| + | Using [[homothecy]] in the diagram above, as well as the [[auxiliary]] triangle, leads to the solution. | ||
== See also == | == See also == | ||
Revision as of 12:37, 16 March 2007
Problem
In right triangle 
with right angle 
, 
and 
. Its legs 
and 
are extended beyond 
and 
. Points 
and 
lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center is tangent to the hypotenuse and to the extension of leg , the circle with center is tangent to the hypotenuse and to the extension of leg , and the circles are externally tangent to each other. The length of the radius either circle can be expressed as 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Solution 1
Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly 
. Let 
. Draw the two perpendicular radii to D and E. Now we have a cyclic quadrilateral. Let the radius be length 
. We see that since the cosine of angle ABC is 
the cosine of angle EBD is 
. Since the measure of the angle opposite to EBD is the complement of this one, its cosine is . Using the law of cosines, we see that 
This tells us that 
.
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 
. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, 
. Solving we find that 
so our answer is 737.
Solution 2
Using homothecy in the diagram above, as well as the auxiliary triangle, leads to the solution.
See also
| 2007 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
