Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(SOlution for problem 9 on this years aime)
 
(wikify)
Line 1: Line 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 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 <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>
+
== Problem ==
−
This tells us that r=4x.
+
In [[right triangle]] <math>ABC</math> with [[right angle]] <math>C</math>, <math>CA = 30</math> and <math>CB = 16</math>.  Its legs <math>CA</math> and <math>CB</math> are extended beyond <math>A</math> and <math>B</math>.  [[Point]]s <math>O_1</math> and <math>O_2</math> lie in the exterior of the triangle and are the centers of two [[circle]]s with equal [[radii]].  The circle with center <math>O_1</math> is tangent to the [[hypotenuse]] and to the extension of leg <math>CA</math>, the circle with center <math>O_2</math> is [[tangent]] to the hypotenuse and to the extension of [[leg]] <math>CB</math>, and the circles are externally tangent to each other.  The length of the radius either circle can be expressed as <math>p/q</math>, where <math>p</math> and <math>q</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>p+q</math>.
 +
 
 +
== 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 EB=DB. 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>.
  
 
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 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 <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, 8x+2.4x+x=34. Solving we find that <math>4x=\frac{680}{57}</math> so our answer is 737.
 +
 
 +
== See also ==
 +
{{AIME box|year=2007|n=I|num-b=8|num-a=10}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 18:58, 15 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

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 (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_9&oldid=14032"