2007 AIME I Problems/Problem 9
Contents
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
Label the points as in the diagram above. If we draw and , we form two right triangles. As and are both tangents to the circle, we see that is an angle bisector. Thus, . Call and . We know that .
If we call , then . Apply the tangent half-angle formula (). We see that . Also, . Thus, , and .
Similarly, we find that .
Therefore, , and .
Solution 2
Use a similar solution to the aforementioned solution. Instead, call , and then proceed by simplifying through identities. We see that . In terms of , we find that . Similarly, we find that .
Substituting, we find that . Under a common denominator, . Trigonometric identities simplify this to . From here, it is possible to simplify:
Our answer is , and .
Solution 3
Let the point where CB's extension hits the circle be G, and the point where the hypotenuse hits that circle be E. Clearly . Let . Draw the two perpendicular radii to G 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 EBG is . Since the measure of the angle opposite to EBG 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 D. Draw the radii to F and D 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 4
By Pythagoras, . Let be the -excenter of triangle . Then the -exradius is given by .
The circle with center is tangent to both and , which means that lies on the external angle bisector of . Therefore, lies on . Similarly, lies on .
Let be the common radius of the circles with centers and . The distances from points and to are both , so is parallel to , which means that triangles and are similar.
The distance from to is , so the distance from to is . Therefore,
.
Hence, the final answer is .
Solution 5
Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, and radius, are found via where is the perimeter.
Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles.
The linear dimensions of the new triangle are times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for :
The answer is .
Solution 6
Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution.
Solution 7
A different approach is to plot the triangle on the Cartesian Plane with at , at , and at . We wish to find the coordinates of and in terms of the radius, which will be expressed as in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to . All points units away from are on the line with slope , and y-intercept
will have x-coordinate and likewise will have y-coordinate plugging this into the equation for the line mentioned in the sentence above gives us:
and
By the distance formula and the fact that the circles and tangent, we have:
which simplifies into the quadratic equation:
And by the quadratic equation, the solutions are: The solution including the "" is extraneous so we have the radius equal to
Which simplifies to . The sum of the numerator and the denominator is
Solution 8 (simple algebra)
It is known that is parallel to AB. Thus, extending and to intersect at H yields similar triangles and BAC, so that , , and . It should be noted that . Also, FHGC is a rectangle, and so AF = and similarly for BG. Because tangents to a circle are equal, the hypotenuse can be expressed in terms of r: Thus, r = , and the answer is
Solution 9 (Simple Algebra)
Let the radius of the circle be r. It can be seen that and are similar to , and the length of the hypotenuses are and , respectively. Then, the entire length of is going to be . The length of the hypotenuse of is 34, so the length of the height to is . Thus, the height to is going to be . is similar to so we have the following: . Cross multiplying and simplifying, we get that r = so the answer is . ~Leonard_my_dude~
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.