2007 AIME I Problems/Problem 9

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

Solution 1

Label the points as in the diagram above. If we draw $\overline{O_1A}$ and $\overline{O_2B}$ , we form two right triangles. As $\overline{AF}$ and $\overline{AD}$ are both tangents to the circle, we see that $\overline{O_1A}$ is an angle bisector. Thus, $\triangle AFO_1 \cong \triangle ADO_1$ . Call $x = AD = AF$ and $y = EB = BG$ . We know that $x + y + 2r = 34$ .

If we call $\angle CAB = \theta$ , then $\angle DAO_1 = \frac{180 - \theta}{2}$ . Apply the tangent half-angle formula ( $\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$ ). We see that $\frac rx = \tan \frac{180 - \theta}{2} = \sqrt{\frac{1 - \cos (180 - \theta)}{1 + \cos (180 - \theta)}}$ $= \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}$ . Also, $\cos \theta = \frac{30}{34} = \frac{15}{17}$ . Thus, $\frac rx = \sqrt{\frac{1 + \frac{15}{17}}{1 - \frac{15}{17}}}$ , and $x = \frac{r}{4}$ .

Similarly, we find that $y = r/\sqrt{\frac{1 + \frac{8}{17}}{1 - \frac{8}{17}}} = \frac{3r}{5}$ .

Therefore, $x + y + 2r = \frac{r}{4} + \frac{3r}{5} + 2r = \frac{57r}{20} = 34 \Longrightarrow r = \frac{680}{57}$ , and $p + q = 737$ .

Solution 2

Use a similar solution to the aforementioned solution. Instead, call $\angle CAB = 2\theta$ , and then proceed by simplifying through identities. We see that $\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)$ . In terms of $r$ , we find that $x = \frac{r}{\cot \theta} = \frac{r\sin \theta}{\cos \theta}$ . Similarly, we find that $y = \frac{r \sin(45 - \theta)}{\cos (45 - \theta)}$ .

Substituting, we find that $r\left(\frac{\sin \theta}{\cos \theta} + \frac{\sin(45 - \theta)}{\cos (45 - \theta)} + 2\right) = 34$ . Under a common denominator, $r\left(\frac{\sin \theta \cos (45 - \theta) + \cos \theta \sin (45 - \theta)}{\cos \theta \cos (45 - \theta)} + 2\right) = 34$ . Trigonometric identities simplify this to $r\left(\frac{\sin\left((\theta) + (45 - \theta)\right)}{\frac 12 \left(\cos (\theta + 45 - \theta) + \cos (\theta - 45 + \theta) \right)} + 2\right) = 34$ . From here, it is possible to simplify:

$r\left(\frac{2 \sin 45}{\cos 45 + \cos 2\theta \cos 45 + \sin 2\theta \sin 45} +2\right) = 34$

$r\left(\frac{2}{\frac{17}{17} + \frac{8}{17} + \frac{15}{17}} + 2\right) = 34$

$r\left(\frac{57}{20}\right) = 34$

Our answer is $34 \cdot \frac{20}{57} = \frac{680}{57}$ , and $p + q = 737$ .

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 $EB=GB$ . Let $EB=x$ . Draw the two perpendicular radii to G 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 EBG is $-\frac{15}{17}$ . Since the measure of the angle opposite to EBG 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 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 $2.4x$ . Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, $\displaystyle 8x+2.4x+x=34$ . Solving we find that $4x=\frac{680}{57}$ so our answer is 737.

Solution 4

By Pythagoras, $AB = 34$ . Let $I_{C}$ be the $C$ -excenter of triangle $ABC$ . Then the $C$ -exradius $r_{C}$ is given by $r_{C}= \frac{K}{s-c}= \frac{240}{40-34}= 40$ .

The circle with center $O_{1}$ is tangent to both $AB$ and $AC$ , which means that $O_{1}$ lies on the external angle bisector of $\angle BAC$ . Therefore, $O_{1}$ lies on $AI_{C}$ . Similarly, $O_{2}$ lies on $BI_{C}$ .

Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$ . The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$ , so $O_{1}O_{2}$ is parallel to $AB$ , which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar.

The distance from $I_{C}$ to $AB$ is $r_{C}= 40$ , so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$ . Therefore,

$\frac{40-r}{40}= \frac{O_{1}O_{2}}{AB}= \frac{2r}{34}\quad \Rightarrow \quad r = \frac{680}{57}$ .

Hence, the final answer is $680+57 = 737$ .

Solution 5

Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\frac{1}{2}\times 16s\times 30s=\frac{1}{2}\times 34s\times h=\frac{1}{2}\times rp,$ where $p$ 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 $\frac{46s}{34s}=\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$ :

$\frac{240s}{17}\times\frac{23}{17} = \frac{240}{17}+12s$

$\displaystyle 20s\times 23 = 20\times 17+s\times 17\times 17$

$s = \frac{340}{171}$

$r = 6s = \frac{680}{57}$

The answer is $737$ .

Solution 6

Using homothecy in the diagram above, as well as the auxiliary triangle, leads to the solution.

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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