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

(→‎Solution 2: trig-heavy solution)
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=== Solution 2 ===
 
=== Solution 2 ===
−
Label the points as they are in the diagram above. Call <math>x = AD = AF</math> and <math>y = EB = BG</math>. We know that <math>x + y + 2r = 34</math>.
+
Label the points as in the diagram above. Call <math>x = AD = AF</math> and <math>y = EB = BG</math>. We know that <math>x + y + 2r = 34</math>.
  
 
If we draw <math>\overline{O_1A}</math> and <math>\overline{O_2B}</math>, we form two [[right triangle]]s. If we call <math>\angle CAB = 2\theta</math>, we see that <math>\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)</math>. In terms of <math>r</math>, we find that <math>x = \frac{r}{\cot \theta} = \frac{r\sin \theta}{\cos \theta}</math>. Similarly, we find that <math>y = \frac{r \sin(45 - \theta)}{\cos (45 - \theta)}</math>.  
 
If we draw <math>\overline{O_1A}</math> and <math>\overline{O_2B}</math>, we form two [[right triangle]]s. If we call <math>\angle CAB = 2\theta</math>, we see that <math>\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)</math>. In terms of <math>r</math>, we find that <math>x = \frac{r}{\cot \theta} = \frac{r\sin \theta}{\cos \theta}</math>. Similarly, we find that <math>y = \frac{r \sin(45 - \theta)}{\cos (45 - \theta)}</math>.  
  
 
Substituting, we find that <math>r\left(\frac{\sin \theta}{\cos \theta} + \frac{\sin(45 - \theta)}{\cos (45 - \theta)} + 2\right) = 34</math>. Under a common denominator, <math>r\left(\frac{\sin \theta \cos (45 - \theta) + \cos \theta \sin (45 - \theta)}{\cos \theta \cos (45 - \theta)} + 2\right) = 34</math>. [[Trigonometric identities]] simplify this to <math>r\left(\frac{\sin\left((\theta) + (45 - \theta)\right)}{\frac 12 \left(\cos (\theta + 45 - \theta) + \cos (\theta - 45 + \theta) \right)} + 2\right) = 34</math>. From here, it is possible to simplify:
 
Substituting, we find that <math>r\left(\frac{\sin \theta}{\cos \theta} + \frac{\sin(45 - \theta)}{\cos (45 - \theta)} + 2\right) = 34</math>. Under a common denominator, <math>r\left(\frac{\sin \theta \cos (45 - \theta) + \cos \theta \sin (45 - \theta)}{\cos \theta \cos (45 - \theta)} + 2\right) = 34</math>. [[Trigonometric identities]] simplify this to <math>r\left(\frac{\sin\left((\theta) + (45 - \theta)\right)}{\frac 12 \left(\cos (\theta + 45 - \theta) + \cos (\theta - 45 + \theta) \right)} + 2\right) = 34</math>. From here, it is possible to simplify:
−
:<math>r\left(\frac{\cos 45}{\cos 45 + \cos 2\theta \cos 45 + \sin 2\theta \sin 45} +2\right) = 34</math>
+
:<math>r\left(\frac{2 \sin 45}{\cos 45 + \cos 2\theta \cos 45 + \sin 2\theta \sin 45} +2\right) = 34</math>
 
:<math>r\left(\frac{2}{\frac{17}{17} + \frac{8}{17} + \frac{15}{17}} + 2\right) = 34</math>
 
:<math>r\left(\frac{2}{\frac{17}{17} + \frac{8}{17} + \frac{15}{17}} + 2\right) = 34</math>
 
:<math>r\left(\frac{57}{20}\right) = 34</math>
 
:<math>r\left(\frac{57}{20}\right) = 34</math>

Revision as of 19:23, 17 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 $p+q$.

AIME I 2007-9.png

Solution

Solution 1

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 2

Label the points as in the diagram above. Call $x = AD = AF$ and $y = EB = BG$. We know that $x + y + 2r = 34$.

If we draw $\overline{O_1A}$ and $\overline{O_2B}$, we form two right triangles. If we call $\angle CAB = 2\theta$, 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

AIME 2007 -9.PNG

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
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