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2009 AIME I Problems/Problem 12

Revision as of 21:36, 24 November 2022 by Eclipse471 (talk | contribs)

Problem

In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$. The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

[asy] size(13cm, 0); pair midpoint(pair coord1, pair coord2) { pair output = ((coord1.x + coord2.x) / 2, (coord1.y + coord2.y) / 2); return output; } real dist(pair coord1, pair coord2) { real xdiff = coord2.x - coord1.x; real ydiff = coord2.y - coord1.y; real num = (xdiff * xdiff) + (ydiff * ydiff); return sqrt(num); } pair C = (0, 0); pair B = (35, 0); pair A = (0, 12); pair O = (210/37 * 12/37, 210/37*35/37); draw(A -- B -- C -- A); pair D = (420/37 * 12/37, 420/37 * 35/37); draw(C -- D); draw(circle(O, 210/37)); pair Btan = intersectionpoints(Circle(O, 210/37), Circle(midpoint(B, O), dist(B, O) / 2))[1]; pair Atan = intersectionpoints(Circle(O, 210/37), Circle(midpoint(A, O), dist(A, O) / 2))[1]; pair I = extension(A, Atan, B, Btan); draw(A -- I -- B); label(A, "A", W); label(C, "C", S); label(B, "B", E); label(I, "I", S); label(D, "D", N); label(Atan, "E", W); label(Btan, "F", SE); draw(O -- Atan); draw(O -- Btan); dot(O); label(O, "O", E); [/asy] Note that $AB=37$. Thus, $CD=\frac{35\cdot 12}{37}=\frac{420}{37}$. We also find that $AD=\frac{12^2}{37}$ and $BD=\frac{35^2}{37}$. From here, we let $\angle AOE=\angle AOD=\alpha,$ $\angle BOF=\angle BOD=\beta$. Thus, $\angle EOF=360^{\circ}-2\alpha-2\beta,$ so $\angle IOE=\angle IOF=180^{\circ}-\alpha-\beta$. Observe that \[\tan \alpha=\frac{AD}{DO}=\frac{\frac{12^2}{37}}{\frac{210}{37}}=\frac{24}{35}\] and \[\tan\beta=\frac{\frac{35^2}{37}}{\frac{210}{37}}=\frac{35}{6}.\] Thus, \begin{align*} \tan(180^{\circ}-\alpha-\beta)&=-\tan(\alpha+\beta)\\ &=-\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\right)\\ &=-\left(\frac{\frac{24}{35}+\frac{35}{6}}{1-\frac{24}{35}\cdot\frac{35}{6}}\right)\\ &=-\left(\frac{\frac{12^2+35^2}{210}}{1-4}\right)\\ &=\frac{\frac{37^2}{210}}{3}\\ &=\frac{37^2}{630}. \end{align*} However, we also know that $\tan(180^{\circ}-\alpha-\beta)=\frac{IF}{OF}=\frac{IE}{OF}.$ Thus, we get \[IF=IE=OF\cdot \tan(180^{\circ}-\alpha-\beta)=\frac{210}{37}\cdot\frac{37^2}{630}=\frac{37}{3}.\] Thus, the perimeter of $\triangle AIB$ is \[2\left(\frac{12^2}{37}+\frac{35^2}{37}+\frac{37}{3}\right)=2\left(\frac{37^2}{37}+\frac{37}{3}\right),\] which gives \[2\left(\frac{37\cdot 4}{3}\right)=\frac{37\cdot 8}{3}.\] Since $AB=37,$ this means that the ratio of the perimeter of $\triangle ABI$ to $AB$ is just $\frac{8}{3},$ so our answer is \[8+3=\boxed{11}.\]

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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