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 "2009 AIME I Problems/Problem 12"

(→‎See also)
(Tag: Blanking)
Line 1: Line 1:
 +
== Problem ==
 +
In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
 +
== 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 <math>AB=37</math>. Thus, <math>CD=\frac{35\cdot 12}{37}=\frac{420}{37}</math>. We also find that <math>AD=\frac{12^2}{37}</math> and <math>BD=\frac{35^2}{37}</math>. From here, we let <math>\angle AOE=\angle AOD=\alpha,</math> <math>\angle BOF=\angle BOD=\beta</math>. Thus, <math>\angle EOF=360^{\circ}-2\alpha-2\beta,</math> so <math>\angle IOE=\angle IOF=180^{\circ}-\alpha-\beta</math>. Observe that
 +
<cmath>\tan \alpha=\frac{AD}{DO}=\frac{\frac{12^2}{37}}{\frac{210}{37}}=\frac{24}{35}</cmath>
 +
and
 +
<cmath>\tan\beta=\frac{\frac{35^2}{37}}{\frac{210}{37}}=\frac{35}{6}.</cmath>
 +
Thus,
 +
<cmath>\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*}</cmath>
 +
However, we also know that <math>\tan(180^{\circ}-\alpha-\beta)=\frac{IF}{OF}=\frac{IE}{OF}.</math> Thus, we get
 +
<cmath>IF=IE=OF\cdot \tan(180^{\circ}-\alpha-\beta)=\frac{210}{37}\cdot\frac{37^2}{630}=\frac{37}{3}.</cmath>
 +
Thus, the perimeter of <math>\triangle AIB</math> is
 +
<cmath>2\left(\frac{12^2}{37}+\frac{35^2}{37}+\frac{37}{3}\right)=2\left(\frac{37^2}{37}+\frac{37}{3}\right),</cmath>
 +
which gives
 +
<cmath>2\left(\frac{37\cdot 4}{3}\right)=\frac{37\cdot 8}{3}.</cmath>
 +
Since <math>AB=37,</math> this means that the ratio of the perimeter of <math>\triangle ABI</math> to <math>AB</math> is just <math>\frac{8}{3},</math> so our answer is
 +
<cmath>8+3=\boxed{11}.</cmath>

Revision as of 19:29, 3 November 2022

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 \[8+3=\boxed{11}.\]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_12&oldid=180003"