Difference between revisions of "2009 AIME I Problems/Problem 12"
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+ | == 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 20:29, 3 November 2022
Problem
In right with hypotenuse
,
,
, and
is the altitude to
. Let
be the circle having
as a diameter. Let
be a point outside
such that
and
are both tangent to circle
. The ratio of the perimeter of
to the length
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution
Note that
. Thus,
. We also find that
and
. From here, we let
. Thus,
so
. Observe that
and
Thus,
However, we also know that
Thus, we get
Thus, the perimeter of
is
which gives
Since
this means that the ratio of the perimeter of
to
is just
so our answer is