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Difference between revisions of "2009 AIME I Problems/Problem 12"

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== Solution ==
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== Solution 1==
 
Let <math>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math>Q</math> is on <math>AI</math>. We know that <math>AD:CD = CD:BD = 12:35</math>.
 
Let <math>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math>Q</math> is on <math>AI</math>. We know that <math>AD:CD = CD:BD = 12:35</math>.
  
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Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>BP = 144, AQ = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>.  
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Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>AQ = 144, BP = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>.  
  
 
Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},</math><math>\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>.
 
Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},</math><math>\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>.
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Hence the perimeter is <math>1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}</math>, and <math>BC</math> is <math>1369</math>. Hence <math>\frac {m}{n} = \frac {8}{3}</math>, <math>m + n = 11</math>.
 
Hence the perimeter is <math>1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}</math>, and <math>BC</math> is <math>1369</math>. Hence <math>\frac {m}{n} = \frac {8}{3}</math>, <math>m + n = 11</math>.
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==Solution 2==
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As in Solution <math>1</math>, let <math>P</math> and <math>Q</math> be the intersections of <math>\omega</math> with <math>BI</math> and <math>AI</math> respectively.  Let <math>IP=IQ=x</math>.
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First, by pythagorean theorem, <math>AB = \sqrt{12^2+35^2} = 37</math>.  Now the area of <math>ABC</math> is <math>1/2*12*35 = 1/2*37*CD</math>, so <math>CD=\frac{420}{37}</math> and the inradius of <math>\triangle ABI</math> is <math>r=\frac{210}{37}</math>.
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Now from <math>\triangle CDB ~ \triangle ACB</math> we find that <math>\frac{BC}{BD} = \frac{AB}{BC}</math> so <math>BD = BC^2/AB = 35^2/37</math> and similarly, <math>AD = 12^2/37</math>.
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Now we have <math>AI = 144/27+x</math>, <math>BI = 1225/37+x</math>.  Now we can compute the area of <math>\triangle ABI</math> in two ways: by heron's formula and by inradius times semiperimeter, which yields
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<math>rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/27)x}</math>
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Solving yields <math>x=37/3</math>.  Plugging this back in, the perimeter of <math>ABI</math> is <math>2s=2(37+x)=2(37+37/3) = 37(8/3)</math> so the ratio of the perimeter to <math>AB</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2009|n=I|num-b=11|num-a=13}}

Revision as of 18:39, 29 March 2009

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 1

Let $O$ be center of the circle and $P$,$Q$ be the two points of tangent such that $P$ is on $BI$ and $Q$ is on $AI$. We know that $AD:CD = CD:BD = 12:35$.

Since the ratios between corresponding lengths of two similar diagrams are equal, we can let $AD = 144, CD = 420$ and $BD = 1225$. Hence $AQ = 144, BP = 1225, AB = 1369$ and the radius $r = OD = 210$.

Since we have $\tan OAB = \frac {35}{24}$ and $\tan OBA = \frac{6}{35}$ , we have $\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},$$\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}$.

Hence $\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}$. let $IP = IQ = x$ , then we have Area$(IBC)$ = $(2x + 1225*2 + 144*2)*\frac {210}{2}$ = $(x + 144)(x + 1225)* \sin {\frac {I}{2}}$. Then we get $x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}$.

Now the equation looks very complex but we can take a guess here. Assume that $x$ is a rational number (If it's not then the answer to the problem would be irrational which can't be in the form of $\frac {m}{n}$) that can be expressed as $\frac {a}{b}$ such that $(a,b) = 1$. Look at both sides; we can know that $a$ has to be a multiple of $1369$ and not of $3$ and it's reasonable to think that $b$ is divisible by $3$ so that we can cancel out the $3$ on the right side of the equation.

Let's see if $x = \frac {1369}{3}$ fits. Since $\frac {1369}{3} + 1369 = \frac {4*1369}{3}$, and $\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}$. Amazingly it fits!

Since we know that $3*1369*144*1225 - 1369*1801*1261 < 0$, the other solution of this equation is negative which can be ignored. Hence $x = 1369/3$.

Hence the perimeter is $1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}$, and $BC$ is $1369$. Hence $\frac {m}{n} = \frac {8}{3}$, $m + n = 11$.

Solution 2

As in Solution $1$, let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively. Let $IP=IQ=x$.

First, by pythagorean theorem, $AB = \sqrt{12^2+35^2} = 37$. Now the area of $ABC$ is $1/2*12*35 = 1/2*37*CD$, so $CD=\frac{420}{37}$ and the inradius of $\triangle ABI$ is $r=\frac{210}{37}$.

Now from $\triangle CDB ~ \triangle ACB$ we find that $\frac{BC}{BD} = \frac{AB}{BC}$ so $BD = BC^2/AB = 35^2/37$ and similarly, $AD = 12^2/37$.

Now we have $AI = 144/27+x$, $BI = 1225/37+x$. Now we can compute the area of $\triangle ABI$ in two ways: by heron's formula and by inradius times semiperimeter, which yields

$rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/27)x}$

Solving yields $x=37/3$. Plugging this back in, the perimeter of $ABI$ is $2s=2(37+x)=2(37+37/3) = 37(8/3)$ so the ratio of the perimeter to $AB$ is $8/3$ and our answer is $8+3=\boxed{011}$

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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