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

 
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== Solution ==
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== Solution 1==
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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>AQ = 144, BP = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>.
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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 <math>\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}</math>. let <math>IP = IQ = x</math> , then we have Area<math>(IBC)</math> = <math>(2x + 1225*2 + 144*2)*\frac {210}{2}</math> = <math>(x + 144)(x + 1225)* \sin {\frac {I}{2}}</math>. Then we get <math>x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}</math>.
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Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> is a rational number
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(If it's not then the answer to the problem would be irrational which can't be in the form of <math>\frac {m}{n}</math>)
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that can be expressed as <math>\frac {a}{b}</math> such that <math>(a,b) = 1</math>. Look at both sides; we can know that <math>a</math> has to be a multiple of <math>1369</math> and not of <math>3</math> and it's reasonable to think that <math>b</math> is divisible by <math>3</math> so that we can cancel out the <math>3</math> on the right side of the equation.
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Let's see if <math>x = \frac {1369}{3}</math> fits. Since <math>\frac {1369}{3} + 1369 = \frac {4*1369}{3}</math>, and <math>\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}</math>. Amazingly it fits!
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Since we know that <math>3*1369*144*1225 - 1369*1801*1261 < 0</math>, the other solution of this equation is negative which can be ignored. Hence <math>x = 1369/3</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>.
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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.
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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 \sim \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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Note <math>IP=IQ=x</math>, <math>BP=BD</math>, and <math>AQ=AD</math>. So we have <math>AI = 144/37+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)(37-1225/37)x}</math>
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<math>210/37(37+x) = 12*35/37 \sqrt{x(37+x)}</math>
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<math>37+x = 2 \sqrt{x(x+37)}</math>
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<math>x^2+74x+1369 = 4x^2 + 148x</math>
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<math>3x^2 + 74x - 1369 = 0</math>
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The quadratic formula now 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>
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==Solution 3==
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As in Solution <math>2</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.
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Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
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Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
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Let <math>x = \overline{AD} = \overline{AQ}</math>. Let <math>y = \overline{BD} = \overline{BP}</math>. Let <math>z = \overline{PI} = \overline{QI}</math>. The semi-perimeter of <math>ABI</math> is <math>x + y + z</math>.
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Since the lengths of the sides of <math>ABI</math> are <math>x + y</math>, <math>y + z</math> and <math>x + z</math>, the square of its area by Heron's formula is <math>(x+y+z)xyz</math>.
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The radius <math>r</math> of <math>\omega</math> is <math>\overline{CD}/2</math>. Therefore <math>r^2 = xy/4</math>. As <math>\omega</math> is the in-circle of <math>ABI</math>, the area of <math>ABI</math> is also <math>r(x+y+z)</math>, and so the square area is <math>r^2(x+y+z)^2</math>.
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Therefore <cmath>(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}</cmath> Dividing both sides by <math>xy(x+y+z)/4</math> we get: <cmath>4z = (x+y+z),</cmath> and so <math>z = (x+y)/3</math>. The semi-perimeter of <math>ABI</math> is therefore <math>\frac{4}{3}(x+y)</math> and the whole perimeter is <math>\frac{8}{3}(x+y)</math>. Now <math>x + y = \overline{AB}</math>, so the ratio of the perimeter of <math>ABI</math> to the hypotenuse <math>\overline{AB}</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math>
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== Solution 4 ==
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We shall yet again let <math>P</math> and <math>Q</math> be the intersections of <math>AI</math> and <math>BI</math> to <math>\omega</math>, respectively. We want to find the perimeter of <math>ABI</math>, which is <math>AD+BD+BQ+QI+IP+PA</math>. We can easily find <math>AD</math> and <math>BD</math> using the fact that <math>ABC</math>, <math>ACD</math>, and <math>BCD</math> are all similar triangles. We get <math>AD=\frac{144}{37}</math> and <math>\frac{1225}{37}</math>. Since <math>AP</math> and <math>AD</math> are tangents to <math>\omega</math>, <math>AP=AD=\frac{144}{37}</math>, and similarly <math>BQ=BD=\frac{1225}{37}</math>. We now wish to find <math>IP</math> and <math>IQ</math>.
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Let the center of the given circle be <math>O</math>. We know that <math>\angle AOP=\angle AOD</math>, <math>\angle BOQ=\angle BOD</math>, and <math>\angle IOQ=\angle IOP</math>. Since all six angles sum to <math>360^{\circ}</math>, <math>\angle AOP+\angle BOQ+\angle IOP=180^{\circ}</math>. If we knew the radius of circle <math>\omega</math> now, then we could find <math>\tan{\angle AOP}</math> and <math>\tan{\angle BOQ}</math>, and then we can use the sum (or difference) of tangents formula to find <math>\tan{\angle IOP}</math>, which reveals <math>IP</math>. This means we should find the radius of <math>\omega</math>. We can easily see that the height of triangle <math>ABC</math> from <math>C</math> has length <math>\frac{12*35}{37}</math>, so the radius of <math>\omega</math> is <math>\frac{210}{37}</math>. Now we shall proceed with the above plan.
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<math>\tan{\angle AOP}=\frac{144}{210}</math>. <math>\tan{\angle BOQ}=\frac{1225}{210}</math>.
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<math>\tan{\angle IOP}=\tan{(180^{\circ}-\angle AOP-\angle BOQ)}=-\tan{(\angle AOP+\angle BOQ)}</math>
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<math>=-\frac{\frac{144}{210}+\frac{1225}{210}}{1-\frac{144}{210}*\frac{1225}{210}}=-\frac{1369}{210-\frac{144*1225}{210}}=\frac{1369}{\frac{144*1225}{210}-210}=\frac{37*37}{35*18}</math>.
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Therefore <math>OP=\frac{210}{37},IP=\frac{37}{3}</math>, and the perimeter of <math>AIB</math> is <math>2*\frac{37}{3}+2*\frac{144}{37}+2*\frac{1225}{37}=37*\frac{8}{3}</math>. Since <math>AB=37</math>, the desired ratio is <math>\frac{8}{3}</math>, and <math>8+3=\boxed{011}</math>.
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== Solution 5 ==
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This solution is not a real solution and is solving the problem with a ruler and compass.
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Draw <math>AC = 4.8, BC = 14, AB = 14.8</math>. Then, drawing the tangents and intersecting them, we get that <math>IA</math> is around <math>6.55</math> and <math>IB</math> is around <math>18.1</math>. We then find the ratio to be around <math>\frac{39.45}{14.8}</math>. Using long division, we find that this ratio is approximately 2.666, which you should recognize as <math>\frac{8}{3}</math>. Since this seems reasonable, we find that the answer is <math>\boxed{11}</math> ~ilp
 
== 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}}
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[[Category: Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:04, 30 December 2020

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.

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

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

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

$210/37(37+x) = 12*35/37 \sqrt{x(37+x)}$

$37+x = 2 \sqrt{x(x+37)}$

$x^2+74x+1369 = 4x^2 + 148x$

$3x^2 + 74x - 1369 = 0$

The quadratic formula now 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 $8+3=\boxed{011}$

Solution 3

As in Solution $2$, let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.

Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.

Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.

Let $x = \overline{AD} = \overline{AQ}$. Let $y = \overline{BD} = \overline{BP}$. Let $z = \overline{PI} = \overline{QI}$. The semi-perimeter of $ABI$ is $x + y + z$. Since the lengths of the sides of $ABI$ are $x + y$, $y + z$ and $x + z$, the square of its area by Heron's formula is $(x+y+z)xyz$.

The radius $r$ of $\omega$ is $\overline{CD}/2$. Therefore $r^2 = xy/4$. As $\omega$ is the in-circle of $ABI$, the area of $ABI$ is also $r(x+y+z)$, and so the square area is $r^2(x+y+z)^2$.

Therefore \[(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}\] Dividing both sides by $xy(x+y+z)/4$ we get: \[4z = (x+y+z),\] and so $z = (x+y)/3$. The semi-perimeter of $ABI$ is therefore $\frac{4}{3}(x+y)$ and the whole perimeter is $\frac{8}{3}(x+y)$. Now $x + y = \overline{AB}$, so the ratio of the perimeter of $ABI$ to the hypotenuse $\overline{AB}$ is $8/3$ and our answer is $8+3=\boxed{011}$

Solution 4

We shall yet again let $P$ and $Q$ be the intersections of $AI$ and $BI$ to $\omega$, respectively. We want to find the perimeter of $ABI$, which is $AD+BD+BQ+QI+IP+PA$. We can easily find $AD$ and $BD$ using the fact that $ABC$, $ACD$, and $BCD$ are all similar triangles. We get $AD=\frac{144}{37}$ and $\frac{1225}{37}$. Since $AP$ and $AD$ are tangents to $\omega$, $AP=AD=\frac{144}{37}$, and similarly $BQ=BD=\frac{1225}{37}$. We now wish to find $IP$ and $IQ$.

Let the center of the given circle be $O$. We know that $\angle AOP=\angle AOD$, $\angle BOQ=\angle BOD$, and $\angle IOQ=\angle IOP$. Since all six angles sum to $360^{\circ}$, $\angle AOP+\angle BOQ+\angle IOP=180^{\circ}$. If we knew the radius of circle $\omega$ now, then we could find $\tan{\angle AOP}$ and $\tan{\angle BOQ}$, and then we can use the sum (or difference) of tangents formula to find $\tan{\angle IOP}$, which reveals $IP$. This means we should find the radius of $\omega$. We can easily see that the height of triangle $ABC$ from $C$ has length $\frac{12*35}{37}$, so the radius of $\omega$ is $\frac{210}{37}$. Now we shall proceed with the above plan.

$\tan{\angle AOP}=\frac{144}{210}$. $\tan{\angle BOQ}=\frac{1225}{210}$.

$\tan{\angle IOP}=\tan{(180^{\circ}-\angle AOP-\angle BOQ)}=-\tan{(\angle AOP+\angle BOQ)}$

$=-\frac{\frac{144}{210}+\frac{1225}{210}}{1-\frac{144}{210}*\frac{1225}{210}}=-\frac{1369}{210-\frac{144*1225}{210}}=\frac{1369}{\frac{144*1225}{210}-210}=\frac{37*37}{35*18}$.

Therefore $OP=\frac{210}{37},IP=\frac{37}{3}$, and the perimeter of $AIB$ is $2*\frac{37}{3}+2*\frac{144}{37}+2*\frac{1225}{37}=37*\frac{8}{3}$. Since $AB=37$, the desired ratio is $\frac{8}{3}$, and $8+3=\boxed{011}$.

Solution 5

This solution is not a real solution and is solving the problem with a ruler and compass.

Draw $AC = 4.8, BC = 14, AB = 14.8$. Then, drawing the tangents and intersecting them, we get that $IA$ is around $6.55$ and $IB$ is around $18.1$. We then find the ratio to be around $\frac{39.45}{14.8}$. Using long division, we find that this ratio is approximately 2.666, which you should recognize as $\frac{8}{3}$. Since this seems reasonable, we find that the answer is $\boxed{11}$ ~ilp

See also

2009 AIME I (ProblemsAnswer KeyResources)
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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