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

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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>.
 
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 ==
+
== 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>.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> 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 <math>\frac {m}{n}</math>)
 +
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.
 +
 
 +
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!
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
==Solution 2==
 +
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.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
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
 +
 
 +
<math>rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}</math>
 +
 
 +
<math>210/37(37+x) = 12*35/37 \sqrt{x(37+x)}</math>
 +
 
 +
<math>37+x = 2 \sqrt{x(x+37)}</math>
 +
 
 +
<math>x^2+74x+1369 = 4x^2 + 148x</math>
 +
 
 +
<math>3x^2 + 74x - 1369 = 0</math>
 +
 
 +
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>
 +
 
 +
==Solution 3==
 +
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.
 +
 
 +
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 <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>.
 +
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>.
 +
 
 +
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>.
 +
 
 +
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>
 +
 
 +
== Solution 4 ==
 
<asy>
 
<asy>
size(13cm, 0);
+
size(300);
pair midpoint(pair coord1, pair coord2) {
+
defaultpen(linewidth(0.4)+fontsize(10));
pair output = ((coord1.x + coord2.x) / 2, (coord1.y + coord2.y) / 2);
+
pen s = linewidth(0.8)+fontsize(8);
return output;
 
}
 
  
real dist(pair coord1, pair coord2) {
+
pair A,B,C,D,O,X;
real xdiff = coord2.x - coord1.x;
+
C=origin;
    real ydiff = coord2.y - coord1.y;
+
A=(0,12);
    real num = (xdiff * xdiff) + (ydiff * ydiff);
+
B=(18,0);
    return sqrt(num);
+
D=foot(C,A,B);
}
+
O = (C+D)/2;
 +
real r = length(D-C)/2;
 +
path c = CR(O, r);
 +
pair OA = (O+A)/2;
 +
real rA = length(A-O)/2;
 +
pair Ap = OP(CR(OA,rA), c);
 +
pair OB = (O+B)/2;
 +
real rB = length(B-O)/2;
 +
pair Bp = OP(CR(OB,rB), c);
 +
X=extension(A,Ap,B,Bp);
 +
draw(A--B--C--A, s);
 +
draw(C--D^^B--O--A^^Ap--O--X, gray+0.25);
 +
draw(c^^A--X--B);
  
pair C = (0, 0);
+
dot("$A$", A, N);
pair B = (35, 0);
+
dot("$B$", B, SE);
pair A = (0, 12);
+
dot("$C$", C, SW);
pair O = (210/37 * 12/37, 210/37*35/37);
+
dot("$D$", D, 0.2*(D-C));
draw(A -- B -- C -- A);
+
dot("$I$", X, 0.5*(X-C));
pair D = (420/37 * 12/37, 420/37 * 35/37);
+
dot("$P$", Ap, 0.3*(Ap-O));
draw(C -- D);
+
dot("$Q$", Bp, 0.3*(Bp-O));
draw(circle(O, 210/37));
+
dot("$O$", O, W);
pair Btan = intersectionpoints(Circle(O, 210/37), Circle(midpoint(B, O), dist(B, O) / 2))[1];
+
label("$\beta$",B,10*dir(157));
pair Atan = intersectionpoints(Circle(O, 210/37), Circle(midpoint(A, O), dist(A, O) / 2))[1];
+
label("$\alpha$",A,5*dir(-55));
pair I = extension(A, Atan, B, Btan);
+
label("$\theta$",X,5*dir(55));
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>
 
</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
+
Let <math>AP=AD=x</math>, let <math>BQ=BD=y</math>, and let <math>IP=IQ=z</math>. Let <math>OD=r</math>. We find <math>AB=37</math>. Let <math>\alpha</math>, <math>\beta</math>, and <math>\theta</math> be the angles <math>OAD</math>, <math>OBD</math>, and <math>OPI</math> respectively. Then <math>\alpha + \beta + \theta = 90^\circ</math>, so <cmath>\theta = 90^\circ - (\alpha+\beta).</cmath>
<cmath>\tan \alpha=\frac{AD}{DO}=\frac{\frac{12^2}{37}}{\frac{210}{37}}=\frac{24}{35}</cmath>
+
The perimeter of <math>\triangle ABI</math> is <math>2(x+y+z)=2(37+z)</math>. The desired ratio is then
and
+
<cmath>\rho = 2\left(1+\frac z{37}\right)</cmath>
<cmath>\tan\beta=\frac{\frac{35^2}{37}}{\frac{210}{37}}=\frac{35}{6}.</cmath>
+
We need to find <math>z</math>. In <math>\triangle OPI</math>, <math>z=r\cot\theta = r\tan (\alpha+\beta)</math>. We get <cmath>\tan\alpha = \frac{OD}{AD} = \frac 12 \frac{CD}{AD} = \frac 12  \tan A = \frac 12 \frac{BC}{AC} = \frac{35}{24}.</cmath> Similarly, <math>\tan\beta = \tfrac 6{35}</math>. Then <cmath>z = r\cdot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}</cmath>
Thus,
+
Computing <math>[ABC]</math> in two ways we get <math>CD = \tfrac{12\cdot 35}{37}</math>, so <math>r=\tfrac{6\cdot 35}{37}</math>. Using this value of <math>r</math> we get <math>z=\tfrac {37}3</math>. Thus <cmath>\rho = 2\left(1+\frac 1{3}\right) = \frac 8{3},</cmath>
<cmath>\begin{align*}
+
and <math>8+3=\boxed{011}</math>.
\tan(180^{\circ}-\alpha-\beta)&=-\tan(\alpha+\beta)\
+
 
&=-\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\right)\
+
== Solution 5 ==
&=-\left(\frac{\frac{24}{35}+\frac{35}{6}}{1-\frac{24}{35}\cdot\frac{35}{6}}\right)\
+
This solution is not a real solution and is solving the problem with a ruler and compass.
&=-\left(\frac{\frac{12^2+35^2}{210}}{1-4}\right)\
+
 
&=\frac{\frac{37^2}{210}}{3}\
+
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
&=\frac{37^2}{630}.
+
 
\end{align*}</cmath>
+
== Solution 6==
However, we also know that <math>\tan(180^{\circ}-\alpha-\beta)=\frac{IF}{OF}=\frac{IE}{OF}.</math> Thus, we get
+
Denoting three tangents has length <math>h_1,h_2,h_3</math> while <math>h_1,h_3</math> lies on <math>AB</math> with <math>h_1>h_3</math>.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>.As we know that the diameter of the circle is the height of <math>\triangle ACB</math> from <math>C</math> to <math>AB</math>. Assume that <math>\tan\alpha=\frac{h_1}{r}</math> and <math>\tan\beta=\frac{h_3}{r}</math> and <math>\tan\omega=\frac{h_2}{r}</math>. But we know that <math>\tan(\alpha+\beta)=-\tan(180-\alpha-\beta)=-\tan\omega</math> According to the basic computation, we can get that <math>\tan(\alpha)=\frac{35}{6}</math>; <math>\tan(\beta)=\frac{24}{35}</math>
<cmath>IF=IE=OF\cdot \tan(180^{\circ}-\alpha-\beta)=\frac{210}{37}\cdot\frac{37^2}{630}=\frac{37}{3}.</cmath>
+
So we know that <math>\tan(\omega)=\frac{1369}{630}</math> according to the tangent addition formula. Hence, it is not hard to find that the length of <math>h_2</math> is <math>\frac{37}{3}</math>. According to basic addition and division, we get the answer is <math>\frac{8}{3}</math> which leads to <math>8+3=\boxed{11}</math> ~bluesoul
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>
 
  
 
== 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}}
 +
[[Category: Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 17:51, 25 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 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

[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8);  pair A,B,C,D,O,X; C=origin; A=(0,12); B=(18,0); D=foot(C,A,B); O = (C+D)/2; real r = length(D-C)/2; path c = CR(O, r); pair OA = (O+A)/2; real rA = length(A-O)/2; pair Ap = OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B);  dot("$A$", A, N); dot("$B$", B, SE); dot("$C$", C, SW); dot("$D$", D, 0.2*(D-C)); dot("$I$", X, 0.5*(X-C)); dot("$P$", Ap, 0.3*(Ap-O)); dot("$Q$", Bp, 0.3*(Bp-O)); dot("$O$", O, W); label("$\beta$",B,10*dir(157)); label("$\alpha$",A,5*dir(-55)); label("$\theta$",X,5*dir(55)); [/asy] Let $AP=AD=x$, let $BQ=BD=y$, and let $IP=IQ=z$. Let $OD=r$. We find $AB=37$. Let $\alpha$, $\beta$, and $\theta$ be the angles $OAD$, $OBD$, and $OPI$ respectively. Then $\alpha + \beta + \theta = 90^\circ$, so \[\theta = 90^\circ - (\alpha+\beta).\] The perimeter of $\triangle ABI$ is $2(x+y+z)=2(37+z)$. The desired ratio is then \[\rho = 2\left(1+\frac z{37}\right)\] We need to find $z$. In $\triangle OPI$, $z=r\cot\theta = r\tan (\alpha+\beta)$. We get \[\tan\alpha = \frac{OD}{AD} = \frac 12 \frac{CD}{AD} = \frac 12  \tan A = \frac 12 \frac{BC}{AC} = \frac{35}{24}.\] Similarly, $\tan\beta = \tfrac 6{35}$. Then \[z = r\cdot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}\] Computing $[ABC]$ in two ways we get $CD = \tfrac{12\cdot 35}{37}$, so $r=\tfrac{6\cdot 35}{37}$. Using this value of $r$ we get $z=\tfrac {37}3$. Thus \[\rho = 2\left(1+\frac 1{3}\right) = \frac 8{3},\] 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

Solution 6

Denoting three tangents has length $h_1,h_2,h_3$ while $h_1,h_3$ lies on $AB$ with $h_1>h_3$.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 $r=\frac{210}{37}$.As we know that the diameter of the circle is the height of $\triangle ACB$ from $C$ to $AB$. Assume that $\tan\alpha=\frac{h_1}{r}$ and $\tan\beta=\frac{h_3}{r}$ and $\tan\omega=\frac{h_2}{r}$. But we know that $\tan(\alpha+\beta)=-\tan(180-\alpha-\beta)=-\tan\omega$ According to the basic computation, we can get that $\tan(\alpha)=\frac{35}{6}$; $\tan(\beta)=\frac{24}{35}$ So we know that $\tan(\omega)=\frac{1369}{630}$ according to the tangent addition formula. Hence, it is not hard to find that the length of $h_2$ is $\frac{37}{3}$. According to basic addition and division, we get the answer is $\frac{8}{3}$ which leads to $8+3=\boxed{11}$ ~bluesoul

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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