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

Latest revision as of 22:35, 3 September 2024

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

First, note that $AB=37$ ; let the tangents from $I$ to $\omega$ have length $x$ . Then the perimeter of $\triangle ABI$ is equal to $2(x+AD+DB)=2(x+37).$ It remains to compute $\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x$ .

Observe $CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}$ , so the radius of $\omega$ is $\dfrac{210}{37}$ . We may also compute $AD=\dfrac{12^{2}}{37}$ and $DB=\dfrac{35^{2}}{37}$ by similar triangles. Let $O$ be the center of $\omega$ ; notice that $\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}$ so it follows $\sin(\angle DAO)=\dfrac{35}{\sqrt{35^{2}+24^{2}}}=\dfrac{35}{\sqrt{1801}}$ while $\cos(\angle DAO)=\dfrac{24}{\sqrt{1801}}$ . By the double-angle formula $\sin(2\theta)=2\sin\theta\cos\theta$ , it turns out that $\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{2\cdot 35\cdot 24}{1801}=\dfrac{1680}{1801}$

Using the area formula $\dfrac{1}{2}ab\sin(C)$ in $\triangle ABI$ , $[ABI]=\left(\dfrac{1}{2}\right)\left(\dfrac{144}{37}+x\right)(37)\left(\dfrac{1680}{1801}\right)=\left(\dfrac{840}{1801}\right)(144+37x).$ But also, using $rs$ , $[ABI]=\left(\dfrac{210}{37}\right)(37+x).$ Now we can get $\dfrac{[ABI]}{210}=\dfrac{4(144+37x)}{1801}=\dfrac{37+x}{37}$ so multiplying everything by $37\cdot 1801=66637$ lets us solve for $x$ : $21312+5476x=66637+1801x.$ We have $x=\dfrac{66637-21312}{5476-1801}=\dfrac{45325}{3675}=\dfrac{37}{3}$ , and now $2+\dfrac{2}{37}x=2+\dfrac{2}{3}=\dfrac{8}{3}$ giving the answer, $\boxed{011}$ .

Solution 2

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 3

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 $8+3=\boxed{011}$

Note: If you don't want to solve the quadratic, you can continue with $37+x = 2 \sqrt{x(x+37)}$ and divide both sides by $\sqrt{x+37}$ to get $\sqrt{37+x} = 2 \sqrt{x}$ . Square both sides to get $37+x = 4x$ and solve to get $x=\frac{37}{3}$ .

Solution 4

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 5

$[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,O,X; C=origin; A=(0,12); B=(18,0); D=foot(C,A,B); O = (C+D)/2; real r = length(D-C)/2; path c = CR(O, r); pair OA = (O+A)/2; real rA = length(A-O)/2; pair Ap = OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B); dot("$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 6

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 7

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

Solution 8 (no trig and intuitive)

Notice $CD$ is the altitude of $\triangle ABC$ , and so $CD = \frac{AC \cdot BC}{AB} = \frac{210}{37}.$ This means in the inradius of the circle is $r=\frac{CD}{2}=\frac{210}{37}.$

Next, by $\triangle ABC \sim \triangle ACD \sim \triangle CBD$ , we have $AD=CD \cdot \frac{12}{35} = \frac{144}{37}$ and $BD = CD \cdot \frac{35}{12} = \frac{1225}{37}$ .

Lastly, denote the tangent point of $AI, BI$ with the circle as $P, Q$ respectively. Then by the properties of inscribed circles, we have that $AP=AD=\frac{144}{37}$ and $BQ=BD=\frac{1225}{37}$ , and $PI=QI=x$ . We want to make an algebraic expression utilizing what we know about ABI.

We use the inradius formula: $r = \frac{[\triangle ABI]}{s}$ . We have $s=37+x$ . With Heron's (very basic calculations) we get $[\triangle ABI] = \sqrt{(x+37)(x)(\frac{1225}{37})(\frac{144}{37})} = \frac{420}{37} \sqrt{x(37+x)}.$ Using the inradius formula, we get $\frac{210}{37} = \frac{\frac{420}{37} \sqrt{x(37+x)}}{37+x}$ which simplifies to $x=\frac{37}{3}$ . The answer is then $\frac{8}{3}. \boxed{11}$ ~ brocolimanx

@@ Line 2: / Line 2: @@
 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 1==
+First, note that <math>AB=37</math>; let the tangents from <math>I</math> to <math>\omega</math> have length <math>x</math>. Then the perimeter of <math>\triangle ABI</math> is equal to <cmath>2(x+AD+DB)=2(x+37).</cmath> It remains to compute <math>\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x</math>.
-== Solution ==
+Observe <math>CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}</math>, so the radius of <math>\omega</math> is <math>\dfrac{210}{37}</math>. We may also compute <math>AD=\dfrac{12^{2}}{37}</math> and <math>DB=\dfrac{35^{2}}{37}</math> by similar triangles. Let <math>O</math> be the center of <math>\omega</math>; notice that <cmath>\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}</cmath> so it follows <cmath>\sin(\angle DAO)=\dfrac{35}{\sqrt{35^{2}+24^{2}}}=\dfrac{35}{\sqrt{1801}}</cmath> while <math>\cos(\angle DAO)=\dfrac{24}{\sqrt{1801}}</math>. By the double-angle formula <math>\sin(2\theta)=2\sin\theta\cos\theta</math>, it turns out that <cmath>\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{2\cdot 35\cdot 24}{1801}=\dfrac{1680}{1801}</cmath>
+Using the area formula <math>\dfrac{1}{2}ab\sin(C)</math> in <math>\triangle ABI</math>, <cmath>[ABI]=\left(\dfrac{1}{2}\right)\left(\dfrac{144}{37}+x\right)(37)\left(\dfrac{1680}{1801}\right)=\left(\dfrac{840}{1801}\right)(144+37x).</cmath> But also, using <math>rs</math>, <cmath>[ABI]=\left(\dfrac{210}{37}\right)(37+x).</cmath> Now we can get <cmath>\dfrac{[ABI]}{210}=\dfrac{4(144+37x)}{1801}=\dfrac{37+x}{37}</cmath> so multiplying everything by <math>37\cdot 1801=66637</math> lets us solve for <math>x</math>: <cmath>21312+5476x=66637+1801x.</cmath> We have <math>x=\dfrac{66637-21312}{5476-1801}=\dfrac{45325}{3675}=\dfrac{37}{3}</math>, and now <cmath>2+\dfrac{2}{37}x=2+\dfrac{2}{3}=\dfrac{8}{3}</cmath> giving the answer, <math>\boxed{011}</math>.
+[[File:AIME 2009-I12 Geogebra Diagram.png]]
+==Solution 2==
 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>BP = 144, AQ = 1225, AB = 1369</math> and the radius <math>r = OD = 210</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>.
-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)}}, \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>.
+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 m/n)
+(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 a/b such that (a,b) = 1, look at both side, we can know that a has to be 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.
+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 3==
+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>
+Note: If you don't want to solve the quadratic, you can continue with <math>37+x = 2 \sqrt{x(x+37)}</math> and divide both sides by <math>\sqrt{x+37}</math> to get <math>\sqrt{37+x} = 2 \sqrt{x}</math>. Square both sides to get <math>37+x = 4x</math> and solve to get <math>x=\frac{37}{3}</math>.
+==Solution 4==
+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 5==
+<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 <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>
+The perimeter of <math>\triangle ABI</math> is <math>2(x+y+z)=2(37+z)</math>. The desired ratio is then
+<cmath>\rho = 2\left(1+\frac z{37}\right)</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>
+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>
+and <math>8+3=\boxed{011}</math>.
+==Solution 6==
+This solution is not a real solution and is solving the problem with a ruler and compass.
+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
+==Solution 7==
+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>
+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
+==Solution 8 (no trig and intuitive)==
+Notice <math>CD</math> is the altitude of <math>\triangle ABC</math>, and so <math>CD = \frac{AC \cdot BC}{AB} = \frac{210}{37}.</math> This means in the inradius of the circle is <math>r=\frac{CD}{2}=\frac{210}{37}.</math>
-Let's try if x = 1369/3 fits. Since 1369/3 + 1369 = 4*1369/3, and 3*1369*(x + 144)(x + 1225)/(1801*1261) = 3*1369*(1801/3)*(1261*4/3)/1801*1261 = 4*1369/3. Amazingly it fits!
+Next, by <math>\triangle ABC \sim \triangle ACD \sim \triangle CBD</math>, we have <math>AD=CD \cdot \frac{12}{35} = \frac{144}{37}</math> and <math>BD = CD \cdot \frac{35}{12} = \frac{1225}{37}</math>.
-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.
+Lastly, denote the tangent point of <math>AI, BI</math> with the circle as <math>P, Q</math> respectively. Then by the properties of inscribed circles, we have that <math>AP=AD=\frac{144}{37}</math> and <math>BQ=BD=\frac{1225}{37}</math>, and <math>PI=QI=x</math>. We want to make an algebraic expression utilizing what we know about ABI.
-Hence the perimeter is 1225*2 + 144*2 + (1369/3)*2 = 1369*(8/3), and BC is 1369. Hence m/n = 8/3, m + n = 11.
+We use the inradius formula: <math>r = \frac{[\triangle ABI]}{s}</math>. We have <math>s=37+x</math>. With Heron's (very basic calculations) we get <math>[\triangle ABI] = \sqrt{(x+37)(x)(\frac{1225}{37})(\frac{144}{37})} = \frac{420}{37} \sqrt{x(37+x)}.</math> Using the inradius formula, we get <cmath>\frac{210}{37} = \frac{\frac{420}{37} \sqrt{x(37+x)}}{37+x}</cmath> which simplifies to <math>x=\frac{37}{3}</math>. The answer is then <math>\frac{8}{3}. \boxed{11}</math>~ brocolimanx
 == See also ==
 {{AIME box|year=2009|n=I|num-b=11|num-a=13}}
+[[Category: Intermediate Geometry Problems]]
+{{MAA Notice}}

2009 AIME I (Problems • Answer Key • Resources)
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