AIME 2009 I · 第 12 题

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

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 (same as solution 3)

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

Side Note

It can be shown that the value of $\frac83$ does not actually depend on the sides of the triangle. That is, if we do this same construction with any right triangle and compute the ratio, we will get $\frac83$. ~Puck_0