# Difference between revisions of "2010 AIME I Problems/Problem 15"

## Problem

In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

## Solution

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### Solution 1

Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$.

By Stewart's Theorem, we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$, we have $432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$

Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $12x$ is an integer. The only such $x$ in the above-stated range is $\frac {22}3$.

Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\frac {22}3$. Then $CM = \frac {23}3$, and our desired ratio $\frac {AM}{CM} = \frac {22}{23}$, giving us an answer of $\boxed{045}$.

### Solution 2

Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$. Let the incenters of $\triangle ABM$ and $\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$. From $r = \sqrt {(s - a)(s - b)(s - c)/s}$, we find that

\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\ & = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}

Let the incircle of $\triangle ABM$ meet $AM$ at $P$ and the incircle of $\triangle BCM$ meet $CM$ at $Q$. Then note that $I_1 P Q I_2$ is a rectangle. Also, $\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\angle AMB$ and $\angle CMB$ respectively and $\angle AMB + \angle CMB = 180^\circ$. By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC)/2 = - x + y + 1$. Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$, so by similar triangles we find that $r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields

\begin{align*} 2y^2 - 30 = 2xy + 5x - 7y \\ 2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}

$$4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\ \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.$$

### Solution 3

Let the incircle of $ABM$ hit $AM$, $AB$, $BM$ at $X_{1},Y_{1},Z_{1}$, and let the incircle of $CBM$ hit $MC$, $BC$, $BM$ at $X_{2},Y_{2},Z_{2}$. Draw the incircle of $ABC$, and let it be tangent to $AC$ at $X$. Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$, and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$.

By standard computations, $AX=\dfrac{AB+AC-BC}{2}=7$ and $CX=\dfrac{BC+AC-AB}{2}=8$. Now, let $AX_{1}=7x$ and $CX_{2}=8x$. We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$. Then, $BZ_{1}=12-7x$. We also have $CY_{2}=CX_{2}=8x$, so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$.

Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$. Also,$X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$. Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).

Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$. To finish, we will compute area ratios. $\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}$. Also, since their inradii are equal, we get $\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}$. Equating and cross multiplying yields the quadratic $3x^{2}-8x+4=0$, so $x=2/3,2$. However, observe that $AX_{1}+CX_{2}=15x<15$, so we take $x=2/3$. Our ratio is therefore $\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}$, giving the answer $\boxed{045}$.

Note: Once we have $MX_1=8-8x$ and $MX_2=7-7x$, it's bit easier to use use the right triangle of $O_1MO_2$ than chasing the area ratio. The inradius of $\triangle{ABC}$ can be calculated to be $r=\sqrt{14}$, and the inradius of $ABM$ and $ACM$ are $r_1=r_2= xr$, so, $$O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2$$ or, $$(15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2$$ $$112(1-x)^2 = 28x^2$$ $$4(1-x)^2 = x^2$$ We get $x=\frac{2}{3}$ or $x=2$.

### Solution 4

Suppose the incircle of $ABM$ touches $AM$ at $X$, and the incircle of $CBM$ touches $CM$ at $Y$. Then

$$r = AX \tan(A/2) = CY \tan(C/2)$$

We have $\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}$, $\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}$

$\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}$, $\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}$,

Therefore $AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.$

And since $AX=\frac{1}{2}(12+AM-BM)$, $CY = \frac{1}{2}(13+CM-BM)$,

$$\frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}$$

$$96+8AM-8BM = 91 +7CM-7BM$$

$$BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)$$

Now,

$\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}$

$$\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}$$ $$6AM^2 - 35AM = 45AM-264$$ $$3AM^2 -40AM+132=0$$ $$(3AM-22)(AM-6)=0$$

So $AM=22/3$ or $6$. But from (1) we know that $5+15(AM-7)>0$, or $AM>7-1/3>6$, so $AM=22/3$, $CM=15-22/3=23/3$, $AM/CM=22/23$.

### Solution 5

Let the common inradius equal r, $BM = x$, $AM = y$, $MC = z$

From the prespective of $\triangle{ABM}$ and $\triangle{BMC}$ we get:

$S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})$ $S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})$

Add two triangles up, we get $\triangle{ABC}$ :

$S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}$

Since $y + z = 15$, we get:

$r = \frac{S_{ABC}}{20 + x}$


By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$, we can get:

$r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2}$ and

$r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}$

$r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x$

$r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}$


Now, we have 2 values equal to r, we can set them equal to each other:

$\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}$


If we let R denote the incircle of ABC, note:

AC = $(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15$ and

$S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R$.

By cross multiplying the equation above, we get:

$400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300$


We can find out x:

$x = 10$.


Now, we can find ratio of y and z:

$\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}$


The answer is $\boxed{045}$.

-Alexlikemath

### Solution 6 (Similar to Solution 1 with easier computation)

Let $CM=x, AM=rx, BM=d$. $x+rx=15\Rightarrow x=\frac{15}{1+r}$.

Similar to Solution 1, we have $$r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}$$ as well as $$12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})$$ $$\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}$$ $$\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2}$$ $$(169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r)$$ $$(r^2+1)(400r)=2r(338r^2-224r+288)$$ $$100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0$$

Since $d=\frac{13r-12}{1-r}>0$, we have $r=\frac{22}{23} \longrightarrow \boxed{045}$.

~ asops

## Sidenote

In the problem, $BM=10$ and the equal inradius of the two triangles happens to be $\frac {2\sqrt{14}}{3}$.