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

$[asy] /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200); /* segments and figures */ draw((0,0)--(15,0)); draw((15,0)--(6.66667,9.97775)); draw((6.66667,9.97775)--(0,0)); draw((7.33333,0)--(6.66667,9.97775)); draw(circle((4.66667,2.49444),2.49444)); draw(circle((9.66667,2.49444),2.49444)); draw((4.66667,0)--(4.66667,2.49444)); draw((9.66667,2.49444)--(9.66667,0)); /* points and labels */ label("r",(10.19662,1.92704),SE); label("r",(5.02391,1.8773),SE); dot((0,0)); label("$A$",(-1.04408,-0.60958),NE); dot((15,0)); label("$C$",(15.41907,-0.46037),NE); dot((6.66667,9.97775)); label("$B$",(6.66525,10.23322),NE); label("$15$",(6.01866,-1.15669),NE); label("$13$",(11.44006,5.50815),NE); label("$12$",(2.28834,5.75684),NE); dot((7.33333,0)); label("$M$",(7.56053,-1.000),NE); label("$H_1$",(3.97942,-1.200),NE); label("$H_2$",(9.54741,-1.200),NE); dot((4.66667,2.49444)); label("$I_1$",(3.97942,2.92179),NE); dot((9.66667,2.49444)); label("$I_2$",(9.54741,2.92179),NE); clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle); [/asy]$

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 $6x$ is an integer because we can divide the polynomial by $2$ . 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

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

and adding these we have

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

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:

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

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

Adding these up, we get:

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

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

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:

We can find out x:

Now, we can find ratio of y and z:

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

Video Solution

https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0

Sidenote

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

2010 AIME I (Problems • Answer Key • Resources)
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