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

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__TOC__
 
== Problem ==
 
== Problem ==
−
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. Let <math>p</math> and <math>q</math> be positive [[relatively prime]] integers such that <math>\frac {AM}{CM} = \frac {p}{q}</math>. Find <math>p + q</math>.
+
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. Then <math>\frac{AM}{CM} = \frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>.
  
 
== Solution ==
 
== Solution ==
−
{{image}}
+
<center><asy> /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */
 +
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);
  
−
Let <math>AM = 2x</math> and <math>BM = 2y</math> so <math>CM = 15 - 2x</math>. Let the [[incenter]]s of <math>\triangle ABM</math> and <math>\triangle BCM</math> be <math>I_1</math> and <math>I_2</math> respectively, and their equal inradii be <math>r</math>. From <math>r = \sqrt {(s - a)(s - b)(s - c)/s}</math>, we find that
+
  /* 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></center>
 +
=== Solution 1 ===
 +
Let <math>AM = x</math>, then <math>CM = 15 - x</math>. Also let <math>BM = d</math> Clearly, <math>\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}</math>. We can also express each area by the rs formula. Then <math>\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}</math>. Equating and cross-multiplying yields <math>25x + 2dx = 15d + 180</math> or <math>d = \frac {25x - 180}{15 - 2x}.</math> Note that for <math>d</math> to be positive, we must have <math>7.2 < x < 7.5</math>.
 +
 
 +
By [[Stewart's Theorem]], we have <math>12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)</math> or <math>432 = 3d^2 + 40x - 3x^2.</math> Brute forcing by plugging in our previous result for <math>d</math>, we have <math>432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.</math> Clearing the fraction and gathering like terms, we get <math>0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.</math>
 +
 
 +
''Aside: Since <math>x</math> must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that <math>6x</math> is an integer because we can divide the polynomial by <math>2</math>. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.''
 +
 
 +
Legitimately solving that quartic, note that <math>x = 0</math> and <math>x = 15</math> should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).</math> The only solution in the desired range is thus <math>\frac {22}3</math>. Then <math>CM = \frac {23}3</math>, and our desired ratio <math>\frac {AM}{CM} = \frac {22}{23}</math>, giving us an answer of <math>\boxed{045}</math>.
 +
 
 +
=== Solution 2 ===
 +
Let <math>AM = 2x</math> and <math>BM = 2y</math> so <math>CM = 15 - 2x</math>. Let the [[incenter]]s of <math>\triangle ABM</math> and <math>\triangle BCM</math> be <math>I_1</math> and <math>I_2</math> respectively, and their equal inradii be <math>r</math>. From <math>r = \sqrt {(s - a)(s - b)(s - c)}/s</math>, we find that
  
 
<cmath>\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \
 
<cmath>\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*}</cmath>
 
& = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}</cmath>
  
−
Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC) = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields
+
Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC)/2 = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields
  
−
<cmath>
+
<cmath>\begin{align*}
 
2y^2 - 30 = 2xy + 5x - 7y \
 
2y^2 - 30 = 2xy + 5x - 7y \
−
2y^2 - 70 = - 2xy - 5x + 7y,
+
2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}
 
</cmath>
 
</cmath>
  
Line 24: Line 68:
 
</cmath>
 
</cmath>
  
−
== See also ==
+
=== Solution 3 ===
 +
Let the incircle of <math>ABM</math> hit <math>AM</math>, <math>AB</math>, <math>BM</math> at <math>X_{1},Y_{1},Z_{1}</math>, and let the incircle of <math>CBM</math> hit <math>MC</math>, <math>BC</math>, <math>BM</math> at <math>X_{2},Y_{2},Z_{2}</math>. Draw the incircle of <math>ABC</math>, and let it be tangent to <math>AC</math> at <math>X</math>. Observe that we have a homothety centered at A sending the incircle of <math>ABM</math> to that of <math>ABC</math>, and one centered at <math>C</math> taking the incircle of <math>BCM</math> to that of <math>ABC</math>. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is <math>AX_{1}/AX=CX_{2}/CX</math>.
 +
 
 +
By standard computations, <math>AX=\dfrac{AB+AC-BC}{2}=7</math> and <math>CX=\dfrac{BC+AC-AB}{2}=8</math>. Now, let <math>AX_{1}=7x</math> and <math>CX_{2}=8x</math>. We will now go around and chase lengths. Observe that <math>BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x</math>. Then, <math>BZ_{1}=12-7x</math>. We also have <math>CY_{2}=CX_{2}=8x</math>, so <math>BY_{2}=13-8x</math> and <math>BZ_{2}=13-8x</math>.
 +
 
 +
Observe now that <math>X_{1}M+MX_{2}=AC-15x=15(1-x)</math>. Also,<math>X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)</math>. Solving, we get <math>X_{1}M=8-8x</math> and <math>MX_{2}=7-7x</math> (as a side note, note that <math>AX_{1}+MX_{2}=X_{1}M+X_{2}C</math>, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
 +
 
 +
Now, we get <math>BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x</math>. To finish, we will compute area ratios. <math>\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}</math>. Also, since their inradii are equal, we get <math>\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}</math>. Equating and cross multiplying yields the quadratic <math>3x^{2}-8x+4=0</math>, so <math>x=2/3,2</math>. However, observe that <math>AX_{1}+CX_{2}=15x<15</math>, so we take <math>x=2/3</math>. Our ratio is therefore <math>\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}</math>, giving the answer <math>\boxed{045}</math>.
 +
 
 +
Note: Once we have <math>MX_1=8-8x</math> and <math>MX_2=7-7x</math>, it's bit easier to use use the right triangle of <math>O_1MO_2</math> than chasing the area ratio. The inradius of <math>\triangle{ABC}</math> can be calculated to be <math>r=\sqrt{14}</math>, and the inradius of <math>ABM</math> and <math>ACM</math> are <math>r_1=r_2= xr</math>, so,
 +
<cmath> O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2</cmath>
 +
or,
 +
<cmath> (15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2 </cmath>
 +
<cmath>112(1-x)^2 = 28x^2</cmath>
 +
<cmath>4(1-x)^2 = x^2</cmath>
 +
We get <math>x=\frac{2}{3}</math> or <math>x=2</math>.
 +
 
 +
=== Solution 4 ===
 +
Suppose the incircle of <math>ABM</math> touches <math>AM</math> at <math>X</math>, and the incircle of <math>CBM</math> touches <math>CM</math> at <math>Y</math>. Then
 +
 
 +
<cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath>
 +
 
 +
We have <math>\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}</math>
 +
 
 +
<math>\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}</math>, <math>\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}</math>,
 +
 
 +
Therefore <math>AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.</math>
 +
 
 +
And since <math>AX=\frac{1}{2}(12+AM-BM)</math>, <math>CY = \frac{1}{2}(13+CM-BM)</math>,
 +
 
 +
<cmath> \frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}</cmath>
 +
 
 +
<cmath> 96+8AM-8BM = 91 +7CM-7BM</cmath>
 +
 
 +
<cmath>BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)</cmath>
 +
 
 +
Now,
 +
 
 +
<math>\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)}</math>
 +
 
 +
<cmath>\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}</cmath>
 +
<cmath>6AM^2 - 35AM = 45AM-264</cmath>
 +
<cmath>3AM^2 -40AM+132=0</cmath>
 +
<cmath>(3AM-22)(AM-6)=0</cmath>
 +
 
 +
So <math>AM=22/3</math> or <math>6</math>. But from (1) we know that <math>5+15(AM-7)>0</math>, or <math>AM>7-1/3>6</math>, so <math>AM=22/3</math>, <math>CM=15-22/3=23/3</math>, <math>AM/CM=22/23</math>.
 +
 
 +
=== Solution 5 ===
 +
 
 +
Let the common inradius equal r, <math>BM = x</math>, <math>AM = y</math>, <math>MC = z</math>
 +
 
 +
From the prespective of <math>\triangle{ABM}</math> and <math>\triangle{BMC}</math> we get:
 +
 
 +
<math>S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})</math>  <math>S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})</math>
 +
 
 +
Add two triangles up, we get <math>\triangle{ABC}</math> :
 +
 
 +
<math>S_{ABC} =  S_{ABM} +  S_{BMC} = r \cdot \frac{25+2x+y+z}{2}</math>
 +
 
 +
Since <math>y + z = 15</math>, we get:
 +
 
 +
<math>r = \frac{S_{ABC}}{20 + x}</math>
 +
 
 +
By drawing an altitude from <math>I_1</math> down to a point <math>H_1</math> and from <math>I_2</math> to <math>H_2</math>, we can get:
 +
 
 +
<math>r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2} </math>  and
 +
 
 +
<math>r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C =  r \cdot \frac{13+z-x}{2}</math>
 +
 
 +
Adding these up, we get:
 +
 
 +
<math>r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x</math>
 +
 
 +
<math>r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math>
 +
 
 +
Now, we have 2 values equal to r, we can set them equal to each other:
 +
 
 +
<math>\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math>
 +
 
 +
If we let R denote the incircle of ABC, note:
 +
 
 +
AC = <math>(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15</math> and
 +
 
 +
<math>S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R</math>.
 +
 
 +
By cross multiplying the equation above, we get:
 +
 
 +
<math>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</math>
 +
 
 +
We can find out x:
 +
 
 +
<math>x = 10</math>.
 +
 
 +
Now, we can find ratio of y and z:
 +
 
 +
<math>\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}</math>
 +
 
 +
The answer is <math>\boxed{045}</math>.
 +
 
 +
-Alexlikemath
 +
 
 +
=== Solution 6 (Similar to Solution 1 with easier computation) ===
 +
 
 +
Let <math>CM=x, AM=rx, BM=d</math>. <math>x+rx=15\Rightarrow x=\frac{15}{1+r}</math>.
 +
 
 +
Similar to Solution 1, we have
 +
<cmath>
 +
r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}
 +
</cmath>
 +
as well as
 +
<cmath>
 +
12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})
 +
</cmath>
 +
<cmath>
 +
\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}
 +
</cmath>
 +
<cmath>
 +
\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2}
 +
</cmath>
 +
<cmath>
 +
(169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r)
 +
</cmath>
 +
<cmath>
 +
(r^2+1)(400r)=2r(338r^2-224r+288)
 +
</cmath>
 +
<cmath>
 +
100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0
 +
</cmath>
 +
 
 +
Since <math>d=\frac{13r-12}{1-r}>0</math>, we have <math>r=\frac{22}{23} \longrightarrow \boxed{045}</math>.
 +
 
 +
~ asops
 +
 
 +
=== Solution 7 (No Stewart's or trig, fast + clever) ===
 +
 
 +
Let <math>BM = d, AM = x, CM = 15 - x</math>. Observe that we have the equation by the incircle formula:
 +
<cmath>\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.</cmath>
 +
Now let <math>X</math> be the point of tangency between the incircle of <math>\triangle ABC</math> and <math>AC</math>. Additionally, let <math>P</math> and <math>Q</math> be the points of tangency between the incircles of <math>\triangle ABM</math> and <math>\triangle CBM</math> with <math>AC</math> respectively. Some easy calculation yields <math>AX = 7, CX = 8</math>. By homothety we have
 +
<cmath>\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.</cmath>
 +
Substituting into the first equation derived earlier it is left to solve
 +
<cmath>\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.</cmath>
 +
Now <math>x = 6</math> yields <math>d = -10</math> which is invalid, hence <math>x = \frac{22}{3}</math> so <math>\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.</math> The requested sum is <math>22 + 23 = \boxed{45}</math>. ~blueprimes
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0
 +
 
 +
== Sidenote ==
 +
In the problem, <math>BM=10</math> and the equal inradius of the two triangles happens to be <math> \frac {2\sqrt{14}}{3}</math>.
 +
 
 +
== See Also ==
 +
 
 +
 
 
{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}
 
{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}
  
−
[[Category:Olympiad Geometry Problems]]
+
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 20:19, 29 September 2024

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 $6x$ 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: 

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

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

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

Adding these up, we get:

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

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

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

Solution 7 (No Stewart's or trig, fast + clever)

Let $BM = d, AM = x, CM = 15 - x$. Observe that we have the equation by the incircle formula: \[\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.\] Now let $X$ be the point of tangency between the incircle of $\triangle ABC$ and $AC$. Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\triangle ABM$ and $\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$. By homothety we have \[\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.\] Substituting into the first equation derived earlier it is left to solve \[\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.\] Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \frac{22}{3}$ so $\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.$ The requested sum is $22 + 23 = \boxed{45}$. ~blueprimes

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 $\frac {2\sqrt{14}}{3}$.

See Also

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