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

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(Solution)
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\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.
 
\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.
 
</cmath>
 
</cmath>
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=== Solution 3 ===
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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>.
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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>.
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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).
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 00:57, 18 May 2010

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. Let $p$ and $q$ be positive relatively prime integers such that $\frac {AM}{CM} = \frac {p}{q}$. 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,-0.908),NE); dot((4.66667,2.49444)); label("$I_1$",(3.97942,2.92179),NE); dot((9.66667,2.49444)); label("$I_2$",(10.04741,2.97153),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 $12x$ 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 $MQ = (MB + MC - BC) = - x + y + 1$. 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

\[2y^2 - 30 = 2xy + 5x - 7y \\ 2y^2 - 70 = - 2xy - 5x + 7y,\]

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 $\boxed{045}$.

See also

  • <url>viewtopic.php?t=338911 Discussion</url>
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