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

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== Solution ==
 
== Solution ==
 
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=== Solution 1 ===
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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 d to be positive, we must have <math>7.2 < x < 7.5</math>.
  
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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>
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''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>12x</math> is an integer. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.''
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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>.
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=== 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
 
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
  
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== See also ==
 
== See also ==
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*<url>viewtopic.php?t=338911 Discussion</url>
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{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}
 
{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}
  
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[[Category:Olympiad Geometry Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 13:28, 17 March 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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

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

See also

  • <url>viewtopic.php?t=338911 Discussion</url>
2010 AIME I (Problems • Answer Key • Resources)
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