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Mock Geometry AIME 2011 Problems/Problem 13

Revision as of 16:29, 21 July 2018 by Mudhaniu (talk | contribs) (Created page with "==Problem 13== In acute triangle <math>ABC,</math> <math>\ell</math> is the bisector of <math>\angle BAC</math>. <math>M</math> is the midpoint of <math>BC</math>. a line thr...")
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Problem 13

In acute triangle $ABC,$ $\ell$ is the bisector of $\angle BAC$. $M$ is the midpoint of $BC$. a line through $M$ parallel to $\ell$ meets $AC,AB$ at $E,F,$ respectively. Given that $AE=1,EF=\sqrt{3}, AB=21,$ the sum of all possible values of $BC$ can be expressed as $\sqrt{a}+\sqrt{b},$ where $a,b$ are positive integers. What is $a+b$?

Solution

Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let $L$ be intersection of angle bisector $\ell$ with $BC$ Let $\angle BAL$ be $\theta$, and $\angle LAC is \theta$ as well, since angle bisector. Since line through $M$ is parallel to $\ell$, $\angle MEC$ is also $\theta$. Let $\angle BLA$ then be $\alpha$, and by parallel lines, $\angle BME$ is also $\alpha$. Doing further angle chasing, we find that $AFE$ is isoceles with base $EF$. Using $30-60-90$ triangle ratio, we find $\theta = 30^\circ$

There are two possible configurations of the triangle, one such that $L$ is to the left of $M$, and vice versa. In the first $A$ falls between $B$ and $F$, with $F$ outside the triangle, and in the second $F$ between $B$ and $A$, with $E$ outside the triangle. Using Law of Sines then:

$\frac{\sin{\alpha}}{BF} = \frac{\sin{\theta}}{MC}$

Plugging in values, we find for acute and obtuse triangles denoted as $[1]$ and $[2]$, respectively,

$[1] \frac{\sin{\alpha}}{22} = \frac{\frac{1}{2}}{\frac{BC}{2}}$, and $[2] \frac{\sin{\alpha}}{20} = \frac{\frac{1}{2}}{\frac{BC}{2}}$

Using Law of Sines again and substituting the expression $\sin{\alpha} = \frac{22}{BC}$ for the $[1]$ and $\sin{\alpha} = \frac{20}{BC}$ for $[2]$,

$[1] \frac{22}{21 \cdot BC} = \frac{\frac{1}{2}}{BL}$, and $[2] \frac{20}{21 \cdot BC} = \frac{\frac{1}{2}}{BL}$

Solving for the ratio of $BL : LC$ on both triangles, and then applying Angle Bisector theorem yields a $21,23$ with included angle $60^\circ$ for $[1]$ and $21,19$ with included angle $60^\circ$ for $[2]$. Solving using Law of Cosine yields answer of $\sqrt{487}$ and $\sqrt{403}$, or $\boxed{890}$.

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