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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 13"

(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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==Solution==
 
==Solution==
  
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Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let <math>L</math> be intersection of angle bisector <math>\ell</math> with <math>BC</math> Let <math>\angle BAL</math> be <math>\theta</math>, and <math>\angle LAC is \theta</math> as well, since angle bisector. Since line through <math>M</math> is parallel to <math>\ell</math>, <math>\angle MEC</math> is also <math>\theta</math>. Let <math>\angle BLA</math> then be <math>\alpha</math>, and by parallel lines, <math>\angle BME</math> is also <math>\alpha</math>. Doing further angle chasing, we find that <math>AFE</math> is isoceles with base <math>EF</math>. Using <math>30-60-90</math> triangle ratio, we find <math>\theta = 30^\circ</math>
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Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let <math>L</math> be intersection of angle bisector <math>\ell</math> with <math>BC</math> Let <math>\angle BAL</math> be <math>\theta</math>, and <math>\angle LAC</math> is <math>\theta</math> as well, since angle bisector. Since line through <math>M</math> is parallel to <math>\ell</math>, <math>\angle MEC</math> is also <math>\theta</math>. Let <math>\angle BLA</math> then be <math>\alpha</math>, and by parallel lines, <math>\angle BME</math> is also <math>\alpha</math>. Doing further angle chasing, we find that <math>AFE</math> is isoceles with base <math>EF</math>. Using <math>30-60-90</math> triangle ratio, we find <math>\theta = 30^\circ</math>
  
 
There are two possible configurations of the triangle, one such that <math>L</math> is to the left of <math>M</math>, and vice versa. In the first <math>A</math> falls between <math>B</math> and <math>F</math>, with <math>F</math> outside the triangle, and in the second <math>F</math> between <math>B</math> and <math>A</math>, with <math>E</math> outside the triangle. Using Law of Sines then:
 
There are two possible configurations of the triangle, one such that <math>L</math> is to the left of <math>M</math>, and vice versa. In the first <math>A</math> falls between <math>B</math> and <math>F</math>, with <math>F</math> outside the triangle, and in the second <math>F</math> between <math>B</math> and <math>A</math>, with <math>E</math> outside the triangle. Using Law of Sines then:

Revision as of 17:30, 21 July 2018

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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