Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 4"
(Created page with "==Problem== In triangle <math>ABC,</math> <math>AB=6, BC=9, \angle ABC=120^{\circ}</math> Let <math>P</math> and <math>Q</math> be points on <math>AC</math> such that <math>BPQ<...") |
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==Solution== | ==Solution== | ||
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+ | Let <math>H</math> be the midpoint of <math>PQ</math>. It follows that <math>BH</math> is perpendicular to <math>PQ</math> and to <math>AC</math>. The area of <math>\Delta ABC</math> can then be calculated two different ways: <math>\frac{1}{2}*AB*BC*\sin{B}</math>, and <math>\frac{BH*AC}{2}</math>. | ||
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− | Let <math>s</math> be the side length of <math>BPQ</math>. The height of an equilateral triangle is given by the formula <math>\frac{s\sqrt3}{2}</math>. Then <math>BH=\frac{s\sqrt{3}}{2}=\frac{ | + | By the Law of Cosines, <math>AC^2=9^2+6^2-2*9*6\cos{120}=171</math> and so <math>AC=3\sqrt{19}</math>. Therefore, <math>[ABC]=\frac{1}{2}*6*9\sin{120}=\frac{3\sqrt{19}BH}{2}</math>. Solving for <math>BH</math> yields <math>BH=\frac{9\sqrt{3}}{\sqrt{19}}</math>. |
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+ | Let <math>s</math> be the side length of <math>BPQ</math>. The height of an equilateral triangle is given by the formula <math>\frac{s\sqrt3}{2}</math>. Then <math>BH=\frac{s\sqrt{3}}{2}=\frac{9\sqrt{3}}{\sqrt{19}}</math>. Solving for <math>s</math> yields <math>s=\frac{18}{\sqrt{19}}</math>. Then the perimeter of the triangle is <math>3s=\frac{54}{\sqrt{19}}</math> and <math>m+n=54+19=\boxed{073}</math>. |
Revision as of 23:57, 1 January 2012
Problem
In triangle
Let
and
be points on
such that
is equilateral. The perimeter of
can be expressed in the form
where
are relatively prime positive integers. Find
Solution
Let be the midpoint of
. It follows that
is perpendicular to
and to
. The area of
can then be calculated two different ways:
, and
.
By the Law of Cosines, and so
. Therefore,
. Solving for
yields
.
Let be the side length of
. The height of an equilateral triangle is given by the formula
. Then
. Solving for
yields
. Then the perimeter of the triangle is
and
.