Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 4"
m |
|||
Line 24: | Line 24: | ||
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>. | 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>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Let <math>\angle A = \alpha</math> and <math>BP = PQ = QB = x</math>. By the Law of Cosines, <math>AC = 3\sqrt{19}</math>. It is easy to see that <math>\angle APB = 120^\circ</math>. Since <math>\angle ABC = 120^\circ</math>, by AA similarity<math>\triangle ABC \sim \triangle APB</math>. From this, we have: <cmath>\frac{AB}{PB} = \frac{AC}{BC}</cmath> <cmath>\frac{6}{x}=\frac{3\sqrt{19}}{9}</cmath> Solving, we find that <math>x = \frac{18}{\sqrt{19}}</math>, so the perimeter is <math>3x = \frac{54}{\sqrt{19}}</math>, and our answer is <math>m+n=\boxed{73}</math> |
Revision as of 18:33, 15 June 2017
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
.
Solution 2
Let and
. By the Law of Cosines,
. It is easy to see that
. Since
, by AA similarity
. From this, we have:
Solving, we find that
, so the perimeter is
, and our answer is