Difference between revisions of "2019 AMC 12A Problems/Problem 22"
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Let <math>S</math> be the point of tangency between <math>\overline{BC}</math> and <math>\gamma</math>, and <math>M</math> be the midpoint of <math>\overline{BC}</math>. Note that <math>AM \perp BS</math> and <math>OS \perp BS</math>. This implies that <math>\angle OAM \cong \angle AOS</math>, and <math>\angle AMP \cong \angle OSP</math>. Thus, <math>\triangle PMA \sim \triangle PSO</math>. | Let <math>S</math> be the point of tangency between <math>\overline{BC}</math> and <math>\gamma</math>, and <math>M</math> be the midpoint of <math>\overline{BC}</math>. Note that <math>AM \perp BS</math> and <math>OS \perp BS</math>. This implies that <math>\angle OAM \cong \angle AOS</math>, and <math>\angle AMP \cong \angle OSP</math>. Thus, <math>\triangle PMA \sim \triangle PSO</math>. | ||
− | If we let <math>s</math> be the side length of <math>\triangle ABC</math>, then it follows that <math>AM = \frac{\sqrt{3}}{2}s</math> and <math>PM = \frac{s}{4}</math>. This implies that <math>AP = \frac{\sqrt{13}}{4}s</math>, so <math>\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}</math>. Furthermore, <math>\frac{AM + SO}{AO} = \frac{AM}{AP}</math> (because | + | If we let <math>s</math> be the side length of <math>\triangle ABC</math>, then it follows that <math>AM = \frac{\sqrt{3}}{2}s</math> and <math>PM = \frac{s}{4}</math>. This implies that <math>AP = \frac{\sqrt{13}}{4}s</math>, so <math>\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}</math>. Furthermore, <math>\frac{AM + SO}{AO} = \frac{AM}{AP}</math> (because <math>\triangle PMA \sim \triangle PSO</math>) so this gives us the equation |
<cmath>\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}</cmath> | <cmath>\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}</cmath> | ||
to solve for the side length <math>s</math>, or <math>AB</math>. Thus, | to solve for the side length <math>s</math>, or <math>AB</math>. Thus, | ||
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<cmath>\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}</cmath> | <cmath>\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}</cmath> | ||
<cmath>s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB</cmath> | <cmath>s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB</cmath> | ||
− | The problem asks for <math>m + n + p + q</ | + | The problem asks for <math>m + n + p + q = </math>80 + 13 + 34 + 3 = \boxed{\textbf{(E) } 130}$. |
+ | |||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=2eASfdhEyUE | https://www.youtube.com/watch?v=2eASfdhEyUE |
Revision as of 22:04, 17 February 2019
Contents
Problem
Circles and
, both centered at
, have radii
and
, respectively. Equilateral triangle
, whose interior lies in the interior of
but in the exterior of
, has vertex
on
, and the line containing side
is tangent to
. Segments
and
intersect at
, and
. Then
can be written in the form
for positive integers
,
,
,
with
. What is
?
Solution
Let be the point of tangency between
and
, and
be the midpoint of
. Note that
and
. This implies that
, and
. Thus,
.
If we let be the side length of
, then it follows that
and
. This implies that
, so
. Furthermore,
(because
) so this gives us the equation
to solve for the side length
, or
. Thus,
The problem asks for
80 + 13 + 34 + 3 = \boxed{\textbf{(E) } 130}$.
Video Solution
https://www.youtube.com/watch?v=2eASfdhEyUE
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.