Difference between revisions of "2019 AMC 12A Problems/Problem 22"
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<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</math> which is <math>80 + 13 + 34 + 3</math> which is <math>130 = \boxed {E}</math>. | The problem asks for <math>m + n + p + q</math> which is <math>80 + 13 + 34 + 3</math> which is <math>130 = \boxed {E}</math>. | ||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=2eASfdhEyUE | ||
==See Also== | ==See Also== |
Revision as of 00:48, 12 February 2019
Contents
[hide]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 of
) so this gives us the equation
to solve for the side length
, or
. Thus,
The problem asks for
which is
which is
.
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.