2019 AMC 12A Problems/Problem 22
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
) so this gives us the equation
to solve for the side length
, or
. Thus,
The problem asks for
.
Video Solution
https://www.youtube.com/watch?v=2eASfdhEyUE
Video Solution by Richard Rusczyk - https://www.youtube.com/watch?v=CaYgfNEUBwA&list=PLyhPcpM8aMvLgfgbaTLDaV_jRYfn1A-x_&index=2 - AMBRIGGS
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 |
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