Difference between revisions of "2022 AIME I Problems/Problem 8"
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We can extend <math>AB</math> and <math>AC</math> to <math>B'</math> and <math>C'</math> such that circle <math>\omega_A</math> is the incircle of <math>\triangle AB'C'</math>. | We can extend <math>AB</math> and <math>AC</math> to <math>B'</math> and <math>C'</math> such that circle <math>\omega_A</math> is the incircle of <math>\triangle AB'C'</math>. | ||
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<asy> | <asy> | ||
unitsize(0.3cm); | unitsize(0.3cm); |
Revision as of 16:18, 18 February 2022
Contents
Problem
Equilateral triangle is inscribed in circle
with radius
Circle
is tangent to sides
and
and is internally tangent to
. Circles
and
are defined analogously. Circles
,
, and
meet in six points
two points for each pair of circles. The three intersection points closest to the vertices of
are the vertices of a large equilateral triangle in the interior of
, and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of
. The side length of the smaller equilateral triangle can be written as
, where
and
are positive integers. Find
.
Diagram
Solution 1
We can extend and
to
and
such that circle
is the incircle of
.
Solution 2
Let bottom left point as the origin, the radius of each circle is , note that three centers for circles are
It is not hard to find that one intersection point lies on since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation
, getting that
, the length is
, leads to the answer
~bluesoul
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.