2021 Fall AMC 12B Problems/Problem 9
Problem
Triangle is equilateral with side length
. Suppose that
is the center of the inscribed
circle of this triangle. What is the area of the circle passing through
,
, and
?
Solution 1 (Cosine Rule)
Construct the circle that passes through ,
, and
, centered at
.
Also notice that and
are the angle bisectors of angle
and
respectively. We then deduce
.
Consider another point on Circle
opposite to point
.
As is an inscribed quadrilateral of Circle
,
.
Afterward, deduce that .
By the Cosine Rule, we have the equation: (where is the radius of circle
)
The area is therefore .
~Wilhelm Z
Solution 2
We have .
Denote by the circumradius of
.
In
, the law of sines implies
Hence, the area of the circumcircle of is
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
As in the previous solution, construct the circle that passes through ,
, and
, centered at
. Let
be the intersection of
and
.
Note that since is the angle bisector of
that
. Also by symmetry,
and
. Thus
so
.
Let be the radius of circle
, and note that
. So
is a right triangle with legs of length
and
and hypotenuse
. By Pythagorus,
. So
.
Thus the area is .
-SharpeMind
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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All AMC 12 Problems and Solutions |
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