2021 Fall AMC 12B Problems/Problem 9
Contents
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
Solution 5 (SIMPLE)
The semiperimeter is units.
The area of the triangle is
units squared. By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius
. Now, the area of the circle is just
. Select
.
~hastapasta
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.