Difference between revisions of "2021 Fall AMC 12B Problems/Problem 9"
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<math>r^2=12</math> | <math>r^2=12</math> | ||
− | The area is therefore <math> \boxed{\textbf{(B)\ 12\pi}}</math>. | + | The area is therefore <math>\boxed{\textbf{(B)\ 12\pi}}</math>. |
~Wilhelm Z | ~Wilhelm Z |
Revision as of 04:07, 25 November 2021
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 $\boxed{\textbf{(B)\ 12\pi}}$ (Error compiling LaTeX. Unknown error_msg).
~Wilhelm Z
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 |
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