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 Pythagoras, . 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 . As , we can form an altitude from point to side at point , forming two 30-60-90 triangles. As , we can solve for . Now, the area of the circle is just . Select .
~hastapasta, bob4108
Solution 6 (Ptolemy)
Call the diameter of the circle . If we extend points and to meet at a point on the circle and call it , then . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, . We know this since and is the hypotenuse of a right triangle, with the longer leg being so . Applying Ptolemy's Theorem on cyclic quadrilateral , we get . Squaring and solving we get so . Therefore, the area of the circle is
Video Solution by TheBeautyofMath
https://www.youtube.com/watch?v=4qgYrCYG-qw&t=1039
~IceMatrix
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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