Difference between revisions of "2021 Fall AMC 12B Problems/Problem 9"
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==Solution 3 == | ==Solution 3 == | ||
− | As | + | As in the previous solution, construct the circle that passes through <math>A</math>, <math>O</math>, and <math>C</math>, centered at <math>X</math>. Let <math>Y</math> be the intersection of <math>\overline{OX}</math> and <math>\overline{AB}</math>. |
Note that since <math>\overline{OA}</math> is the angle bisector of <math>\angle BAC</math> that <math>\angle OAC=30^\circ</math>. Also by symmetry, <math>\overline{OX}</math> <math>\perp</math> <math>\overline{AB}</math> and <math>AY = 3</math>. Thus <math>\tan(30^\circ) = \frac{OY}{3}</math> so <math>OY = \sqrt{3}</math>. | Note that since <math>\overline{OA}</math> is the angle bisector of <math>\angle BAC</math> that <math>\angle OAC=30^\circ</math>. Also by symmetry, <math>\overline{OX}</math> <math>\perp</math> <math>\overline{AB}</math> and <math>AY = 3</math>. Thus <math>\tan(30^\circ) = \frac{OY}{3}</math> so <math>OY = \sqrt{3}</math>. | ||
− | Let <math>r</math> be the radius of circle <math>X</math>, and note that <math>AX = OX = r</math>. So <math>\triangle AYX</math> is a right triangle with legs | + | Let <math>r</math> be the radius of circle <math>X</math>, and note that <math>AX = OX = r</math>. So <math>\triangle AYX</math> is a right triangle with legs of length <math>3</math> and <math>r - \sqrt{3}</math> and hypotenuse <math>r</math>. By Pythagorus, <math>3^2 + (r - \sqrt{3})^2 = r^2</math>. So <math>r = 2\sqrt{3}</math>. |
+ | |||
+ | Thus the area is <math> \pi r^2 = \boxed{\textbf{(B)}\ 12\pi}</math>. | ||
Revision as of 19:41, 9 January 2022
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 |
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.