Difference between revisions of "2021 Fall AMC 12B Problems/Problem 9"
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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 of length <math>3</math> and <math>r - \sqrt{3}</math> and hypotenuse <math>r</math>. By | + | 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 Pythagoras, <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>. | Thus the area is <math> \pi r^2 = \boxed{\textbf{(B)}\ 12\pi}</math>. | ||
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==Solution 5 (SIMPLE) == | ==Solution 5 (SIMPLE) == | ||
− | The semiperimeter is <math>9</math> units. | + | The semiperimeter is <math>\frac{6+6+6}{2}=9</math> units. |
− | The area of the triangle is <math> | + | The area of the triangle is <math>9\sqrt{3}</math> units squared. By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius <math>r=\sqrt{3}</math>. As <math>\angle{AOC}=120^\circ</math>, we can form an altitude from point <math>O</math> to side <math>AC</math> at point <math>M</math>, forming two 30-60-90 triangles. As <math>CM=MA=3</math>, we can solve for <math>OC=2\sqrt{3}</math>. Now, the area of the circle is just <math>\pi*(2*\sqrt{3})^2 = 12\pi</math>. Select <math>\boxed{B}</math>. |
+ | |||
+ | ~hastapasta, bob4108 | ||
+ | |||
+ | ==Solution 6 (Ptolemy) == | ||
+ | |||
+ | Call the diameter of the circle <math>d</math>. If we extend points <math>A</math> and <math>C</math> to meet at a point on the circle and call it <math>E</math>, then <math>\bigtriangleup OAE=\bigtriangleup OCE</math> . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, <math>CE=AE=\sqrt{d^2-12}</math>. We know this since <math>OC=OA=OB</math> and <math>OC</math> is the hypotenuse of a <math>30-60-90</math> right triangle, with the longer leg being <math>\frac{6}{2}=3</math> so <math>OC=2\sqrt{3}</math>. Applying Ptolemy's Theorem on cyclic quadrilateral <math>OCEA</math>, we get <math>2({\sqrt{d^2-12}})\cdot{2\sqrt{3}}=6d</math>. Squaring and solving we get <math>d^2=48 \Longrightarrow (2r)^2=48</math> so <math>r^2=12</math>. Therefore, the area of the circle is <math>\boxed{12\pi}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] | ||
+ | |||
+ | |||
+ | ==Video Solution (Just 3 min!)== | ||
+ | https://youtu.be/kkm1d09bVpM | ||
+ | |||
+ | <i>~Education, the Study of Everything</i> | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://www.youtube.com/watch?v=4qgYrCYG-qw&t=1039 | ||
+ | |||
+ | ~IceMatrix | ||
− | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=10|num-b=8}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=10|num-b=8}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:33, 10 July 2023
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 (Just 3 min!)
~Education, the Study of Everything
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 |
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.