2021 Fall AMC 12B Problems/Problem 9: Difference between revisions
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==Solution 1 (Cosine Rule) == | ==Solution 1 (Cosine Rule) == | ||
Construct the circle that passes through <math>A</math>, <math>O</math>, and <math>C</math>, centered at <math>X</math>. | |||
Then connect <math>OX</math>, and notice that <math>OX</math> is the perpendicular bisector of <math>AC</math>. Let the intersection of <math>OX</math> with <math>AC</math> be <math>D</math>. | |||
Also notice that <math>AO</math> and <math>CO</math> are the angle bisectors of angle <math>BAC</math> and <math>BCA</math> respectively. We then deduce <math>AOC=120^\circ</math>. | |||
Consider another point <math>M</math> on Circle <math>X</math> opposite to point <math>O</math>. | |||
As <math>AOCM</math> an inscribed quadrilateral of Circle <math>X</math>, <math>AMC=180^\circ-120^\circ=60^\circ</math>. | |||
Afterward, deduce that <math>AXC=2·AMC=120^\circ</math>. | |||
By the Cosine Rule, we have the equation: (where <math>r</math> is the radius of circle <math>X</math>) | |||
<math>2r^2(1-\cos(120^\circ))=6^2</math> | |||
<math>r^2=12</math> | |||
The area is therefore <math>12\pi \Rightarrow \boxed{\textbf{(B)}}</math>. | |||
~Wilhelm Z | ~Wilhelm Z | ||
Revision as of 21:51, 23 November 2021
Solution 1 (Cosine Rule)
Construct the circle that passes through 
, 
, and 
, centered at 
.
Then connect 
, and notice that is the perpendicular bisector of 
. Let the intersection of with be 
.
Also notice that 
and 
are the angle bisectors of angle 
and 
respectively. We then deduce 
.
Consider another point 
on Circle opposite to point .
As 
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
