Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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+ | == Solution 3 == | ||
+ | Denote by <math>O</math> the center of the circle that is tangent to line <math>AB</math> at <math>B</math> and to line <math>AC</math> at <math>C</math>. | ||
+ | |||
+ | Because this circle is tangent to line <math>AB</math> at <math>B</math>, <math>OB \perp AB</math> and <math>OB = 5 \sqrt{2}</math>. | ||
+ | Because this circle is tangent to line <math>AC</math> at <math>C</math>, <math>OC \perp AC</math> and <math>OC = 5 \sqrt{2}</math>. | ||
+ | |||
+ | Because <math>AB = AC</math>, <math>OB = OC</math>, <math>AO = AO</math>, <math>\triangle ABO \cong \triangle ACO</math>. | ||
+ | Hence, <math>\angle BAO = \angle CAO</math>. | ||
+ | |||
+ | Let <math>AO</math> and <math>BC</math> meet at point <math>D</math>. | ||
+ | Because <math>AB = AC</math>, <math>\angle BAO = \angle CAO</math>, <math>AD = AD</math>, <math>\triangle ABD \cong \triangle ACD</math>. | ||
+ | |||
+ | Hence, <math>BD = CD</math> and <math>\angle ADB = \angle ADC = 90^\circ</math>. | ||
+ | |||
+ | Denote <math>\theta = \angle BAO</math>. Hence, <math>\angle BAC = 2 \theta</math>. | ||
+ | |||
+ | Denote by <math>R</math> the circumradius of <math>\triangle ABC</math>. | ||
+ | In <math>\triangle ABC</math>, following from the law of sines, <math>2 R = \frac{BC}{\sin \angle BAC}</math>. | ||
+ | |||
+ | Therefore, the area of the circumcircle of <math>\triangle ABC</math> is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \pi R^2 & = \pi \left( \frac{BC}{2 \sin \angle BAC} \right)^2 \\ | ||
+ | & = \pi \left( \frac{2 BD}{2 \sin \angle BAC} \right)^2 \\ | ||
+ | & = \pi \left( \frac{BD}{\sin 2 \theta} \right)^2 \\ | ||
+ | & = \pi \left( \frac{AB \sin \theta }{\sin 2 \theta} \right)^2 \\ | ||
+ | & = \pi \left( \frac{AB \sin \theta }{2 \sin \theta \cos \theta} \right)^2 \\ | ||
+ | & = \pi \left( \frac{AB }{2 \cos \theta} \right)^2 \\ | ||
+ | & = \pi \left( \frac{AO}{2} \right)^2 \\ | ||
+ | & = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\ | ||
+ | & = 26 \pi . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(C) }}26 \pi</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:12, 25 November 2021
Contents
Problem
Isosceles triangle has
, and a circle with radius
is tangent to line
at
and to line
at
. What is the area of the circle that passes through vertices
,
, and
Solution 1 (Similar Triangles)
Because circle is tangent to
at
. Because O is the circumcenter of
is the perpendicular bisector of
, and
, so therefore
by AA similarity. Then we have
. We also know that
because of the perpendicular bisector, so the hypotenuse of
.
This is the radius of the circumcircle of
, so the area of this circle is
Solution in Progress
~KingRavi
Solution 2 (Geometric Assumption)
Let the center of the first circle be By Pythagorean Theorem,
Now, notice that since
is
degrees, so arc
is
degrees and
is the diameter. Thus, the radius is
so the area is
- kante314
Video Solution by The Power of Logic
~math2718281828459
Solution 3
Denote by the center of the circle that is tangent to line
at
and to line
at
.
Because this circle is tangent to line at
,
and
.
Because this circle is tangent to line
at
,
and
.
Because ,
,
,
.
Hence,
.
Let and
meet at point
.
Because
,
,
,
.
Hence, and
.
Denote . Hence,
.
Denote by the circumradius of
.
In
, following from the law of sines,
.
Therefore, the area of the circumcircle of is
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.