Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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− | Because circle <math>I</math> is tangent to <math>\overline{AB}</math> at <math>B, \angle{ABI} | + | Because circle <math>I</math> is tangent to <math>\overline{AB}</math> at <math>B, \angle{ABI} \cong 90^{\circ}</math>. Because O is the circumcenter of <math>\bigtriangleup ABC, \overline{OD}</math> is the perpendicular bisector of <math>\overline{AB}</math>, and <math>\angle{BAI} \cong \angle{DAO}</math>, so therefore <math>\bigtriangleup ADO \sim \bigtriangleup ABI</math> by AA similarity. |
Solution in Progress | Solution in Progress |
Revision as of 01:52, 24 November 2021
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
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
Solution 2 (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.
Solution in Progress
~KingRavi
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 |
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