Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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MRENTHUSIASM (talk | contribs) (The two recent editions are not quite right. Please read carefully before editing, thanks. :)) |
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draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C)); | draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C)); | ||
</asy> | </asy> | ||
− | By the Inscribed Angle Theorem, we conclude that <math>\overline{AO_1}</math> is the diameter of <math>\odot O_2.</math> By the Pythagorean Theorem on right <math>\triangle ABO_1,</math> we have <cmath>AO_1 = \sqrt{AB^2 + | + | By the Inscribed Angle Theorem, we conclude that <math>\overline{AO_1}</math> is the diameter of <math>\odot O_2.</math> By the Pythagorean Theorem on right <math>\triangle ABO_1,</math> we have <cmath>AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.</cmath> |
Therefore, the area of <math>\odot O_2</math> is <math>\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{\textbf{(C) }26\pi}.</math> | Therefore, the area of <math>\odot O_2</math> is <math>\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{\textbf{(C) }26\pi}.</math> | ||
− | ~MRENTHUSIASM ~kante314 | + | ~MRENTHUSIASM ~kante314 |
==Solution 2 (Similar Triangles)== | ==Solution 2 (Similar Triangles)== |
Revision as of 23:57, 17 September 2022
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 (Cyclic Quadrilateral)
Let be the circle with radius
that is tangent to
at
and to
at
Note that
Since the opposite angles of quadrilateral
are supplementary, quadrilateral
is cyclic.
Let be the circumcircle of quadrilateral
It follows that
is also the circumcircle of
as shown below:
By the Inscribed Angle Theorem, we conclude that
is the diameter of
By the Pythagorean Theorem on right
we have
Therefore, the area of
is
~MRENTHUSIASM ~kante314
Solution 2 (Similar Triangles)
Because circle
is tangent to
at
. Because
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
is
This is the radius of the circumcircle of
, so the area of this circle is
.
Solution in Progress
~KingRavi
Solution 3 (Trigonometry)
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
, we have
and
.
Because this circle is tangent to line at
, we have
and
.
Because ,
,
, we get
. Hence,
.
Let and
meet at point
.
Because
,
,
, we get
. Hence,
and
.
Denote . Hence,
.
Denote by the circumradius of
.
In
, following from the law of sines,
.
Therefore, the area of the circumcircle of is
~Steven Chen (www.professorchenedu.com)
Solution 4 (Geometric Mean)
Let be the center of the circle with radius
. Let the intersection of
and
be
. And let
and
. Since
is a kite
bisects
so
.
by the Geometric Mean Theorem.
substituting
gives
factoring and dividing both equations gives
. Substituting
back into the 2nd equation and factoring gives
so
. The radius of the circumcircle of a triangle is
where
are the sides of the triangle and
is the area. So we just need to find
and plugging our variables and simplifying gives
. Plugging all the variable into the formula gives us
.
~LuisFonseca123
Video Solution by The Power of Logic
~math2718281828459
Video Solution by TheBeautyofMath
https://youtu.be/zq3UPu4nwsE?t=1674
~IceMatrix
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.