Difference between revisions of "2021 AIME II Problems/Problem 14"
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draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); | draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); | ||
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==Solution 1== | ==Solution 1== |
Revision as of 14:13, 1 October 2021
Contents
Problem
Let be an acute triangle with circumcenter
and centroid
. Let
be the intersection of the line tangent to the circumcircle of
at
and the line perpendicular to
at
. Let
be the intersection of lines
and
. Given that the measures of
and
are in the ratio
the degree measure of
can be written as
where
and
are relatively prime positive integers. Find
.
Diagram
~MRENTHUSIASM
Solution 1
In this solution, all angle measures are in degrees.
Let be the midpoint of
so that
and
are collinear. Let
and
Note that:
- Since
quadrilateral
is cyclic by the Converse of the Inscribed Angle Theorem.
It follows that
as they share the same intercepted arc
- Since
quadrilateral
is cyclic by the supplementary opposite angles.
It follows that
as they share the same intercepted arc
Together, we conclude that by AA, so
Next, we express in terms of
By angle addition, we have
Substituting back gives
from which
For the sum of the interior angles of we get
Finally, we obtain
from which the answer is
~Constance-variance (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Let be the midpoint of
. Because
,
and
are cyclic, so
is the center of the spiral similarity sending
to
, and
. Because
, it's easy to get
from here.
~Lcz
Solution 3 (Easy and Simple)
Firstly, let be the midpoint of
. Then,
. Now, note that since
, quadrilateral
is cyclic. Also, because
,
is also cyclic. Now, we define some variables: let
be the constant such that
and
. Also, let
and
(due to the fact that
and
are cyclic). Then,
Now, because
is tangent to the circumcircle at
,
, and
. Finally, notice that
. Then,
Thus,
and
However, from before,
, so
. To finish the problem, we simply compute
so our final answer is
.
~advanture
Solution 4 (Guessing in the Last 3 Minutes, Unreliable)
Notice that looks isosceles, so we assume it's isosceles. Then, let
and
Taking the sum of the angles in the triangle gives
so
so the answer is
Video Solution 1
https://www.youtube.com/watch?v=zFH1Z7Ydq1s
Video Solution 2
https://www.youtube.com/watch?v=7Bxr2h4btWo
~Osman Nal
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.