Difference between revisions of "2021 AIME II Problems/Problem 14"
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If <math>\angle AOM = 2 \angle ACB + \angle ABC,</math> then <math>\angle ABC = \angle CAB, AC = BC.</math> | If <math>\angle AOM = 2 \angle ACB + \angle ABC,</math> then <math>\angle ABC = \angle CAB, AC = BC.</math> | ||
+ | <pre style="color: blue"> | ||
+ | Solution: | ||
+ | </pre> | ||
+ | Since <math>\angle OAX = \angle OGX = \frac {\pi}{2},</math> quadrilateral <math>XAGO</math> is cyclic by the Converse of the Inscribed Angle Theorem. | ||
+ | |||
+ | It follows that <math>\angle GXO = \angle GAO,</math> as they share the same intersept <math>\overset{\Large\frown} {GO}.</math> | ||
+ | Since <math>\angle OGY = \angle OMY = \frac {\pi}{2},</math> quadrilateral <math>OGYM</math> is cyclic by the supplementary opposite angles. | ||
+ | |||
+ | It follows that <math>\angle GYO = \angle OMG,</math> as they share the same intercept <math>\overset{\Large\frown} {GO}.</math> | ||
+ | |||
+ | In triangles <math>\triangle XOY</math> and <math>\triangle AOM,</math> two pairs of angles are equal, which means that the third angles <math>\angle XOY = \angle AOM</math> are also equal. | ||
+ | <math>\angle ABC : \angle BCA : \angle AOM = 13 : 2 : 17,</math> so <math>\angle AOM = \angle ABC + 2 \angle BCA.</math> | ||
+ | According to the lemma, <math>\triangle ABC</math> is isosceles, | ||
+ | <cmath>\angle ABC : \angle BCA : \angle BAC = 13 : 2 : 13.</cmath> | ||
+ | <cmath>\angle BAC = \frac{13} {13 + 2 + 13} \cdot 180 = \frac {585}{7}.</cmath> | ||
==Video Solution 1== | ==Video Solution 1== |
Revision as of 09:09, 30 May 2022
Contents
[hide]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
Solution 5 (Why isosceles)
Lemma:
Let be an acute triangle with circumcenter
Let
be the midpoint of
so
If then
We define as the sum of
this angle can be greater than
Proof:
as an inscribed and half central, based on the
as inscribed and central, based on the
If then
Solution:
Since quadrilateral
is cyclic by the Converse of the Inscribed Angle Theorem.
It follows that as they share the same intersept
Since
quadrilateral
is cyclic by the supplementary opposite angles.
It follows that as they share the same intercept
In triangles and
two pairs of angles are equal, which means that the third angles
are also equal.
so
According to the lemma,
is isosceles,
Video Solution 1
https://www.youtube.com/watch?v=zFH1Z7Ydq1s
~Mathematical Dexterity
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 |
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