Difference between revisions of "2021 AIME II Problems/Problem 14"
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==Solution 5 (Why isosceles)== | ==Solution 5 (Why isosceles)== | ||
− | + | [[File:2021 AIME II 14.png|230px|right]] | |
− | [[File:2021 AIME II 14.png| | + | <i><b>Lemma</b></i> |
+ | |||
Let <math>\triangle ABC</math> be an acute triangle with circumcenter <math>O.</math> Let <math>M</math> be the midpoint of <math>BC</math> so <math>MO\perp BC.</math> | Let <math>\triangle ABC</math> be an acute triangle with circumcenter <math>O.</math> Let <math>M</math> be the midpoint of <math>BC</math> so <math>MO\perp BC.</math> | ||
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We define <math>\angle AOM</math> as the sum of <math>\angle AOB + \angle BOM,</math> this angle can be greater than <math>\pi.</math> | We define <math>\angle AOM</math> as the sum of <math>\angle AOB + \angle BOM,</math> this angle can be greater than <math>\pi.</math> | ||
− | + | <i><b>Proof</b></i> | |
− | < | ||
− | <math>\angle AOB = 2\angle ACB</math> as | + | <math>\angle BAC = \angle BOM</math> as they share the same intercept <math>\overset{\Large\frown} {BC}</math> (an inscribed angle and half of central angle). |
+ | |||
+ | <math>\angle AOB = 2\angle ACB</math> as they share the same intercept <math>\overset{\Large\frown} {AB}.</math> | ||
<cmath>\angle AOM = \angle AOB + \angle BOM = 2 \angle ACB + \angle CAB.</cmath> | <cmath>\angle AOM = \angle AOB + \angle BOM = 2 \angle ACB + \angle CAB.</cmath> | ||
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> | ||
− | + | ||
+ | <i><b>Solution</b></i> | ||
+ | [[File:2021 AIME II proble 14.png|230px|right]] | ||
Since <math>\angle OAX = \angle OGX = \frac {\pi}{2},</math> quadrilateral <math>XAGO</math> is cyclic by the Converse of the Inscribed Angle Theorem. | 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> | 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. | 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> | 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. | 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> | <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, | According to the lemma, <math>\triangle ABC</math> is isosceles, | ||
<cmath>\angle ABC : \angle BCA : \angle BAC = 13 : 2 : 13.</cmath> | <cmath>\angle ABC : \angle BCA : \angle BAC = 13 : 2 : 13.</cmath> | ||
− | <cmath>\angle BAC = \frac{13} {13 + 2 + 13} \cdot 180^ | + | <cmath>\angle BAC = \frac{13} {13 + 2 + 13} \cdot 180^\circ = \frac {585^\circ}{7}.</cmath> |
Revision as of 02:31, 10 June 2022
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
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 they share the same intercept
(an inscribed angle and half of central angle).
as they share the same intercept
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,
~vvsss, www.deoma-cmd.ru
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.