Difference between revisions of "2021 AIME II Problems/Problem 14"
MRENTHUSIASM (talk | contribs) m (→Solution 2) |
m (→Solution 4 (Why Isosceles)) |
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==Diagram== | ==Diagram== | ||
− | + | <asy> | |
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(375); | ||
− | ~MRENTHUSIASM | + | pair A, B, C, O, G, X, Y; |
+ | A = origin; | ||
+ | B = (1,0); | ||
+ | C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); | ||
+ | O = circumcenter(A,B,C); | ||
+ | G = centroid(A,B,C); | ||
+ | Y = intersectionpoint(G--G+(100,0),B--C); | ||
+ | X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); | ||
+ | markscalefactor=3/160; | ||
+ | draw(rightanglemark(O,G,X),red); | ||
+ | dot("$A$",A,1.5*dir(180+585/7),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(-585/7),linewidth(4)); | ||
+ | dot("$C$",C,1.5N,linewidth(4)); | ||
+ | dot("$O$",O,1.5N,linewidth(4)); | ||
+ | dot("$G$",G,1.5S,linewidth(4)); | ||
+ | dot("$Y$",Y,1.5E,linewidth(4)); | ||
+ | dot("$X$",X,1.5W,linewidth(4)); | ||
+ | draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
==Solution 1== | ==Solution 1== | ||
− | |||
− | |||
− | |||
− | |||
− | |||
In this solution, all angle measures are in degrees. | In this solution, all angle measures are in degrees. | ||
− | Let <math>M</math> be the midpoint of <math>\overline{BC}</math> so that <math> | + | Let <math>M</math> be the midpoint of <math>\overline{BC}</math> so that <math>\overline{OM}\perp\overline{BC}</math> and <math>A,G,M</math> are collinear. Let <math>\angle ABC=13k,\angle BCA=2k</math> and <math>\angle XOY=17k.</math> |
− | + | Note that: | |
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
− | <li>Since <math>\angle OGX = \angle OAX = 90,</math> quadrilateral <math>OGAX</math> is cyclic by the Converse of the Inscribed Angle Theorem.<p>It follows that <math>\angle OAG = \angle OXG,</math> as they share the same intercepted arc <math>OG.</math></li><p> | + | <li>Since <math>\angle OGX = \angle OAX = 90,</math> quadrilateral <math>OGAX</math> is cyclic by the Converse of the Inscribed Angle Theorem.<p>It follows that <math>\angle OAG = \angle OXG,</math> as they share the same intercepted arc <math>\widehat{OG}.</math></li><p> |
− | <li>Since <math>\angle OGY = \angle OMY = 90,</math> quadrilateral <math>OGYM</math> is cyclic by the supplementary opposite angles.<p>It follows that <math>\angle OMG = \angle OYG,</math> as they share the same intercepted arc <math>OG.</math></li><p> | + | <li>Since <math>\angle OGY = \angle OMY = 90,</math> quadrilateral <math>OGYM</math> is cyclic by the supplementary opposite angles.<p>It follows that <math>\angle OMG = \angle OYG,</math> as they share the same intercepted arc <math>\widehat{OG}.</math></li><p> |
</ol> | </ol> | ||
Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math> | Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math> | ||
Line 27: | Line 43: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\angle AOM &= \angle AOB + \angle BOM \\ | \angle AOM &= \angle AOB + \angle BOM \\ | ||
− | &= | + | &= 2\angle BCA + \frac12\angle BOC \hspace{10mm} &&\text{by Inscribed Angle Theorem and Perpendicular Bisector Property} \\ |
− | &= 2\angle BCA + | + | &= 2\angle BCA + \angle BAC. &&\text{by Inscribed Angle Theorem} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Substituting back gives <math>17k=2(2k)+\angle BAC,</math> from which <math>\angle BAC=13k.</math> | Substituting back gives <math>17k=2(2k)+\angle BAC,</math> from which <math>\angle BAC=13k.</math> | ||
Line 41: | Line 57: | ||
Finally, we obtain <math>\angle BAC=13k=\frac{585}{7},</math> from which the answer is <math>585+7=\boxed{592}.</math> | Finally, we obtain <math>\angle BAC=13k=\frac{585}{7},</math> from which the answer is <math>585+7=\boxed{592}.</math> | ||
− | ~Constance-variance ( | + | ~Constance-variance ~MRENTHUSIASM |
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>M</math> be the midpoint of <math>BC</math>. Because <math>\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o</math>, <math>AXOG</math> and <math>OMYG</math> are cyclic, so <math>O</math> is the center of the spiral similarity sending <math>AM</math> to <math>XY</math>, and <math>\angle{XOY}=\angle{AOM}</math>. Because <math>\angle{AOM}=2\angle{BCA}+\angle{BAC}</math>, it's easy to get <math>\frac{585}{7} \implies \boxed{592}</math> from here. | ||
+ | |||
+ | ~Lcz | ||
+ | |||
+ | ==Solution 3 (Easy and Simple)== | ||
+ | Firstly, let <math>M</math> be the midpoint of <math>BC</math>. Then, <math>\angle OMB = 90^o</math>. Now, note that since <math>\angle OGX = \angle XAO = 90^o</math>, quadrilateral <math>AGOX</math> is cyclic. Also, because <math>\angle OMY + \angle OGY = 180^o</math>, <math>OMYG</math> is also cyclic. Now, we define some variables: let <math>\alpha</math> be the constant such that <math>\angle ABC = 13\alpha, \angle ACB = 2\alpha, </math> and <math>\angle XOY = 17\alpha</math>. Also, let <math>\beta = \angle OMG = \angle OYG</math> and <math>\theta = \angle OXG = \angle OAG</math> (due to the fact that <math>AGOX</math> and <math>OMYG</math> are cyclic). Then, <cmath>\angle XOY = 180 - \beta - \theta = 17\alpha \implies \beta + \theta = 180 - 17\alpha.</cmath> Now, because <math>AX</math> is tangent to the circumcircle at <math>A</math>, <math>\angle XAC = \angle CBA = 13\alpha</math>, and <math>\angle CAO = \angle OAX - \angle CAX = 90 - 13\alpha</math>. Finally, notice that <math>\angle AMB = \angle OMB - \angle OMG = 90 - \beta</math>. Then, <cmath>\angle BAM = 180 - \angle ABC - \angle AMB = 180 - 13\alpha - (90 - \beta) = 90 + \beta - 13\alpha.</cmath> Thus, <cmath>\angle BAC = \angle BAM + \angle MAO + \angle OAC = 90 + \beta - 13\alpha + \theta + 90 - 13\alpha = 180 - 26\alpha + (\beta + \theta),</cmath> and <cmath>180 = \angle BAC + 13\alpha + 2\alpha = 180 - 11\alpha + \beta + \theta \implies \beta + \theta = 11\alpha.</cmath> However, from before, <math>\beta+\theta = 180 - 17 \alpha</math>, so <math>11 \alpha = 180 - 17 \alpha \implies 180 = 28 \alpha \implies \alpha = \frac{180}{28}</math>. To finish the problem, we simply compute <cmath>\angle BAC = 180 - 15 \alpha = 180 \cdot \left(1 - \frac{15}{28}\right) = 180 \cdot \frac{13}{28} = \frac{585}{7},</cmath> so our final answer is <math>585+7=\boxed{592}</math>. | ||
+ | |||
+ | ~advanture | ||
+ | |||
+ | ==Solution 4 (Why Isosceles)== | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(375); | ||
+ | |||
+ | pair A, B, C, O, G, X, Y; | ||
+ | A = origin; | ||
+ | B = (1,0); | ||
+ | C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); | ||
+ | O = circumcenter(A,B,C); | ||
+ | G = centroid(A,B,C); | ||
+ | Y = intersectionpoint(G--G+(100,0),B--C); | ||
+ | X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); | ||
+ | pair O1=circumcenter(O,G,A); | ||
+ | real r1=length(O1-O); | ||
+ | markscalefactor=3/160; | ||
+ | filldraw(O--X--Y--cycle, rgb(255,255,0)); | ||
+ | draw(rightanglemark(O,G,X),red); | ||
+ | draw(A--O--B,fuchsia+0.4); | ||
+ | draw(Arc(O1,r1,-40,50),royalblue+0.5); | ||
+ | draw(circumcircle(O,G,Y), heavygreen+0.5); | ||
+ | dot("$A$",A,1.5*dir(180+585/7),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(-585/7),linewidth(4)); | ||
+ | dot("$C$",C,1.5N,linewidth(4)); | ||
+ | dot("$O$",O,1.5N,linewidth(4)); | ||
+ | dot("$G$",G,1.5S,linewidth(4)); | ||
+ | dot("$Y$",Y,1.5E,linewidth(4)); | ||
+ | dot("$X$",X,1.5W,linewidth(4)); | ||
+ | draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); | ||
+ | </asy> | ||
+ | <math>\angle OAX = \angle OGX = 90^\circ \implies</math> quadrilateral <math>XAGO</math> is cyclic <math>\implies</math> | ||
+ | |||
+ | <math>\angle GXO = \angle GAO,</math> as they share the same intersept <math>\overset{\Large\frown} {GO}.</math> | ||
+ | |||
+ | <math>\angle OGY = \angle OMY = 90^\circ \implies</math> quadrilateral <math>OGYM</math> is cyclic <math>\implies</math> | ||
+ | |||
+ | <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 <i><b>Claim</b></i>, <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^\circ = \frac {585^\circ}{7} \implies 585 + 7 = \boxed{592}.</cmath> | ||
+ | |||
+ | [[File:AIME-II-2021-14.png|230px|right]] | ||
+ | <i><b>Claim</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> | ||
+ | |||
+ | If <math>\angle AOM = 2\angle ACB + \angle ABC,</math> then <math>AC = BC.</math> | ||
+ | |||
+ | We define <math>\angle AOM</math> as the sum of <math>\angle AOB + \angle BOM,</math> this angle can be greater than <math>180^\circ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <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> | |
− | + | If <math>\angle AOM = 2 \angle ACB + \angle ABC,</math> then <math>\angle ABC = \angle CAB, AC = BC.</math> | |
− | + | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Extend <math>XA</math> and meet line <math>CB</math> at <math>P</math>. Extend <math>AG</math> to meet <math>BC</math> at <math>F</math>. Since <math>AF</math> is the median from <math>A</math> to <math>BC</math>, <math>A,G,F</math> are collinear. Furthermore, <math>OF</math> is perpendicular to <math>BC</math> | ||
+ | |||
+ | Draw the circumcircle of <math>\triangle{XPY}</math>, as <math>OA\bot XP, OG\bot XY, OF\bot PY</math>, <math>A,G,F</math> are collinear, <math>O</math> lies on <math>(XYP)</math> as <math>AGF</math> is the Simson line of <math>O</math> with respect to <math>\triangle{XPY}</math>. Thus, <math>\angle{P}=180-17x, \angle{PAB}=\angle{C}=2x, 180-15x=13x, x=\frac{45}{7}</math>, the answer is <math>180-15\cdot \frac{45}{7}=\frac{585}{7}</math> which is <math>\boxed{592}</math>. | ||
+ | |||
+ | ~bluesoul | ||
==Video Solution 1== | ==Video Solution 1== | ||
https://www.youtube.com/watch?v=zFH1Z7Ydq1s | https://www.youtube.com/watch?v=zFH1Z7Ydq1s | ||
+ | |||
+ | ~Mathematical Dexterity | ||
==Video Solution 2== | ==Video Solution 2== | ||
Line 56: | Line 156: | ||
~Osman Nal | ~Osman Nal | ||
− | ==See | + | ==Video Solution by Interstigation== |
+ | https://www.youtube.com/watch?v=yIWe7ME6fpA | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2021|n=II|num-b=13|num-a=15}} | {{AIME box|year=2021|n=II|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:50, 25 December 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 ~MRENTHUSIASM
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 (Why Isosceles)
quadrilateral
is cyclic
as they share the same intersept
quadrilateral
is cyclic
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 Claim, is isosceles,
Claim
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
vladimir.shelomovskii@gmail.com, vvsss
Solution 5
Extend and meet line
at
. Extend
to meet
at
. Since
is the median from
to
,
are collinear. Furthermore,
is perpendicular to
Draw the circumcircle of , as
,
are collinear,
lies on
as
is the Simson line of
with respect to
. Thus,
, the answer is
which is
.
~bluesoul
Video Solution 1
https://www.youtube.com/watch?v=zFH1Z7Ydq1s
~Mathematical Dexterity
Video Solution 2
https://www.youtube.com/watch?v=7Bxr2h4btWo
~Osman Nal
Video Solution by Interstigation
https://www.youtube.com/watch?v=yIWe7ME6fpA
~Interstigation
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.