Difference between revisions of "2021 AIME II Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (→Solution 2) |
MRENTHUSIASM (talk | contribs) m (Rearranged the solutions based on education values. I pushed the guessing solution to the end. Also, since spiral similarity is a bit rare, I made it Solution 2. I did not edit any content of the solution. If anyone is unhappy with it, please PM me.) |
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==Solution 1== | ==Solution 1== | ||
− | + | In this solution, all angle measures are in degrees. | |
− | |||
− | |||
− | + | 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> | |
− | Let <math>M</math> be the midpoint of <math>\overline{BC}</math> so that <math> | ||
− | + | Note that: | |
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
− | <li>Since <math>\angle OGX = \angle OAX = 90 | + | <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 OGY = \angle OMY = 90 | + | <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> |
</ol> | </ol> | ||
− | Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, | + | Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math> |
− | Next, we | + | Next, we express <math>\angle BAC</math> in terms of <math>k.</math> By angle addition, we have |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\angle AOM &= \angle AOB + \angle BOM \\ | \angle AOM &= \angle AOB + \angle BOM \\ | ||
Line 30: | Line 27: | ||
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> | ||
− | For the sum of the interior angles of <math>\ | + | For the sum of the interior angles of <math>\triangle ABC,</math> we get |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \angle ABC + \angle BCA + \angle BAC &= 180 | + | \angle ABC + \angle BCA + \angle BAC &= 180 \\ |
− | 13k+2k+13k&=180 | + | 13k+2k+13k&=180 \\ |
28k&=180 \\ | 28k&=180 \\ | ||
k&=\frac{45}{7}. | k&=\frac{45}{7}. | ||
Line 41: | Line 38: | ||
~Constance-variance (Fundamental Logic) | ~Constance-variance (Fundamental Logic) | ||
− | ~MRENTHUSIASM ( | + | ~MRENTHUSIASM (Reconstruction) |
+ | |||
+ | ==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 | + | ==Solution 4 (Guessing in the Last 3 Minutes, Unreliable)== |
Notice that <math>\triangle ABC</math> looks isosceles, so we assume it's isosceles. Then, let <math>\angle BAC = \angle ABC = 13x</math> and <math>\angle BCA = 2x.</math> Taking the sum of the angles in the triangle gives <math>28x=180,</math> so <math>13x = \frac{13}{28} \cdot 180 = \frac{585}{7}</math> so the answer is <math>\boxed{592}.</math> | Notice that <math>\triangle ABC</math> looks isosceles, so we assume it's isosceles. Then, let <math>\angle BAC = \angle ABC = 13x</math> and <math>\angle BCA = 2x.</math> Taking the sum of the angles in the triangle gives <math>28x=180,</math> so <math>13x = \frac{13}{28} \cdot 180 = \frac{585}{7}</math> so the answer is <math>\boxed{592}.</math> | ||
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~Osman Nal | ~Osman Nal | ||
− | ==See | + | ==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}} |
Revision as of 05:41, 3 July 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 (by Geometry Expressions)
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.