Difference between revisions of "2021 AIME II Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | + | 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. | |
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| + | ~Lcz | ||
==See also== | ==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 19:04, 22 March 2021
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 
.
Solution
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
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. 
