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 |
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