Difference between revisions of "2024 AIME II Problems/Problem 10"
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Note that this problem is extremely similar to [[2019 CIME I Problems/Problem 14|2019 CIME I/14]]. | Note that this problem is extremely similar to [[2019 CIME I Problems/Problem 14|2019 CIME I/14]]. | ||
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+ | ==Solution 2== | ||
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+ | Denote <math>AB=a, AC=b, BC=c</math>. By the given condition, <math>\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6</math>, where <math>A</math> is the area of <math>\triangle{ABC}</math>. | ||
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+ | Moreover, since <math>OI\bot AI</math>, the second intersection of the line <math>AI</math> and <math>(ABC)</math> is the reflection of <math>A</math> about <math>I</math>, denote that as <math>D</math>. By Fact 5, <math>DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c</math>. | ||
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+ | Thus, we have <math>\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c</math>. Now, we have <math>\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}</math> | ||
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+ | ~Bluesoul | ||
==See also== | ==See also== |
Revision as of 18:58, 8 February 2024
Contents
Problem
Let have circumcenter
and incenter
with
, circumradius
, and inradius
. Find
.
Solution
By Euler's formula , we have
. Thus, by the Pythagorean theorem,
. Let
; notice
is isosceles and
which is enough to imply that
is the midpoint of
, and
itself is the midpoint of
where
is the
-excenter of
. Therefore,
and
Note that this problem is extremely similar to 2019 CIME I/14.
Solution 2
Denote . By the given condition,
, where
is the area of
.
Moreover, since , the second intersection of the line
and
is the reflection of
about
, denote that as
. By Fact 5,
.
Thus, we have . Now, we have
~Bluesoul
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.