2011 AMC 12A Problems/Problem 25
Problem
Triangle has
,
,
, and
. Let
,
, and
be the orthocenter, incenter, and circumcenter of
, respectively. Assume that the area of pentagon
is the maximum possible. What is
?
Solution
Let ,
,
for convenience.
It's well-known that ,
, and
(verifiable by angle chasing). Then, as
, it follows that
and consequently pentagon
is cyclic. Observe that
is fixed, whence the circumcircle of cyclic pentagon
is also fixed. Similarly, as
(both are radii), it follows that
and also
is fixed. Since
is maximal, it suffices to maximize
.
Verify that ,
by angle chasing; it follows that
since
by Triangle Angle Sum. Similarly,
(isosceles base angles are equal), whence
Since
\angle IBO
IH=IO$by Inscribed Angles.
There are two ways to proceed.
Letting$ (Error compiling LaTeX. Unknown error_msg)O'R
BCOIH
[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]
R
2\sin(60-C)+\sin(2C-60)$with Jensen's Inequality.
A more elegant way is shown below.
'''Lemma:'''$ (Error compiling LaTeX. Unknown error_msg)[BOIH]HB=HI$.
'''Proof by contradiction:''' Suppose$ (Error compiling LaTeX. Unknown error_msg)[BOIH]HB\neq HI
H'
BI
I'
H'O
[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]
H'
BI
H
BI
[BH'I'O]>[BOIH']>[BOIH]
H'
I'
[BOIH]
\blacksquare
HB=HI
IH=IO$ from above:
-Solution by thecmd999
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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