Difference between revisions of "2011 AMC 12A Problems/Problem 25"
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'''Proof by contradiction:''' Suppose <math>[BOIH]</math> is maximized when <math>HB\neq HI</math>. Let <math>H'</math> be the midpoint of minor arc <math>BI</math> be and <math>I'</math> the midpoint of minor arc <math>H'O</math>. Then <math>[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]</math> since the altitude from <math>H'</math> to <math>BI</math> is greater than that from <math>H</math> to <math>BI</math>; similarly <math>[BH'I'O]>[BOIH']>[BOIH]</math>. Taking <math>H'</math>, <math>I'</math> to be the new orthocenter, incenter, respectively, this contradicts the maximality of <math>[BOIH]</math>, so our claim follows. <math>\blacksquare</math> | '''Proof by contradiction:''' Suppose <math>[BOIH]</math> is maximized when <math>HB\neq HI</math>. Let <math>H'</math> be the midpoint of minor arc <math>BI</math> be and <math>I'</math> the midpoint of minor arc <math>H'O</math>. Then <math>[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]</math> since the altitude from <math>H'</math> to <math>BI</math> is greater than that from <math>H</math> to <math>BI</math>; similarly <math>[BH'I'O]>[BOIH']>[BOIH]</math>. Taking <math>H'</math>, <math>I'</math> to be the new orthocenter, incenter, respectively, this contradicts the maximality of <math>[BOIH]</math>, so our claim follows. <math>\blacksquare</math> | ||
− | With our lemma(<math>HB=HI</math>) and <math>IH=IO</math> from above: <cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath> | + | With our lemma(<math>HB=HI</math>) and <math>IH=IO</math> from above: |
+ | <cmath>\angle OCI=\angle ICH=\angle HCB</cmath>, since they are inscribed angles intersecting congruent chords. | ||
+ | <cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath> | ||
-Solution by '''thecmd999''' | -Solution by '''thecmd999''' |
Revision as of 17:50, 28 September 2013
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
,
by Inscribed Angles.
There are two ways to proceed.
Letting and
be the circumcenter and circumradius, respectively, of cyclic pentagon
, the most straightforward is to write
, whence
and, using the fact that
is fixed, maximize
with Jensen's Inequality.
A more elegant way is shown below.
Lemma: is maximized only if
.
Proof by contradiction: Suppose is maximized when
. Let
be the midpoint of minor arc
be and
the midpoint of minor arc
. Then
since the altitude from
to
is greater than that from
to
; similarly
. Taking
,
to be the new orthocenter, incenter, respectively, this contradicts the maximality of
, so our claim follows.
With our lemma() and
from above:
, since they are inscribed angles intersecting congruent chords.
-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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