Difference between revisions of "2011 AMC 12A Problems/Problem 25"
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== Solution == | == Solution == | ||
− | + | Let <math>\angle CAB=A</math>, <math>\angle ABC=B</math>, <math>\angle BCA=C</math> for convenience. | |
− | 2) | + | It's well-known that <math>\angle BOC=2A</math>, <math>\angle BIC=90+\frac{A}{2}</math>, and <math>\angle BHC=180-A</math> (indeed, all are verifiable by angle chasing). Then, as <math>A=60</math>, it follows that <math>\angle BOC=\angle BIC=\angle BHC=120</math> and consequently pentagon <math>BCOIH</math> is cyclic. Observe that <math>BC=1</math> is fixed, whence the circumcircle of cyclic pentagon <math>BCOIH</math> is also fixed. Similarly, as <math>OB=OC</math>, it follows that <math>O</math> is the midpoint of minor arc <math>BC</math>, so it's fixed as well. This implies that <math>[BCO]</math> is fixed, and since <math>[BCOIH]=[BCO]+[BOIH]</math> is maximal, it suffices to maximize <math>[BOIH]</math>. |
− | + | Verify that <math>\angle IBC=\frac{B}{2}</math>, <math>\angle HBC=90-C</math> by angle chasing; it follows that <math>\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</math> since <math>A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90</math> by Triangle Angle Sum. Similarly, <math>\angle OBC=90-A=30</math>, whence <math>\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</math> and consequently <math>IH=IO</math> by Inscribed Angles. | |
− | + | There are several ways to proceed. Letting <math>O'</math> and <math>R</math> be the circumcenter and circumradius, respectively, of cyclic pentagon <math>BCOIH</math>, the most straightforward is to write <math>[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]</math>, whence <cmath>[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))</cmath> and, using the fact that <math>R</math> is fixed, maximize <math>2\sin(60-C)+\sin(2C-60)</math> with Jensen's Inequality. A much more elegant way is shown below. | |
− | '''Lemma:''' | + | '''Lemma:''' <math>[BOIH]</math> is maximized only if <math>HB=HI</math>. |
− | <math> | + | '''Proof:''' Suppose for the sake of contradiction that <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>, whence the claim follows. <math>\blacksquare</math> |
− | + | It's necessary to show the existence of a maximum <math>[BOIH]</math> (although the wording of the problem gives it to you for free), which is not hard. Either way, since <math>HB=HI</math> by our lemma and <math>IH=IO</math> from above, it follows that <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''' | |
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== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}} | {{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:16, 22 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
(indeed, all are 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
, it follows that
is the midpoint of minor arc
, so it's fixed as well. This implies that
is fixed, and since
is maximal, it suffices to maximize
.
Verify that ,
by angle chasing; it follows that
since
by Triangle Angle Sum. Similarly,
, whence
and consequently
by Inscribed Angles.
There are several 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 much more elegant way is shown below.
Lemma: is maximized only if
.
Proof: Suppose for the sake of contradiction that 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
, whence the claim follows.
It's necessary to show the existence of a maximum (although the wording of the problem gives it to you for free), which is not hard. Either way, since
by our lemma and
from above, it follows that
-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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