Difference between revisions of "2011 AMC 12A Problems/Problem 25"
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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> (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>(both are radii), it follows that <math>O</math> and also <math>[BCO]</math> is fixed. Since <math>[BCOIH]=[BCO]+[BOIH]</math> is maximal, it suffices to maximize <math>[BOIH]</math>. | 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> (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>(both are radii), it follows that <math>O</math> and also <math>[BCO]</math> is fixed. 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=(180-120)/2=30</math> (isosceles base angles are equal), whence <cmath>\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</cmath> | + | 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=(180-120)/2=30</math> (isosceles base angles are equal), whence <cmath>\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</cmath> |
+ | Since <math>\angle IBH=</math>\angle IBO<math>, </math>IH=IO<math> by Inscribed Angles. | ||
There are two ways to proceed. | There are two ways to proceed. | ||
− | Letting <math>O'< | + | 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 more elegant way is shown below. | A more elegant way is shown below. | ||
− | '''Lemma:''' <math>[BOIH]< | + | '''Lemma:''' </math>[BOIH]<math> is maximized only if </math>HB=HI<math>. |
− | '''Proof by contradiction:''' Suppose <math>[BOIH]< | + | '''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< | + | With our lemma(</math>HB=HI<math>) and </math>IH=IO$ 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> |
-Solution by '''thecmd999''' | -Solution by '''thecmd999''' |
Revision as of 16:06, 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
\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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.