Difference between revisions of "2011 AMC 12A Problems/Problem 25"
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\textbf{(E)}\ 90^{\circ} </math> | \textbf{(E)}\ 90^{\circ} </math> | ||
− | == Solution == | + | == Solution 1 (MAA) == |
− | < | + | By the Inscribed Angle Theorem, <cmath>\angle BOC = 2\angle BAC = 120^\circ .</cmath>Let <math>D</math> and <math>E</math> be the feet of the altitudes of <math>\triangle ABC</math> from <math>B</math> and <math>C</math>, respectively. In <math>\triangle ACE</math> we get <math>\angle ACE = 30^\circ</math>, and as exterior angle <cmath>\angle BHC = 90^\circ + \angle ACE = 120^\circ .</cmath>Because the lines <math>BI</math> and <math>CI</math> are bisectors of <math>\angle CBA</math> and <math>\angle ACB</math>, respectively, it follows that<cmath>\angle BIC = 90^\circ + \tfrac 12\angle A = 120^\circ .</cmath>Thus the points <math>B, C, O, I</math>, and <math>H</math> are all on a circle. |
− | + | <asy> | |
+ | import geometry; | ||
+ | size(200); | ||
+ | defaultpen(fontsize(12)+0.8); | ||
− | 1) | + | pair O,A,B,C,D,E,I,H; |
+ | real h=2*sqrt(3); | ||
+ | O=(0,1/h); B=(-0.5,0); C=(0.5,0); | ||
+ | path c1=CR(O,length(O-B)); | ||
+ | // A=IP(c1,O--O+5*dir(120)); | ||
+ | A=IP(c1,B--B+5*dir(80)); | ||
+ | I=incenter(A,B,C); | ||
+ | H=orthocenter(A,B,C); | ||
+ | D=extension(A,C,B,H); E=extension(A,B,C,H); | ||
+ | path c2=circumcircle(O,I,H); | ||
+ | pair o2=circumcenter(O,I,H); | ||
+ | draw(c1, royalblue); draw(A--B--C--A); draw(B--D^^C--E, dotted); | ||
+ | draw(arc(o2,length(O-o2),10,170), dotted+red); draw(B--I--C--O--B, black+0.3); | ||
+ | pen p =black+3.25; | ||
+ | dot("$O$", O, N,p); dot("$A$", A, dir(110),p); dot("$B$", B, dir(210),p); dot("$C$", C, dir(-30),p); dot("$I$", I, N,p); dot("$H$", H, N,p); dot("$D$", D, D-H,p); dot("$E$", E, (E-C),p); | ||
+ | </asy> | ||
+ | Further, since | ||
+ | <cmath>\angle OCI = \angle OCB - \angle ICB = 30^\circ - \tfrac 12\angle C </cmath> | ||
+ | <cmath>\angle ICH = \angle ACE - \angle ACI = 30^\circ - \tfrac 12\angle C </cmath> | ||
+ | we have <math>OI=IH</math>. | ||
− | + | Because <math>[BCOIH]=[BCO]+[BOIH]</math>, it is sufficient to maximize the area of quadrilateral <math>BOIH</math>. If <math>P_1</math>, <math>P_2</math> are two points in an arc of circle <math>BO</math> with <math>BP_1<BP_2</math>, then the maximum area of <math>BOP_1P_2</math> occurs when <math>BP_1=P_1P_2=P_2O</math>. Indeed, if <math>BP_1\neq P_1P_2</math>, then replacing <math>P_1</math> by the point <math>P_1’</math> located halfway in the arc of the circle <math>BP_2</math> yields a triangle <math>BP_1’P_2</math> with larger area than <math>\triangle BP_1P_2</math>, and the area of <math>\triangle BOP_2</math> remains the same. Similarly, if <math>P_1P_2\neq P_2O</math>. | |
− | + | Therefore the maximum is achieved when <math>OI=IH=HB</math>, that is, when <cmath>\angle OCI = \angle ICH = \angle HCB = \tfrac 13 \angle OCB = 10^\circ.</cmath>Thus <math>\angle ACB = 40^\circ</math> and <math>\angle CBA = 80^\circ</math>. | |
− | + | == Solution 2== | |
− | < | + | Let <math>\angle CAB=A</math>, <math>\angle ABC=B</math>, <math>\angle BCA=C</math> for convenience. |
− | |||
− | <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, hence 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), hence <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=\angle IBO</math>, <math>IH=IO</math> by Inscribed Angles. | ||
− | + | There are two ways to proceed. | |
− | |||
− | |||
− | <math> | + | 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. | |
− | |||
− | <math> | + | '''Lemma:''' <math>[BOIH]</math> is maximized only if <math>HB=HI</math>. |
− | <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, along with the fact that inscribed angles that intersect the same length chords are equal, |
− | + | <cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath> | |
− | + | ==Video Solution by Osman Nal== | |
− | + | https://www.youtube.com/watch?v=O3amRG9zEHE&ab_channel=OsmanNal | |
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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}} |
Latest revision as of 13:05, 28 June 2022
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 1 (MAA)
By the Inscribed Angle Theorem, Let
and
be the feet of the altitudes of
from
and
, respectively. In
we get
, and as exterior angle
Because the lines
and
are bisectors of
and
, respectively, it follows that
Thus the points
, and
are all on a circle.
Further, since
we have
.
Because , it is sufficient to maximize the area of quadrilateral
. If
,
are two points in an arc of circle
with
, then the maximum area of
occurs when
. Indeed, if
, then replacing
by the point
located halfway in the arc of the circle
yields a triangle
with larger area than
, and the area of
remains the same. Similarly, if
.
Therefore the maximum is achieved when , that is, when
Thus
and
.
Solution 2
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, hence 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), hence
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, along with the fact that inscribed angles that intersect the same length chords are equal,
Video Solution by Osman Nal
https://www.youtube.com/watch?v=O3amRG9zEHE&ab_channel=OsmanNal
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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