Difference between revisions of "2008 USAMO Problems/Problem 2"
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== Solution == | == Solution == | ||
=== Solution 1 (synthetic) === | === Solution 1 (synthetic) === | ||
− | |||
<center><asy> | <center><asy> | ||
/* setup and variables */ | /* setup and variables */ | ||
Line 14: | Line 13: | ||
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); | pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); | ||
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); | D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); | ||
− | D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O, | + | D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s))); |
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); | D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); | ||
D(C--D(MP("F",F,NW,s))); | D(C--D(MP("F",F,NW,s))); | ||
D(B--O--C,linetype("4 4")+linewidth(0.7)); | D(B--O--C,linetype("4 4")+linewidth(0.7)); | ||
+ | D(M--N,linetype("4 4")+linewidth(0.7)); | ||
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); | D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); | ||
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); | D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); | ||
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Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar. ''End Lemma 1.'' | Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar. ''End Lemma 1.'' | ||
+ | <center><asy> | ||
+ | /* setup and variables */ | ||
+ | size(280); | ||
+ | pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); | ||
+ | pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ | ||
+ | /* construction and drawing */ | ||
+ | pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); | ||
+ | D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); | ||
+ | D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s))); | ||
+ | D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); | ||
+ | D(C--D(MP("F",F,NW,s))); | ||
+ | D(B--O--C,linetype("4 4")+linewidth(0.7)); | ||
+ | D(F--N); D(O--M); | ||
+ | D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); | ||
+ | |||
+ | /* commented from above asy | ||
+ | D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); | ||
+ | D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); | ||
+ | D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); | ||
+ | picture p = new picture; | ||
+ | draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); | ||
+ | clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); | ||
+ | */ | ||
+ | </asy></center> | ||
By the similarity in Lemma 1, <math>FE: EO = NE: EM\implies FE: EN = OE: NM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence | By the similarity in Lemma 1, <math>FE: EO = NE: EM\implies FE: EN = OE: NM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence | ||
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Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired. | Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired. | ||
− | === Solution 2 (isogonal conjugates) === | + | === Solution 2 (synthetic) === |
+ | Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle. {{incomplete|solution}} | ||
+ | |||
+ | === Solution 3 (isogonal conjugates) === | ||
<center><asy> | <center><asy> | ||
/* setup and variables */ | /* setup and variables */ | ||
Line 95: | Line 122: | ||
Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle. | Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle. | ||
− | === Solution | + | === Solution 4 (inversion) === |
{{image}} | {{image}} | ||
<center><asy> | <center><asy> | ||
Line 101: | Line 128: | ||
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); | pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); | ||
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */ | pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */ | ||
+ | real r = 1.2; /* inversion radius */ | ||
+ | |||
/* construction and drawing */ | /* construction and drawing */ | ||
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); | pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); | ||
Line 106: | Line 135: | ||
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); | D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); | ||
D(D(MP("D",D,SE,s))--MP("P",P,W,s)); | D(D(MP("D",D,SE,s))--MP("P",P,W,s)); | ||
− | D(B--D(MP("F",F,s) | + | D(B--D(MP("F",F,s))); |
− | |||
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); | D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); | ||
− | /* | + | D(CR(A,r)); |
+ | pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s)); | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
</asy></center> | </asy></center> | ||
We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}} | We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}} | ||
− | === Solution | + | === Solution 5 (trigonometric) === |
Use the [[Law of Sines]] to show that <math>\angle BFA = \angle AFC</math>. It follows that <math>\triangle BFA \sim \triangle AFC</math>. Noting the [[spiral similarity]] from <math>P</math> to <math>N</math>, {{incomplete|solution}} | Use the [[Law of Sines]] to show that <math>\angle BFA = \angle AFC</math>. It follows that <math>\triangle BFA \sim \triangle AFC</math>. Noting the [[spiral similarity]] from <math>P</math> to <math>N</math>, {{incomplete|solution}} | ||
− | === Solution | + | === Solution 6 (analytical) === |
+ | We let <math>A</math> be at the [[origin]], <math>B</math> be at the point <math>(a,0)</math>, and <math>C</math> be at the point <math>(b,c):\ b<a</math>. Then the equation of the perpendicular bisector of <math>\overline{AB}</math> is <math>x = a/2</math>, and {{incomplete|solution}} | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 17:53, 2 May 2008
Problem
(Zuming Feng) Let be an acute, scalene triangle, and let
,
, and
be the midpoints of
,
, and
, respectively. Let the perpendicular bisectors of
and
intersect ray
in points
and
respectively, and let lines
and
intersect in point
, inside of triangle
. Prove that points
,
,
, and
all lie on one circle.
Contents
Solution
Solution 1 (synthetic)
![[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s))); D(B--O--C,linetype("4 4")+linewidth(0.7)); D(M--N,linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); /* D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); */ [/asy]](http://latex.artofproblemsolving.com/5/9/c/59c1afd86c69597587b940e9645c56d85342ba62.png)
Without loss of generality . The intersection of
and
is
, the circumcenter of
.
Let and
. Note
lies on the perpendicular bisector of
, so
. So
. Similarly,
, so
. Notice that
intercepts the minor arc
in the circumcircle of
, which is double
. Hence
, so
is cyclic.
Lemma 1: is directly similar to
since
,
,
are collinear,
is cyclic, and
. Also
because
, and
is the medial triangle of
so
. Hence
.
Notice that since
.
. Then
Hence
.
Hence is similar to
by AA similarity. It is easy to see that they are oriented such that they are directly similar. End Lemma 1.
![[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s))); D(B--O--C,linetype("4 4")+linewidth(0.7)); D(F--N); D(O--M); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); /* commented from above asy D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); */ [/asy]](http://latex.artofproblemsolving.com/9/2/c/92cb9683263caf4e6841a1554bf52c74dca6b4d1.png)
By the similarity in Lemma 1, .
so
by SAS similarity. Hence
Using essentially the same angle chasing, we can show that
is directly similar to
. It follows that
is directly similar to
. So
Hence
, so
is cyclic. In other words,
lies on the circumcircle of
. Note that
, so
is cyclic. In other words,
lies on the circumcircle of
.
,
,
,
, and
all lie on the circumcircle of
. Hence
,
,
, and
lie on a circle, as desired.
Solution 2 (synthetic)
Hint: consider intersection with
; show that the resulting intersection lies on the desired circle. Template:Incomplete
Solution 3 (isogonal conjugates)
![[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); D(D(MP("D",D,SE,s))--MP("P",P,W,s)); D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7)); D(MP("O'",circumcenter(A,P,N),NW,s)); D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); [/asy]](http://latex.artofproblemsolving.com/e/f/4/ef48251bf50ebbfa10db0114d80f3a0f7efe3276.png)
Construct on
such that
. Then
. Then
, so
, or
. Then
, so
. Then we have
and
. So
and
are isogonally conjugate. Thus
. Then
.
If is the circumcenter of
then
so
is cyclic. Then
.
Then . Then
is a right triangle.
Now by the homothety centered at with ratio
,
is taken to
and
is taken to
. Thus
is taken to the circumcenter of
and is the midpoint of
, which is also the circumcenter of
, so
all lie on a circle.
Solution 4 (inversion)
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
![[asy] size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */ real r = 1.2; /* inversion radius */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); D(D(MP("D",D,SE,s))--MP("P",P,W,s)); D(B--D(MP("F",F,s))); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); D(CR(A,r)); pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s)); [/asy]](http://latex.artofproblemsolving.com/d/c/0/dc0a7c4a28ed6b542c57a0f4e8641860c217f3f7.png)
We consider an inversion by an arbitrary radius about . We want to show that
and
are collinear. Notice that
and
lie on a circle with center
, and similarly for the other side. We also have that
form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that
is a parallelogram, indicating that
is the midpoint of
. Template:Incomplete
Solution 5 (trigonometric)
Use the Law of Sines to show that . It follows that
. Noting the spiral similarity from
to
, Template:Incomplete
Solution 6 (analytical)
We let be at the origin,
be at the point
, and
be at the point
. Then the equation of the perpendicular bisector of
is
, and Template:Incomplete
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2008 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>