Difference between revisions of "2008 USAMO Problems/Problem 2"
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Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle. {{incomplete|solution}} | 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) === | + | === Solution 3 (trigonometric) === |
+ | By the [[Law of Sines]], <math>\frac {\sin\angle BAM}{\sin\angle CAM} = \frac {\sin B}{\sin C} = \frac bc = \frac {b/AF}{c/AF} = \frac {\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}</math>. Since <math>\angle ABF = \angle ABD = \angle BAD = \angle BAM</math> and similarly <math>\angle ACF = \angle CAM</math>, we cancel to get <math>\sin\angle AFC = \sin\angle AFB</math>. Obviously, <math>\angle AFB + \angle AFC > 180^\circ</math> so <math>\angle AFC = \angle AFB</math>. | ||
+ | |||
+ | Then <math>\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF</math> and <math>\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC</math>. Subtracting these two equations, <math>\angle FAB - \angle FCA = \angle FCA - \angle FAB</math> so <math>\angle BAF = \angle AFC</math>. Therefore, <math>\triangle ABF\sim\triangle CAF</math>, so a spiral similarity centered at <math>F</math> takes <math>B</math> to <math>A</math> and <math>A</math> to <math>C</math>. Therefore, it takes the midpoint of <math>\overline{BA}</math> to the midpoint of <math>\overline{AC}</math>, or <math>P</math> to <math>N</math>. So <math>\angle APF = \angle CNF = 180^\circ - \angle ANF</math> and <math>APFN</math> is cyclic. | ||
+ | |||
+ | === Solution 4 (isogonal conjugates) === | ||
<center><asy> | <center><asy> | ||
/* setup and variables */ | /* setup and variables */ | ||
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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 5 (symmedians) === |
+ | Median <math>AM</math> of a triangle <math>ABC</math> implies <math>\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}</math>. | ||
+ | Trig ceva for <math>F</math> shows that <math>AF</math> is a symmedian. | ||
+ | Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done. {{incomplete|solution}} | ||
+ | |||
+ | === Solution 6 (inversion) === | ||
{{image}} | {{image}} | ||
<center><asy> | <center><asy> | ||
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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 7 (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}} | 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}} | ||
Revision as of 17:20, 11 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 (trigonometric)
By the Law of Sines, . Since
and similarly
, we cancel to get
. Obviously,
so
.
Then and
. Subtracting these two equations,
so
. Therefore,
, so a spiral similarity centered at
takes
to
and
to
. Therefore, it takes the midpoint of
to the midpoint of
, or
to
. So
and
is cyclic.
Solution 4 (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 5 (symmedians)
Median of a triangle
implies
.
Trig ceva for
shows that
is a symmedian.
Then
is a median, use the lemma again to show that
, and similarly
, so you're done. Template:Incomplete
Solution 6 (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 7 (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>