2008 USAMO Problems/Problem 2
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
[hide]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.
Lemma 2
![[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)
Without Loss of Generality, assume . It is sufficient to prove that
, as this would immediately prove that
are concyclic.
By applying the Menelaus' Theorem in the Triangle
for the transversal
, we have (in magnitude)
Here, we used that
, as
is the midpoint of
. Now, since
and
, we have
Now, note that
bisects the exterior
and
bisects exterior
, making
the
-excentre of
. This implies that
bisects interior
, making
, as was required.
Solution 3 (synthetic)
Hint: consider intersection with
; show that the resulting intersection lies on the desired circle. Template:Incomplete
Solution 4 (synthetic)
This solution utilizes the phantom point method. Clearly, APON are cyclic because . Let the circumcircles of triangles
and
intersect at
and
.
Lemma. If are points on circle
with center
, and the tangents to
at
intersect at
, then
is the symmedian from
to
.
This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.
It is easy to see (the intersection of ray
and the circumcircle of
) is colinear with
and
, and because line
is the diameter of that circle,
, so
is the point
in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that
satisfies the original construction for
, showing
; we are done. Template:Incomplete
Solution 5 (trigonometric)
By the Law of Sines, . Since
and similarly
, we cancel to get
. Obviously,
so
.
Then and
. Subtracting these two equations,
so
. Therefore,
(by AA similarity), 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 6 (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 7 (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 8 (inversion)
![[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 9 (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 |