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 (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--F--A,linetype("4 4")+linewidth(0.7)); 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/1/6/6/1662df361e5cd5b3bc1f7c504835be99d6bd2f21.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 2 (inversion)
Invert by an arbitrary radius about A. We want to show that P', F', and N' are collinear. Notice that D', A, and P' lie on a circle with center B', and similarly for the other side. We also have that B', D', F', A form a cyclic quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.) Template:Incomplete
Solution 3 (trigonometric)
Solution 4 (synthetic)
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>