2000 USAMO Problems/Problem 5
Problem
Let be a triangle and let
be a circle in its plane passing through
and
Suppose there exist circles
such that for
is externally tangent to
and passes through
and
where
for all
. Prove that
Solution
Let the circumcenter of be
, and let the center of
be
.
and
are externally tangent at the point
, so
are collinear.
is the intersection of the perpendicular bisectors of
, and each of the centers
lie on the perpendicular bisector of the side of the triangle that determines
. It follows from
that
.
![[asy] size(300); pathpen = linewidth(0.7); pen t = linetype("2 2"); pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */ pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("O",O,NW)); D(MP("O_1",O1,E)); D(MP("O_2",O2)); D(O--B--O1--C--O--A--O2--C, linetype("4 4") + linewidth(0.7)); D(O1--O--O2,linetype("4 4") + linewidth(0.6)); D(CP(O1,C),t);D(CP(O2,C),t);D(circumcircle(A,B,C),t); [/asy]](http://latex.artofproblemsolving.com/3/7/1/3717a532a6f30403d5d32173be48cce905c005af.png)
Since , and the perpendicular bisector of
are fixed, the angle
determines the position of
(since
lies on the perpendicular bisector). Let
; then,
and
together imply that
.
Now (due to collinearility). Hence, we have the recursion
, and so
. Thus,
.
implies that
, and circles
and
are the same circle since they have the same center and go through the same two points.
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |