1988 USAMO Problems/Problem 4
Problem
is a triangle with incenter
. Show that the circumcenters of
,
, and
lie on a circle whose center is the circumcenter of
.
Solution
Solution 1
Let the circumcenters of ,
, and
be
,
, and
, respectively. It then suffices to show that
,
,
,
,
, and
are concyclic.
We shall prove that quadrilateral is cyclic first. Let
,
, and
. Then
and
. Therefore minor arc
in the circumcircle of
has a degree measure of
. This shows that
, implying that
. Therefore quadrilateral
is cyclic.
This shows that point is on the circumcircle of
. Analagous proofs show that
and
are also on the circumcircle of
, which completes the proof.
Solution 2
Let denote the midpoint of arc
. It is well known that
is equidistant from
,
, and
(to check, prove
), so that
is the circumcenter of
. Similar results hold for
and
, and hence
,
, and
all lie on the circumcircle of
.
Solution 3
Extend to point
on
. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle
. In other words,
, so
is on
. Similarly, we can show that
and
are on
, and thus,
are all concyclic. It follows that the circumcenters are equal.
See Also
1988 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.