Difference between revisions of "2004 IMO Problems/Problem 1"
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Latest revision as of 00:51, 19 November 2023
Problem
Let be an acute-angled triangle with
. The circle with diameter
intersects the sides
and
at
and
respectively. Denote by
the midpoint of the side
. The bisectors of the angles
and
intersect at
. Prove that the circumcircles of the triangles
and
have a common point lying on the side
.
Solution
Let ,
, and
. Call
the circle with diameter
and
the circumcircle of
.
Our ultimate goal is to show that . To show why this solves the problem, assume this statement holds true. Call
the intersection point of the circumcircle of
with side
. Then,
, and
. Since
,
, implying
also lies on the circumcircle of
, thereby solving the problem.
We now prove that . Note that
and
are radii of
, so
is isosceles. The bisector of
is thus the perpendicular bisector of
. Since
lies on the bisector of
,
. Angle computations yield that
from
and from
.
The bisector of hits
at the midpoint of the arc
not containing
. This point must lie on the perpendicular bisector of segment
, which is the bisector of
. It follows that
is indeed the midpoint of arc
, so
,
,
,
, are concyclic. Since
and
subtend the same arc
,
=
. With
being the bisector of
, we have
We know that
. so we have
. Since
, and
, we have
The problem is solved.
We have
. Noting that
, we then have
which is indeed the measure of
. This implies that
lies on the bisector of
, and from this, it is clear that
must lie on the interior of segment
. Not proving that
had to lie in the interior of
was a reason that a large portion of students who submitted a solution received a 1-point deduction.
See also
2004 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |