2012 IMO Problems/Problem 1
Problem
Given triangle the point
is the centre of the excircle opposite the vertex
This excircle is tangent to the side
at
, and to the lines
and
at
and
, respectively. The lines
and
meet at
, and the lines
and
meet at
Let
be the point of intersection of the lines
and
, and let
be the point of intersection of the lines
and
Prove that
is the midpoint of
.
Solution
First, because
and
are both tangents from
to the excircle
. Then
. Call the
the intersection between
and
. Similarly, let the intersection between the perpendicular line segments
and
be
. We have
and
. We then have,
. So
. We also have
. Then
. Notice that
. Then,
.
. Similarly,
. Draw the line segments
and
.
and
are congruent and
and
are congruent. Quadrilateral
is cyclic because
. Quadrilateral
is also cyclic because
. The circumcircle of
also contains the points
and
because there is a circle around the quadrilaterals
and
. Therefore, pentagon
is also cyclic. Finally, quadrilateral
is cyclic because
. Again,
is common in both the cyclic pentagon
and cyclic quadrilateral
, so the circumcircle of
also contains the points
,
, and
. Therefore, hexagon
is cyclic. Since
and
are both right angles,
is the diameter of the circle around cyclic hexagon
. Therefore,
and
are both right angles.
and
are congruent by ASA congruency, and so are
and
. We have
,
,
, and
. Since
and
are tangents from
to the circle
,
. Then, we have
, which becomes
, which is
, or
. This means that
is the midpoint of
.
QED
--Aopsqwerty 21:19, 19 July 2012 (EDT)
Solution 2
For simplicity, let written alone denote the angles of triangle
, and
,
,
denote its sides.
Let be the radius of the A-excircle. Because
, we have
isosceles and so
by the Exterior Angle Theorem. Then because
, we have
, again by the Exterior Angle Theorem.
Notice that and
, and so
after converting tangents to sine and cosine. Thus,
It follows that
. By the Law of Sines on triangle
and
and the double-angle formula for sine, we have
Therefore, triangle
is congruent to a right triangle with hypotenuse length
and one angle of measure
by SAS Congruence, and so
. It then follows that triangles
and
are congruent by
, and so
. Thus,
lies on the perpendicular bisector of
. Similarly,
lies on the perpendicular bisector of
, and so
is the circumcenter of
. In particular,
lies on the perpendicular bisector of
, and so, because
is perpendicular to
,
must be the midpoint of
, as desired.
--Suli 17:53, 8 February 2015 (EST)
Solution 3
Same as Solution 2, except noticing that (letting be the semi-perimeter):
--Suli 18:21, 8 February 2015 (EST)
Solution 4
As before in Solution 2, we find that But it is clear that
bisects
, so
and hence
is cyclic. In particular,
, and continue as in Solution 2.