2013 IMO Problems/Problem 4
Contents
Problem
Let be an acute triangle with orthocenter
, and let
be a point on the side
, lying strictly between
and
. The points
and
are the feet of the altitudes from
and
, respectively. Denote by
is [sic] the circumcircle of
, and let
be the point on
such that
is a diameter of
. Analogously, denote by
the circumcircle of triangle
, and let
be the point such that
is a diameter of
. Prove that
and
are collinear.
Hint
Draw a good diagram, or use the one below. What do you notice? (In particular, what do you want to be true? How do you prove it true?)
Solution 1
Let be the intersection of
and
other than
.
Lemma 1: is on
.
Proof: We have because they intercept semicircles. Hence,
, so
is a straight line.
Lemma 2: is on
.
Proof: Let the circumcircles of and
be
and
, respectively, and, as
is cyclic (from congruent
), let its circumcircle be
. Then each pair of circles' radical axises,
and
, must concur at the intersection of
and
, which is
.
Lemma 3: is perpendicular to
.
Proof: This is immediate from .
Let meet
at
, which is also the foot of the altitude to that side. Hence,
Lemma 4: Quadrilateral is cyclic.
Proof: We know that is cyclic because
and
, opposite and right angles, sum to
. Furthermore, we are given that
is cyclic. Hence, by Power of a Point,
The converse of Power of a Point then proves cyclic.
Hence, , and so
is perpendicular to
as well. Combining this with Lemma 3's statement, we deduce that
are collinear. But, as
is on
(from Lemma 1),
are collinear. This completes the proof.
--Suli 13:51, 25 August 2014 (EDT)
Solution 2
Probably a simpler solution than above.
As above, let By Miquel
is cyclic. Then since
we know, because
and
that
thus
are collinear.
There are a few ways to finish.
"(a)"
so
are collinear, as desired
"(b)" Since is cyclic we know
which means
so
is on the radical axis,
hence
so
lies on this line as well and we may conclude
--mathguy623 03:07, 12 August 2016 (EDT)