2021 IMO Problems/Problem 4
Contents
Problem
Let be a circle with centre
, and
a convex quadrilateral such that each of
the segments
and
is tangent to
. Let
be the circumcircle of the triangle
.
The extension of
beyond
meets
at
, and the extension of
beyond
meets
at
.
The extensions of
and
beyond
meet
at
and
, respectively. Prove that
Video Solutions
https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems]
https://www.youtube.com/watch?v=U95v_xD5fJk
Solution
Let be the centre of
.
For the result follows simply. By Pitot's Theorem we have
so that,
The configuration becomes symmetric about
and the result follows immediately.
Now assume WLOG . Then
lies between
and
in the minor arc
and
lies between
and
in the minor arc
.
Consider the cyclic quadrilateral
.
We have
and
. So that,
Since
is the incenter of quadrilateral
,
is the angular bisector of
. This gives us,
Hence the chords
and
are equal.
So
is the reflection of
about
.
Hence,
and now it suffices to prove
Let
and
be the tangency points of
with
and
respectively. Then by tangents we have,
. So
.
Similarly we get,
. So it suffices to prove,
Consider the tangent
to
with
. Since
and
are reflections about
and
is a circle centred at
the tangents
and
are reflections of each other. Hence
By a similar argument on the reflection of
and
we get
and finally,
as required.
~BUMSTAKA
Solution2
Denote tangents to the circle
at
,
tangents to the same circle at
;
tangents at
and
tangents at
. We can get that
.Since
Same reason, we can get that
We can find that
. Connect
separately, we can create two pairs of congruent triangles. In $\trangle{XIF},\triangle{YIM}$ (Error compiling LaTeX. Unknown error_msg), since
After getting that
, we can find that
. Getting that
, same reason, we can get that
.
Now the only thing left is that we have to prove
. Since
we can subtract and get that
,means
and we are done
~bluesoul