Difference between revisions of "2021 IMO Problems/Problem 4"
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+ | ==Video Solutions== | ||
+ | https://www.youtube.com/watch?v=vUftJHRaNx8 [Video contains solutions to all day 2 problems] | ||
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+ | https://www.youtube.com/watch?v=U95v_xD5fJk | ||
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+ | https://youtu.be/WkdlmduOnRE | ||
==Solution== | ==Solution== | ||
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~BUMSTAKA | ~BUMSTAKA | ||
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Revision as of 22:59, 30 July 2021
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://www.youtube.com/watch?v=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