Difference between revisions of "2021 IMO Problems/Problem 4"
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==Solution2== | ==Solution2== | ||
Denote <math>AD</math> tangents to the circle <math>I</math> at <math>M</math>, <math>CD</math> tangents to the same circle at <math>N</math>; <math>XB</math> tangents at <math>F</math> and <math>ZB</math> tangents at <math>J</math>. We can get that <math>AD=AM+MD;CD=DN+CN</math>.Since <math>AM=AF,XA=XF-AM;ZC=ZJ-CN</math> Same reason, we can get that <math>DT=TN-DM;DY=YM-DM</math> | Denote <math>AD</math> tangents to the circle <math>I</math> at <math>M</math>, <math>CD</math> tangents to the same circle at <math>N</math>; <math>XB</math> tangents at <math>F</math> and <math>ZB</math> tangents at <math>J</math>. We can get that <math>AD=AM+MD;CD=DN+CN</math>.Since <math>AM=AF,XA=XF-AM;ZC=ZJ-CN</math> Same reason, we can get that <math>DT=TN-DM;DY=YM-DM</math> | ||
− | We can find that <math>AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM</math>. Connect <math>IM,IT,IT,IZ,IX,IN,IJ,IF</math> separately, we can create two pairs of congruent triangles. In <math>\ | + | We can find that <math>AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM</math>. Connect <math>IM,IT,IT,IZ,IX,IN,IJ,IF</math> separately, we can create two pairs of congruent triangles. In <math>\triangle{XIF},\triangle{YIM}</math>, since <math>\widehat{AI}=\widehat{AI},\angle{FXI}=\angle{MYI}</math> After getting that <math>\angle{IFX}=\angle{IMY};\angle{FXI}=\angle{MYI};IF=IM</math>, we can find that <math>\triangle{IFX}\cong \triangle{IMY}</math>. Getting that <math>YM=XF</math>, same reason, we can get that <math>ZJ=TN</math>. |
Now the only thing left is that we have to prove <math>TX=YZ</math>. Since <math>\widehat{IX}=\widehat{IY};\widehat{IT}=\widehat{IZ}</math> we can subtract and get that <math>\widehat{XT}=\widehat{YZ}</math>,means <math>XT=YZ</math> and we are done | Now the only thing left is that we have to prove <math>TX=YZ</math>. Since <math>\widehat{IX}=\widehat{IY};\widehat{IT}=\widehat{IZ}</math> we can subtract and get that <math>\widehat{XT}=\widehat{YZ}</math>,means <math>XT=YZ</math> and we are done | ||
~bluesoul | ~bluesoul |
Revision as of 13:05, 19 December 2021
Contents
[hide]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
, 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