Difference between revisions of "2021 IMO Problems/Problem 4"
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==Solution 3 (Visual)== | ==Solution 3 (Visual)== | ||
+ | [[File:2021 IMO 4.png|450px|right]] | ||
+ | <i><b>Lemma 1</b></i> | ||
− | |||
Let <math>O</math> be the center of <math>\Omega.</math> Then point <math>T(</math> point <math>X)</math> is symmetryc to <math>Z(Y)</math> with respect <math>IO.</math> | Let <math>O</math> be the center of <math>\Omega.</math> Then point <math>T(</math> point <math>X)</math> is symmetryc to <math>Z(Y)</math> with respect <math>IO.</math> | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
+ | |||
Let <math>\angle BAD =2\alpha, \angle ABC =2\beta, \angle BCD =2\gamma, \angle ADC =2\delta.</math> | Let <math>\angle BAD =2\alpha, \angle ABC =2\beta, \angle BCD =2\gamma, \angle ADC =2\delta.</math> | ||
+ | |||
We find measure of some arcs: | We find measure of some arcs: | ||
<cmath>\overset{\Large\frown} {IT}= 2\angle ICT = 2\gamma,</cmath> | <cmath>\overset{\Large\frown} {IT}= 2\angle ICT = 2\gamma,</cmath> | ||
Line 51: | Line 54: | ||
<math>\overset{\Large\frown} {IX}= 2\pi - \overset{\Large\frown} {IY} - \overset{\Large\frown} {XY} = 2\alpha =\overset{\Large\frown} {IY}\implies</math> symmetry <math>X</math> and <math>Y.</math> | <math>\overset{\Large\frown} {IX}= 2\pi - \overset{\Large\frown} {IY} - \overset{\Large\frown} {XY} = 2\alpha =\overset{\Large\frown} {IY}\implies</math> symmetry <math>X</math> and <math>Y.</math> | ||
<cmath>\overset{\Large\frown} {TZ}= 2\angle DCZ = 2\pi – 4\gamma,</cmath> | <cmath>\overset{\Large\frown} {TZ}= 2\angle DCZ = 2\pi – 4\gamma,</cmath> | ||
− | <math>\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} { | + | <math>\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} {IT} - \overset{\Large\frown} {TZ}= 2\gamma= \overset{\Large\frown} {IT}\implies</math> symmetry <math>T</math> and <math>Z.</math> |
+ | |||
+ | <i><b>Lemma 2</b></i> | ||
+ | |||
+ | Let circles <math>\omega</math> centered at <math>I</math> and <math>\Omega</math> centered at <math>O</math> be given. Let points <math>A</math> and <math>A'</math> lies on <math>\Omega</math> and symmetrical with respect <math>OI.</math> Let <math>AC</math> and <math>A'B</math> be tangents to <math>\omega</math>. Then <math>AC = A'B.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>AI = A'I, IB = IC, \angle ACI = \angle A'BI = 90^\circ \implies \triangle AIC = \triangle A'IB \implies A'B = AC.</math> |
Revision as of 19:37, 8 July 2022
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
Solution 3 (Visual)
Lemma 1
Let be the center of
Then point
point
is symmetryc to
with respect
Proof
Let
We find measure of some arcs:
symmetry
and
symmetry
and
Lemma 2
Let circles centered at
and
centered at
be given. Let points
and
lies on
and symmetrical with respect
Let
and
be tangents to
. Then
Proof