Difference between revisions of "2021 IMO Problems/Problem 4"
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− | + | == Problem == | |
+ | |||
+ | Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of | ||
the segments <math>AB, BC, CD</math> and <math>DA</math> is tangent to <math>\Gamma</math>. Let <math>\Omega</math> be the circumcircle of the triangle <math>AIC</math>. | the segments <math>AB, BC, CD</math> and <math>DA</math> is tangent to <math>\Gamma</math>. Let <math>\Omega</math> be the circumcircle of the triangle <math>AIC</math>. | ||
The extension of <math>BA</math> beyond <math>A</math> meets <math>\Omega</math> at <math>X</math>, and the extension of <math>BC</math> beyond <math>C</math> meets <math>\Omega</math> at <math>Z</math>. | The extension of <math>BA</math> beyond <math>A</math> meets <math>\Omega</math> at <math>X</math>, and the extension of <math>BC</math> beyond <math>C</math> meets <math>\Omega</math> at <math>Z</math>. | ||
Line 5: | Line 7: | ||
<cmath>AD + DT + T X + XA = CD + DY + Y Z + ZC</cmath> | <cmath>AD + DT + T X + XA = CD + DY + Y Z + ZC</cmath> | ||
+ | == Video Solutions == | ||
+ | https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems] | ||
− | + | https://www.youtube.com/watch?v=U95v_xD5fJk | |
− | Let <math>O</math> be the centre of <math>\Omega</math> | + | https://youtu.be/WkdlmduOnRE |
− | For <math>AB=BC</math> the result follows simply. By Pitot's Theorem we have <cmath>AB + CD = BC + AD</cmath> so that, < | + | |
+ | == Solution 1 == | ||
+ | |||
+ | Let <math>O</math> be the centre of <math>\Omega</math>. | ||
+ | |||
+ | For <math>AB=BC</math> the result follows simply. By Pitot's Theorem we have <cmath>AB + CD = BC + AD</cmath> so that, <math>AD = CD.</math> The configuration becomes symmetric about <math>OI</math> and the result follows immediately. | ||
Now assume WLOG <math>AB < BC</math>. Then <math>T</math> lies between <math>A</math> and <math>X</math> in the minor arc <math>AX</math> and <math>Z</math> lies between <math>Y</math> and <math>C</math> in the minor arc <math>YC</math>. | Now assume WLOG <math>AB < BC</math>. Then <math>T</math> lies between <math>A</math> and <math>X</math> in the minor arc <math>AX</math> and <math>Z</math> lies between <math>Y</math> and <math>C</math> in the minor arc <math>YC</math>. | ||
Consider the cyclic quadrilateral <math>ACZX</math>. | Consider the cyclic quadrilateral <math>ACZX</math>. | ||
− | We have <math>\angle CZX = \angle CAB</math> and <math>\angle IAC = \angle IZC</math>. So that, <cmath>\angle CZX - \angle IZC = \angle CAB - \angle IAC</cmath> <cmath>\angle IZX = \angle IAB</cmath> | + | We have <math>\angle CZX = \angle CAB</math> and <math>\angle IAC = \angle IZC</math>. So that, <cmath>\angle CZX - \angle IZC = \angle CAB - \angle IAC</cmath> <cmath>\angle IZX = \angle IAB</cmath> Since <math>I</math> is the incenter of quadrilateral <math>ABCD</math>, <math>AI</math> is the angular bisector of <math>\angle DBA</math>. This gives us, <cmath>\angle IZX = \angle IAB = \angle IAD = \angle IAY</cmath> Hence the chords <math>IX</math> and <math>IY</math> are equal. |
So <math>Y</math> is the reflection of <math>X</math> about <math>OI</math>. | So <math>Y</math> is the reflection of <math>X</math> about <math>OI</math>. | ||
+ | Hence, <cmath>TX = YZ</cmath> and now it suffices to prove <cmath>AD + DT + XA = CD + DY + ZC</cmath> | ||
+ | Let <math>P, Q, N</math> and <math>M</math> be the tangency points of <math>\Gamma</math> with <math>AB, BC, CD</math> and <math>DA</math> respectively. Then by tangents we have, <math>AD = AM + MD = AP + ND</math>. So <math>AD + DT + XA = AP + ND + DT + XA = XP + NT</math>. | ||
+ | Similarly we get, <math>CD + DY + ZC = ZQ + YM</math>. So it suffices to prove, <cmath>XP + NT = ZQ + YM</cmath> | ||
+ | Consider the tangent <math>XJ</math> to <math>\Gamma</math> with <math>J \ne P</math>. Since <math>X</math> and <math>Y</math> are reflections about <math>OI</math> and <math>\Gamma</math> is a circle centred at <math>I</math> the tangents <math>XJ</math> and <math>YM</math> are reflections of each other. Hence <cmath>XP = XJ = YM</cmath> By a similar argument on the reflection of <math>T</math> and <math>Z</math> we get <math>NT = ZQ</math> and finally, | ||
+ | <cmath> XP + NT = ZQ + YM</cmath> as required. | ||
+ | <math>QED</math> | ||
+ | |||
+ | ~BUMSTAKA | ||
+ | == Solution 2 == | ||
+ | Denote <math>AD</math> tangents to the circle <math>I</math> at <math>N</math>, <math>CD</math> tangents to the same circle at <math>M</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>\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 | ||
+ | ~bluesoul | ||
+ | |||
+ | == Solution 3 (Visual) == | ||
+ | [[File:2021 IMO 4b.png|420px|right]] | ||
+ | [[File:2021 IMO 4.png|420px|right]] | ||
+ | [[File:2021 IMO 4a.png|420px|right]] | ||
+ | We use the equality of the tangent segments and symmetry. | ||
+ | |||
+ | Using <i><b>Claim 1</b></i> we get <math>\overset{\Large\frown} {TX}</math> symmetric to <math>\overset{\Large\frown} {ZY}</math> with respect <math>IO.</math> | ||
+ | |||
+ | Therefore <math>\hspace{10mm} TX = ZY.</math> | ||
+ | |||
+ | Let <math>P, Q, N</math> and <math>M</math> be the tangency points of <math>\Gamma</math> with <math>AB, BC, CD,</math> and <math>DA,</math> respectively. | ||
+ | |||
+ | Using <i><b>Claim 2</b></i> we get <math>TM = QZ, PX = NY.</math> | ||
+ | |||
+ | <cmath>AD + DT + XA = AN+ND + TM – MD +XP-PA =</cmath> | ||
+ | <cmath>= XP + TM = QZ + NY = MC + ZC + MD + DY =</cmath> | ||
+ | <cmath>=CD + ZC + DY.</cmath> | ||
+ | <i><b>Claim 1</b></i> | ||
+ | |||
+ | Let <math>O</math> be the center of <math>\Omega.</math> Then point <math>T</math> is symmetric to <math>Z</math> with respect <math>IO,</math> point <math>X</math> is symmetric to <math>Y</math> with respect <math>IO.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 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: | ||
+ | <cmath>\overset{\Large\frown} {IT}= 2\angle ICT = 2\gamma,</cmath> | ||
+ | <cmath>\overset{\Large\frown} {IY}= 2\angle IAY = 2\alpha,</cmath> | ||
+ | <cmath>\overset{\Large\frown} {XY}= 2\angle XAY = 2\pi - 4\alpha,</cmath> | ||
+ | <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> | ||
+ | <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>Claim 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 <math>A</math> be symmetric to <math>A'</math> 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> | ||
+ | <cmath>AI = A'I, IB = IC, \angle ACI = \angle A'BI = 90^\circ \implies</cmath> | ||
+ | <cmath>\triangle AIC = \triangle A'IB \implies A'B = AC.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | == See also == | ||
+ | {{IMO box|year=2021|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 09:30, 18 June 2023
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 1
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
Solution 2
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)
We use the equality of the tangent segments and symmetry.
Using Claim 1 we get symmetric to
with respect
Therefore
Let and
be the tangency points of
with
and
respectively.
Using Claim 2 we get
Claim 1
Let be the center of
Then point
is symmetric to
with respect
point
is symmetric to
with respect
Proof
Let
We find measure of some arcs:
symmetry
and
symmetry
and
Claim 2
Let circles centered at
and
centered at
be given. Let points
and
lies on
and
be symmetric to
with respect
Let
and
be tangents to
. Then
Proof
vladimir.shelomovskii@gmail.com, vvsss
See also
2021 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |