Difference between revisions of "2021 IMO Problems/Problem 4"
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− | ==Problem== | + | == Problem == |
Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of | Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of | ||
Line 7: | 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] | ||
− | ==Solution== | + | https://www.youtube.com/watch?v=U95v_xD5fJk |
+ | |||
+ | https://youtu.be/WkdlmduOnRE | ||
+ | |||
+ | == Solution 1 == | ||
Let <math>O</math> be the centre of <math>\Omega</math>. | Let <math>O</math> be the centre of <math>\Omega</math>. | ||
Line 26: | Line 32: | ||
~BUMSTAKA | ~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 |