# Difference between revisions of "2021 IMO Problems/Problem 4"

(Created page with "<math>Problem:</math> 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,...") |
|||

(10 intermediate revisions by 2 users not shown) | |||

Line 1: | Line 1: | ||

− | + | ==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== | ||

+ | |||

+ | 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>. | ||

+ | 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>. | ||

+ | 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 |

## Revision as of 00:56, 2 August 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://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