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

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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> | 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>. | + | 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>. | + | 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. | ||

+ | <cmath>QED</cmath> |

## Revision as of 05:31, 23 July 2021

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

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.