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

Line 15: | Line 15: | ||

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. | 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>. | ||

+ | Similarly we get <cmath>\angle IXZ = \angle ICT</cmath> and so the chords <math>IZ</math> and <math>IT</math> are equal. Hence <math>Z</math> is the reflection of <math>T</math> about <math>OI</math>. | ||

+ | This gives us <math>YZ</math> = <math>TX</math> immediately and now it suffices to prove, <cmath>AD + DT + XA = CD + DY + ZC</cmath>. |

## Revision as of 06:08, 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 . Similarly we get and so the chords and are equal. Hence is the reflection of about . This gives us = immediately and now it suffices to prove, .