Difference between revisions of "2020 IMO Problems/Problem 1"
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| − | + | == Problem == | |
| − | <cmath>\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC</cmath> | + | Consider the convex quadrilateral <math>ABCD</math>. The point <math>P</math> is in the interior of <math>ABCD</math>. The following ratio equalities hold: |
| − | Prove that the following three lines meet in a point: the internal bisectors of angles <math>\angle ADP</math> and | + | <cmath>\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC.</cmath> Prove that the following three lines meet in a point: the internal bisectors of angles <math>\angle ADP</math> and <math>\angle PCB</math> and the perpendicular bisector of segment <math>\overline{AB}</math>. |
| − | <math>\angle PCB</math> and the perpendicular bisector of segment <math>AB</math>. | + | |
== Video solution == | == Video solution == | ||
Revision as of 11:23, 14 May 2021
Problem
Consider the convex quadrilateral 
. The point 
is in the interior of . The following ratio equalities hold:
![\[\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC.\]](http://latex.artofproblemsolving.com/2/6/e/26e0fa91015e07f3f022d7b879a86c21313be9f1.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
Video solution
https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems]
