Difference between revisions of "2020 IMO Problems/Problem 1"
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− | Problem | + | == Problem == |
− | The following ratio equalities hold: | + | 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: |
− | + | <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>. | |
− | Prove that the following three lines meet in a point: the internal bisectors of angles | + | |
− | + | ==solution 1== | |
+ | |||
+ | Let the perpendicular bisector of <math>AP,BP</math> meet at point <math>O</math>, those two lined meet at <math>AD,BC</math> at <math>N,M</math> respectively. | ||
+ | |||
+ | As the problem states, denote that <math>\angle{PBC}=\alpha, \angle{BAP}=2\alpha, \angle {BPC}=3\alpha</math>. We can express another triple with <math>\beta</math> as well. Since the perpendicular line of <math>BP</math> meets <math>BC</math> at point <math>M</math>, <math>BM=MP, \angle {BPM}=\alpha, \angle {PMC}=2\alpha</math>, which means that points <math>A,P,M,B</math> are concyclic since <math>\angle{PAB}=\angle{PMC}</math> | ||
+ | |||
+ | Similarly, points <math>A,N,P,B</math> are concyclic as well, which means five points <math>A,N,P,M,B</math> are concyclic., <math>ON=OP=OM</math> | ||
+ | |||
+ | Moreover, since <math>\angle{CPM}=\angle{CMP}</math>, <math>CP=CM</math> so the angle bisector if the angle <math>MCP</math> must be the perpendicular line of <math>MP</math>, so as the angle bisector of <math>\angle{ADP}</math>, which means those three lines must be concurrent at the circumcenter of the circle containing five points <math>A,N,P,M,B</math> as desired | ||
+ | |||
+ | ~ bluesoul | ||
+ | |||
== Video solution == | == Video solution == | ||
https://youtu.be/rWoA3wnXyP8 | https://youtu.be/rWoA3wnXyP8 | ||
+ | |||
+ | https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems] | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2020|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 13:29, 25 February 2022
Contents
Problem
Consider the convex quadrilateral . The point is in the interior of . The following ratio equalities hold: Prove that the following three lines meet in a point: the internal bisectors of angles and and the perpendicular bisector of segment .
solution 1
Let the perpendicular bisector of meet at point , those two lined meet at at respectively.
As the problem states, denote that . We can express another triple with as well. Since the perpendicular line of meets at point , , which means that points are concyclic since
Similarly, points are concyclic as well, which means five points are concyclic.,
Moreover, since , so the angle bisector if the angle must be the perpendicular line of , so as the angle bisector of , which means those three lines must be concurrent at the circumcenter of the circle containing five points as desired
~ bluesoul
Video solution
https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems]
See Also
2020 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |