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 | + | |
− | + | ||
== Video solution == | == Video solution == | ||
https://youtu.be/rWoA3wnXyP8 | https://youtu.be/rWoA3wnXyP8 | ||
+ | |||
https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems] | https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems] | ||
+ | == Short Video solution(中文解说)in Chinese and subtitle in English == | ||
+ | https://youtu.be/WhJTaJjtjM8 | ||
+ | |||
+ | ==solution 1== | ||
+ | |||
+ | Let the perpendicular bisector of <math>AP,BP</math> meet at point <math>O</math>, those two lines 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 and "Shen Kislay kai" | ||
+ | ~ edits by Pearl2008 | ||
+ | |||
+ | ==Solution 2 (Three perpendicular bisectors)== | ||
+ | [[File:2020 IMO 1a.png|450px|right]] | ||
+ | The essence of the proof is the replacement of the bisectors of angles by the perpendicular bisectors of the sides of the cyclic pentagon. | ||
+ | |||
+ | Let <math>O</math> be the circumcenter of <math>\triangle ABP, \angle PAD = \alpha, OE</math> is the perpendicular bisector of <math>AP,</math> and point <math>E</math> lies on <math>AD.</math> Then | ||
+ | |||
+ | <cmath>\angle APE = \alpha, \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies</cmath> | ||
+ | <math>\hspace{33mm} ABPE</math> is cyclic. | ||
+ | <cmath>\angle PED = 2\alpha = \angle DPE \implies</cmath> | ||
+ | the bisector of the <math>\angle ADP</math> is the perpendicular bisector of the side <math>EP</math> of the cyclic <math>ABPE</math> that passes through the center <math>O.</math> | ||
+ | |||
+ | A similar reasoning can be done for <math>OF,</math> the perpendicular bisector of <math>BP.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2020|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 23:20, 13 November 2024
Contents
[hide]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 .
Video solution
https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems]
Short Video solution(中文解说)in Chinese and subtitle in English
solution 1
Let the perpendicular bisector of meet at point , those two lines 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 and "Shen Kislay kai" ~ edits by Pearl2008
Solution 2 (Three perpendicular bisectors)
The essence of the proof is the replacement of the bisectors of angles by the perpendicular bisectors of the sides of the cyclic pentagon.
Let be the circumcenter of is the perpendicular bisector of and point lies on Then
is cyclic. the bisector of the is the perpendicular bisector of the side of the cyclic that passes through the center
A similar reasoning can be done for the perpendicular bisector of
vladimir.shelomovskii@gmail.com, vvsss
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 |