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Difference between revisions of "2020 IMO Problems/Problem 1"
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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:
 
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>.
 
<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>.
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==solution 1==
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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.
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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>
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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>
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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
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~ bluesoul
  
 
== Video solution ==
 
== Video solution ==

Revision as of 13:29, 25 February 2022

Problem

Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold: \[\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC.\] Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $\overline{AB}$.

solution 1

Let the perpendicular bisector of $AP,BP$ meet at point $O$, those two lined meet at $AD,BC$ at $N,M$ respectively.

As the problem states, denote that $\angle{PBC}=\alpha, \angle{BAP}=2\alpha, \angle {BPC}=3\alpha$. We can express another triple with $\beta$ as well. Since the perpendicular line of $BP$ meets $BC$ at point $M$, $BM=MP, \angle {BPM}=\alpha, \angle {PMC}=2\alpha$, which means that points $A,P,M,B$ are concyclic since $\angle{PAB}=\angle{PMC}$

Similarly, points $A,N,P,B$ are concyclic as well, which means five points $A,N,P,M,B$ are concyclic., $ON=OP=OM$

Moreover, since $\angle{CPM}=\angle{CMP}$, $CP=CM$ so the angle bisector if the angle $MCP$ must be the perpendicular line of $MP$, so as the angle bisector of $\angle{ADP}$, which means those three lines must be concurrent at the circumcenter of the circle containing five points $A,N,P,M,B$ as desired

~ bluesoul

Video solution

https://youtu.be/rWoA3wnXyP8

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
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