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2020 IMO Problems/Problem 1: Difference between revisions

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<math>\textbf{Problem 1}</math>. 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:
== 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 ==


Line 8: Line 9:


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

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 $\overline{AB}$.


Video solution

https://youtu.be/rWoA3wnXyP8

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 $AP,BP$ meet at point $O$, those two lines 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 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 $O$ be the circumcenter of $\triangle ABP, \angle PAD = \alpha, OE$ is the perpendicular bisector of $AP,$ and point $E$ lies on $AD.$ Then

\[\angle APE = \alpha, \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies\] $\hspace{33mm} ABPE$ is cyclic. \[\angle PED = 2\alpha = \angle DPE \implies\] the bisector of the $\angle ADP$ is the perpendicular bisector of the side $EP$ of the cyclic $ABPE$ that passes through the center $O.$

A similar reasoning can be done for $OF,$ the perpendicular bisector of $BP.$

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
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