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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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== Video solution ==
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 +
https://youtu.be/rWoA3wnXyP8
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https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems]
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== Short Video solution（中文解说）in Chinese and subtitle in English  ==
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https://youtu.be/WhJTaJjtjM8
  
 
==solution 1==
 
==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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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>
 
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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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
 
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  
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~ bluesoul and "Shen Kislay kai"
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~ edits by Pearl2008
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==Solution 2 (Three perpendicular bisectors)==
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[[File:2020 IMO 1a.png|450px|right]]
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The essence of the proof is the replacement of the bisectors of angles by the perpendicular bisectors of the sides of the cyclic pentagon.
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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
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<cmath>\angle APE = \alpha,  \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies</cmath>
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<math>\hspace{33mm} ABPE</math> is cyclic.
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<cmath>\angle PED = 2\alpha = \angle DPE \implies</cmath>
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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>
  
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== Video solution ==
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A similar reasoning can be done for <math>OF,</math> the perpendicular bisector of <math>BP.</math>
  
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https://youtu.be/rWoA3wnXyP8
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'''vladimir.shelomovskii@gmail.com, vvsss'''
  
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https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems]
 
  
 
==See Also==
 
==See Also==

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)

2020 IMO 1a.png

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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