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Difference between revisions of "2020 IMO Problems/Problem 1"

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==Solution 2 (Three perpendicular bisectors)==
 
==Solution 2 (Three perpendicular bisectors)==
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[[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.
 
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 circumcircle center of <math>\triangle ABP, \angle PAD = \alpha, OE</math> is the perpendicular bisector of <math>AP,</math> and point <math>E</math> lies on side <math>AD.</math> Then
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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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<math>\angle APE = \alpha,  \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies ABPE</math> is cyclic.
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<cmath>\angle APE = \alpha,  \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies</cmath>
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<math>\angle PED = 2\alpha = \angle DPE = \alpha \implies</math> 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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<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>
  
 
A similar reasoning can be done for <math>OF,</math> the perpendicular bisector of <math>BP.</math>
 
A similar reasoning can be done for <math>OF,</math> the perpendicular bisector of <math>BP.</math>

Revision as of 13:15, 29 July 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

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, www.deoma–cmd.ru

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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