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

Revision as of 14:08, 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)

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 circumcircle center of $\triangle ABP, \angle PAD = \alpha, OE$ is the perpendicular bisector of $AP,$ and point $E$ lies on side $AD.$ Then

$\angle APE = \alpha, \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies ABPE$ is cyclic. $\angle PED = 2\alpha = \angle DPE = \alpha \implies$ 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]

@@ Line 14: / Line 14: @@
 ~ bluesoul
+==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 <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
+<math>\angle APE = \alpha,  \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies ABPE</math> is cyclic.
+<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>
+A similar reasoning can be done for <math>OF,</math> the perpendicular bisector of <math>BP.</math>
+'''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru'''
 == Video solution ==

2020 IMO (Problems) • Resources
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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