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>. | ||
+ | |||
+ | ==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 | ||
+ | ~ 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''' | ||
== Video solution == | == Video solution == | ||
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https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems] | https://youtu.be/bDHtM1wijbY [Shorter solution, video covers all day 1 problems] | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2020|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 08:29, 28 September 2023
Contents
Problem
Consider the convex quadrilateral . The point
is in the interior of
. The following ratio equalities hold:
Prove that the following three lines meet in a point: the internal bisectors of angles
and
and the perpendicular bisector of segment
.
solution 1
Let the perpendicular bisector of meet at point
, those two lines meet at
at
respectively.
As the problem states, denote that . We can express another triple with
as well. Since the perpendicular line of
meets
at point
,
, which means that points
are concyclic since
Similarly, points are concyclic as well, which means five points
are concyclic.,
Moreover, since ,
so the angle bisector if the angle
must be the perpendicular line of
, so as the angle bisector of
, which means those three lines must be concurrent at the circumcenter of the circle containing five points
as desired
~ bluesoul ~ 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 be the circumcenter of
is the perpendicular bisector of
and point
lies on
Then
is cyclic.
the bisector of the
is the perpendicular bisector of the side
of the cyclic
that passes through the center
A similar reasoning can be done for the perpendicular bisector of
vladimir.shelomovskii@gmail.com, vvsss
Video solution
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 |