Difference between revisions of "2020 IMO Problems/Problem 1"
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==solution 1== | ==solution 1== | ||
| − | Let the perpendicular bisector of <math>AP,BP</math> meet at point <math>O</math>, those two | + | 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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~ bluesoul | ~ bluesoul | ||
| + | ~ edits by Pearl2008 | ||
==Solution 2 (Three perpendicular bisectors)== | ==Solution 2 (Three perpendicular bisectors)== | ||
Revision as of 07:29, 28 September 2023
Contents
Problem
Consider the convex quadrilateral 
. The point 
is in the interior of . The following ratio equalities hold:
![\[\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC.\]](http://latex.artofproblemsolving.com/2/6/e/26e0fa91015e07f3f022d7b879a86c21313be9f1.png)
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
![\[\angle APE = \alpha, \angle PEA = \pi - 2\alpha, \angle ABP = 2\alpha \implies\]](http://latex.artofproblemsolving.com/9/3/8/9386ebefa540f593e809b5ac78a95455323fbd53.png)
is cyclic.
![\[\angle PED = 2\alpha = \angle DPE \implies\]](http://latex.artofproblemsolving.com/1/0/0/1006714997f9b7928bd3a1209442319635576a61.png)
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 | ||
