Difference between revisions of "2020 IMO Problems/Problem 1"
(→solution 1) |
(→Solution 2 (Three perpendicular bisectors)) |
||
Line 16: | Line 16: | ||
==Solution 2 (Three perpendicular bisectors)== | ==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. | 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 | + | |
+ | 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> | 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
Contents
[hide]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 lined 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
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, www.deoma–cmd.ru
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 |