Difference between revisions of "2021 IMO Problems/Problem 3"
(→Problem) |
(→Solution) |
||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
+ | [[File:2021 IMO 3.png|450px|right]] | ||
<i><b>Lemma</b></i> | <i><b>Lemma</b></i> | ||
Let <math>AK</math> be bisector of the triangle <math>ABC</math>, point <math>D</math> lies on <math>AK.</math> The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>. The point <math>E'</math> is symmetric to <math>E</math> with respect to <math>AK.</math> The point <math>L</math> on the segment <math>AK</math> satisfies <math>E'L||BC.</math> | Let <math>AK</math> be bisector of the triangle <math>ABC</math>, point <math>D</math> lies on <math>AK.</math> The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>. The point <math>E'</math> is symmetric to <math>E</math> with respect to <math>AK.</math> The point <math>L</math> on the segment <math>AK</math> satisfies <math>E'L||BC.</math> | ||
Then <math>EL</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A</math> and <cmath>\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.</cmath> | Then <math>EL</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A</math> and <cmath>\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.</cmath> | ||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Symmetry of points <math>E</math> and <math>E'</math> with respect bisector <math>AK</math> implies <math>\angle AEL = \angle AE'L.</math> | ||
+ | <cmath>\angle DCK = \angle E'DL, \angle DKC = \angle E'LD \implies \triangle DCK \sim \triangle E'DL \implies \frac {E'L}{KD}= \frac {DL}{KC}.</cmath> | ||
+ | <cmath>\triangle ALE' \sim \triangle AKB \implies \frac {E'L}{BK}= \frac {AL}{AK}\implies \frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.</cmath> | ||
==Video solution== | ==Video solution== | ||
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] |
Revision as of 20:43, 9 July 2022
Problem
Let be an interior point of the acute triangle
with
so that
. The point
on the segment
satisfies
, the point
on the segment
satisfies
, and the point
on the line
satisfies
. Let
and
be the circumcentres of the triangles
and
respectively. Prove that the lines
,
, and
are concurrent.
Solution
Lemma
Let be bisector of the triangle
, point
lies on
The point
on the segment
satisfies
. The point
is symmetric to
with respect to
The point
on the segment
satisfies
Then
and
are antiparallel with respect to the sides of an angle
and
Proof
Symmetry of points and
with respect bisector
implies
Video solution
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]