Difference between revisions of "2023 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
+ | [[File:2023 IMO 2o0.png|400px|right]] | ||
+ | Denote the point diametrically opposite to a point <math>S</math> through <math>S' \implies AS'</math> is the internal angle bisector of <math>\angle BAC</math>. | ||
+ | |||
+ | Denote the crosspoint of <math>BS</math> and <math>AS'</math> through <math>H, \angle ABS = \varphi.</math> | ||
+ | |||
+ | <cmath>AE \perp BC, SS' \perp BC \implies \overset{\Large\frown} {AS} = \overset{\Large\frown} {ES'} = 2\varphi \implies</cmath> | ||
+ | |||
+ | <cmath>\angle EAS' = \varphi = \angle ABS \implies \angle DAH = \angle ABH \implies</cmath> | ||
+ | <cmath>\triangle AHD \sim \triangle BAH \implies \frac {AH}{BH} = \frac {DH}{AH} \implies AH^2 = BH \cdot DH.</cmath> | ||
+ | To finishing the solution we need only to prove that <math>PH = AH.</math> | ||
+ | |||
+ | Denote <math>F = SS' \cap AC \implies \angle CBS = \frac {\overset{\Large\frown} {CS}}{2} = \frac {\overset{\Large\frown} {BS}}{2} = \frac {\overset{\Large\frown} {AB}}{2} + \frac {\overset{\Large\frown} {AS}}{2} =</math> | ||
+ | <math>=\angle FCB + \varphi \implies \angle FBS = \angle ABC \implies H</math> is incenter of <math>\triangle ABF.</math> | ||
+ | |||
+ | Denote <math>T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H</math> is the orthocenter of <math>\triangle TSS'.</math> | ||
+ | |||
+ | Denote <math>G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,</math> and <math>G</math> are concyclic. | ||
+ | |||
+ | <math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math> | ||
+ | |||
+ | points <math>B, G,</math> and <math>F</math> are colinear <math>\implies GF</math> is symmetric to <math>AF</math> with respect <math>TF.</math> | ||
+ | |||
+ | We use the lemma and complete the proof. | ||
+ | |||
+ | ==Solutions== | ||
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems] | https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems] |
Revision as of 09:53, 23 July 2023
Problem
Let be an acute-angled triangle with
. Let
be the circumcircle of
. Let
be the midpoint of the arc
of
containing
. The perpendicular from
to
meets
at
and meets
again at
. The line through
parallel to
meets line
at
. Denote the circumcircle of triangle
by
. Let
meet
again at
. Prove that the line tangent to
at
meets line
on the internal angle bisector of
.
Solution
Denote the point diametrically opposite to a point through
is the internal angle bisector of
.
Denote the crosspoint of and
through
To finishing the solution we need only to prove that
Denote
is incenter of
Denote is the orthocenter of
Denote and
are concyclic.
points and
are colinear
is symmetric to
with respect
We use the lemma and complete the proof.
Solutions
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]