Difference between revisions of "2023 IMO Problems/Problem 2"
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<math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math> | <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 | + | points <math>B, G,</math> and <math>F</math> are collinear <math>\implies GF</math> is symmetric to <math>AF</math> with respect <math>TF.</math> |
We use the lemma and complete the proof. | We use the lemma and complete the proof. | ||
+ | |||
+ | <i><b>Lemma 1</b></i> | ||
+ | [[File:2023 IMO 2b Lemma.png|300px|right]] | ||
+ | Let acute triangle <math>\triangle ABC, AB > AC</math> be given. | ||
+ | |||
+ | Let <math>H</math> be the orthocenter of <math>\triangle ABC, BHD</math> be the height. | ||
+ | |||
+ | Let <math>\Omega</math> be the circle <math>BCD. BC</math> is the diameter of <math>\Omega.</math> | ||
+ | |||
+ | The point <math>E</math> is symmetric to <math>D</math> with respect to <math>AH.</math> | ||
+ | The line <math>BE</math> meets <math>\Omega</math> again at <math>F \neq B</math>. | ||
+ | |||
+ | Prove that <math>HF = HD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\omega</math> be the circle centered at <math>H</math> with radius <math>HD.</math> | ||
+ | The <math>\omega</math> meets <math>\Omega</math> again at <math>F' \neq D, HD = HF'.</math> | ||
+ | Let <math>\omega</math> meets <math>BF'</math> again at <math>E' \neq F'</math>. | ||
+ | We use Reim’s theorem for <math>\omega, \Omega</math> and lines <math>CDD</math> and <math>BE'F'</math> and get <math>E'D || BC</math> (the idea was recommended by Leonid Shatunov). | ||
+ | <math>AH \perp BC \implies AH \perp E'D \implies</math> | ||
+ | The point <math>E'</math> is symmetric to <math>D</math> with respect to <math>AH \implies E' = E \implies F' = F \implies HF = HD.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Solutions== | ==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 10:01, 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 collinear
is symmetric to
with respect
We use the lemma and complete the proof.
Lemma 1
Let acute triangle be given.
Let be the orthocenter of
be the height.
Let be the circle
is the diameter of
The point is symmetric to
with respect to
The line
meets
again at
.
Prove that
Proof
Let be the circle centered at
with radius
The
meets
again at
Let
meets
again at
.
We use Reim’s theorem for
and lines
and
and get
(the idea was recommended by Leonid Shatunov).
The point
is symmetric to
with respect to
vladimir.shelomovskii@gmail.com, vvsss
Solutions
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]