Difference between revisions of "2023 IMO Problems/Problem 2"
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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. | 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. | ||
− | + | [[File:2023 IMO 2 lemma.png|300px|right]] | |
<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> | ||
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<i><b>Lemma 1</b></i> | <i><b>Lemma 1</b></i> | ||
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[[File:2023 IMO 2b Lemma.png|300px|right]] | [[File:2023 IMO 2b Lemma.png|300px|right]] | ||
Let acute triangle <math>\triangle ABC, AB > AC</math> be given. | Let acute triangle <math>\triangle ABC, AB > AC</math> be given. | ||
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The point <math>E</math> is symmetric to <math>D</math> with respect to <math>AH.</math> | The point <math>E</math> is symmetric to <math>D</math> with respect to <math>AH.</math> | ||
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The line <math>BE</math> meets <math>\Omega</math> again at <math>F \neq B</math>. | The line <math>BE</math> meets <math>\Omega</math> again at <math>F \neq B</math>. | ||
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Let <math>\omega</math> be the circle centered at <math>H</math> with radius <math>HD.</math> | Let <math>\omega</math> be the circle centered at <math>H</math> with radius <math>HD.</math> | ||
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The <math>\omega</math> meets <math>\Omega</math> again at <math>F' \neq D, HD = HF'.</math> | The <math>\omega</math> meets <math>\Omega</math> again at <math>F' \neq D, HD = HF'.</math> | ||
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Let <math>\omega</math> meets <math>BF'</math> again at <math>E' \neq F'</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> ( | + | |
+ | 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> | ||
+ | |||
+ | (this idea was recommended by <i><b>Leonid Shatunov</b></i>). | ||
+ | |||
<math>AH \perp BC \implies AH \perp E'D \implies</math> | <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> | 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> | ||
Revision as of 10:09, 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
(this 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]