2023 IMO Problems/Problem 2: Difference between revisions

Revision as of 00:57, 19 November 2023

Problem

Let $ABC$ be an acute-angled triangle with $AB < AC$ . Let $\Omega$ be the circumcircle of $ABC$ . Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$ . The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$ . The line through $D$ parallel to $BC$ meets line $BE$ at $L$ . Denote the circumcircle of triangle $BDL$ by $\omega$ . Let $\omega$ meet $\Omega$ again at $P \neq B$ . Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$ .

Video Solution

https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]

https://youtu.be/I4FoXeVnSpM?si=BMxwsH4thsZ-7uLF [Video contains IMO 2023 P2 motivation + discussion] (~little-fermat)

Solution

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Denote the point diametrically opposite to a point $S$ through $S' \implies AS'$ is the internal angle bisector of $\angle BAC$ .

Denote the crosspoint of $BS$ and $AS'$ through $H, \angle ABS = \varphi.$

$AE \perp BC, SS' \perp BC \implies \overset{\Large\frown} {AS} = \overset{\Large\frown} {ES'} = 2\varphi \implies$

$\angle EAS' = \varphi = \angle ABS \implies \angle DAH = \angle ABH \implies$ $\triangle AHD \sim \triangle BAH \implies \frac {AH}{BH} = \frac {DH}{AH} \implies AH^2 = BH \cdot DH.$ To finishing the solution we need only to prove that $PH = AH.$

Denote $F = SS' \cap AC \implies \angle CBS = \frac {\overset{\Large\frown} {CS}}{2} = \frac {\overset{\Large\frown} {BS}}{2} = \frac {\overset{\Large\frown} {AB}}{2} + \frac {\overset{\Large\frown} {AS}}{2} =$ $=\angle FCB + \varphi \implies \angle FBS = \angle ABS \implies H$ is incenter of $\triangle ABF.$

Denote $T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H$ is the orthocenter of $\triangle TSS'.$

Denote $G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,$ and $G$ are concyclic.

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$\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies$

points $B, G,$ and $F$ are collinear $\implies GF$ is symmetric to $AF$ with respect $TF.$

We use the lemma and complete the proof.

Lemma 1

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Let acute triangle $\triangle ABC, AB > AC$ be given.

Let $H$ be the orthocenter of $\triangle ABC, BHD$ be the height.

Let $\Omega$ be the circle $BCD. BC$ is the diameter of $\Omega.$

The point $E$ is symmetric to $D$ with respect to $AH.$

The line $BE$ meets $\Omega$ again at $F \neq B$ .

Prove that $HF = HD.$

Proof

Let $\omega$ be the circle centered at $H$ with radius $HD.$

The $\omega$ meets $\Omega$ again at $F' \neq D, HD = HF'.$

Let $\omega$ meets $BF'$ again at $E' \neq F'$ .

We use Reim’s theorem for $\omega, \Omega$ and lines $CDD$ and $BE'F'$ and get $E'D || BC$

(this idea was recommended by Leonid Shatunov).

$AH \perp BC \implies AH \perp E'D \implies$

The point $E'$ is symmetric to $D$ with respect to $AH \implies E' = E \implies F' = F \implies HF = HD.$

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 67: / Line 67: @@
 '''vladimir.shelomovskii@gmail.com, vvsss'''
+==See Also==
+{{IMO box|year=2023|num-b=1|num-a=3}}

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2023 IMO Problems/Problem 2: Difference between revisions

Revision as of 00:57, 19 November 2023

Contents

Problem

Video Solution

Solution

See Also