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Trivial fun Equilateral

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Source: Own , Mock Thailand Mathematic Olympiad P1

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#1 h Apr 30, 2025, 9:05 AM

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Let $ABC$ be a scalene triangle with point $P$ and $Q$ on the plane such that $\triangle BPC , \triangle CQB$ is an equilateral . Let $AB$ intersect $CP$ and $CQ$ at $X$ and $Z$ respectively and $AC$ intersect $BP$ and $BQ$ at $Y$ and $W$ respectively .
Prove that $XY\parallel ZW$

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moony_

#2 h Apr 30, 2025, 11:06 AM • 1 Y

Y by Retemoeg

PQ, XW, YZ are concurrent cuz pappus theorem for points X, B, Z and Y, C, W
=> Desargues theorem for triangles XPY and WQZ and YP || ZQ, XP || WQ => ZW || XY too

cool problem >w<

This post has been edited 1 time. Last edited by moony_, Apr 30, 2025, 2:52 PM

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moony_

#3 h Apr 30, 2025, 11:08 AM

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generalization btw: BPCQ - parallelogram

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#4 h Apr 30, 2025, 12:07 PM

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apothikefsi

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cursed_tangent1434

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cursed_tangent1434

#5 h May 1, 2025, 6:01 AM • 1 Y

Y by reni_wee

Alternative solution via trigonometry. We WLOG consider the case where $ABC$ is obtuse with $AC > AB> AC$ to avoid configurational confusion. Now, note that by the Law of Sines on $\triangle AWB$ ,
$\frac{AW}{AB} = \frac{\sin \angle ABW}{\sin \angle AWB} = \frac{\sin (120-B)}{ \sin (60-C)}$ Similarly,
$\frac{AZ}{AC} = \frac{\sin (60+C)}{\sin (B-60)}$ $\frac{AY}{AB} = \frac{\sin (B-60)}{\sin (C+60)}$ and
$\frac{AX}{AC} = \frac{\sin (60-C)}{\sin (120-B)}$ Now it is easy to see that,
$\frac{AW}{AZ} = \frac{\sin (120-B)}{ \sin (60-C)} \cdot \frac{\sin (B-60)}{\sin (60+C)} \cdot \frac{AB}{AC}$ and
$\frac{AY}{AX} = \frac{\sin (B-60)}{\sin (C+60)} \cdot \frac{\sin (120-B)}{\sin (60-C)} \cdot \frac{AB}{AC}$ from which we may conclude that
$\frac{AW}{AZ} = \frac{AY}{AX}$ implying that $XY \parallel WZ$ as desired.

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#6 h May 1, 2025, 5:50 PM • 1 Y

Y by cursed_tangent1434

Easy Problem. Barely MOHS 5.
WLOG Let $AC > AB$ . As $\triangle BPC$ and $\triangle BQC$ are equilateral, $BP \parallel QC, PC \parallel BQ \implies \triangle AYB \sim \triangle ACZ, \triangle XAC \sim \triangle BAW$ .
Let us handle these 2 separately. We get 2 equations as follows,
$\frac{AB}{AZ} = \frac {AY}{AC} \implies AZ \cdot AY = AB \cdot AC$ $\frac{AW}{AC} = \frac {AB}{AX} \implies AW \cdot AX = AB \cdot AC$ $\implies AW \cdot AX =AZ \cdot AY \implies \triangle XAY \sim \triangle ZAW$
From which we can get that $XY \parallel WZ$

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This post has been edited 1 time. Last edited by reni_wee, May 1, 2025, 7:55 PM

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