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Oksutok   0
20 minutes ago
1.(Świerczkowski's Theorem) A tetrahedral chain is a sequence of regular tetrahedra where any two consecutive tetrahedra are glued together face to face. Show that there is no closed tetrahedral chain.
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Oksutok
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cute geo
Royal_mhyasd   4
N an hour ago by Royal_mhyasd
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
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Royal_mhyasd
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Royal_mhyasd
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Trivial fun Equilateral
ItzsleepyXD   5
N May 1, 2025 by reni_wee
Source: Own , Mock Thailand Mathematic Olympiad P1
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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ItzsleepyXD
Apr 30, 2025
reni_wee
May 1, 2025
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Trivial fun Equilateral
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Source: Own , Mock Thailand Mathematic Olympiad P1
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ItzsleepyXD
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ItzsleepyXD
#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_
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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_
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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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Tsikaloudakis
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Tsikaloudakis
#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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reni_wee
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reni_wee
#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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