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• -- Let A B C be the orthic and A B C the medial triangles of ABC. Denote: Ab, Ac = the reflections of A in CC , BB , resp. Bc, Ba = the reflections of B in
Message 1 of 8 , Dec 31 2:18 PM
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--
Let A'B'C' be the orthic and A"B"C" the medial
triangles of ABC.

Denote:

Ab, Ac = the reflections of A" in CC', BB', resp.
Bc, Ba = the reflections of B" in AA', CC', resp.
Ca, Cb = the reflections of C" in BB', AA', resp.

CONJECTURE:

The Nine Point Circles of the triangles
A"AbAc, B"BcBa, C"CaCb concur.

HAPPY NEW YEAR 2004 !

Greetings from Athens

Antreas
• Dear Antreas ... In both cases, your assertions are true and the locus of the common point of the three NP-circles is the line joining H to X(143) = the
Message 1 of 8 , Jan 1, 2004
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Dear Antreas
> Let ABC be a triangle, and A'B'C' its Orthic Triangle.
>
> 1. Let Pa, Pb, Pc be points on AH, BH, CH, resp. such that:
>
> APa / AH = BPb / BH = CPc / CH = t
>
> Denote
>
> Ab, Ac = the reflections of Pa in CC', BB', resp.
> Bc, Ba = the reflections of Pb in AA', CC', resp.
> Ca, Cb = the reflections of Pc in BB', AA', resp.
>
> The Nine Point Circles of the Triangles
> PaAbAc, PbBcBa, PcCaCb are concurrent (??)
>
> Which is the locus of the point P of concurrence,
> as t varies?
>
> [If t = 0 (ie Pa = A, Pb = B, Pc = C), then P = X(1986) (BW)]
>
> 2. Let Pa, Pb, Pc be points on A'H, B'H, C'H, resp. such that:
>
> A'Pa / A'H = B'Pb / B'H = C'Pc / C'H = t'
>
> Denote
>
> Ab, Ac = the reflections of Pa in CC', BB', resp.
> Bc, Ba = the reflections of Pb in AA', CC', resp.
> Ca, Cb = the reflections of Pc in BB', AA', resp.
>
> The Nine Point Circles of the Triangles
> PaAbAc, PbBcBa, PcCaCb are concurrent (??)
>
> Which is the locus of the point P of concurrence,
> as t' varies?
>
> [If t' = 0 (ie Pa = A', Pb = B', Pc = C'), then P = X(1112) (JPE) ]

In both cases, your assertions are true and the locus of the common
point of the three NP-circles is the line joining H to X(143) = the
NP-center of the orthic triangle

Friendly. Jean-Pierre
• Dear Antreas ... They go through X(974) Happy and peaceful 2004 to every Hyacinthist Friendly. Jean-Pierre
Message 1 of 8 , Jan 1, 2004
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Dear Antreas
> Let A'B'C' be the orthic and A"B"C" the medial
> triangles of ABC.
>
> Denote:
>
> Ab, Ac = the reflections of A" in CC', BB', resp.
> Bc, Ba = the reflections of B" in AA', CC', resp.
> Ca, Cb = the reflections of C" in BB', AA', resp.
>
> CONJECTURE:
>
> The Nine Point Circles of the triangles
> A"AbAc, B"BcBa, C"CaCb concur.

They go through X(974)
Happy and peaceful 2004 to every Hyacinthist
Friendly. Jean-Pierre
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