## Re: [EMHL] Lines bisecting perimeter & area

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• Dear Hyacinthists, ... Related to the lines which bisects perimeter and area of triangle ABC I have the following result. Let be b c a. If (a+b+c)^2-8bc
Message 1 of 2 , Jul 7, 2002
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Dear Hyacinthists,
-------------------------------------------------
[RG]:
> A `proof without words' in the current Math Mag
> prompts the following question. Answer well known?
>
> What is the envelope of straight lines which
> simultaneously bisect the perimeter and the
> area of an arbitrary fixed traingle? R.

[ND]:
> More precisely
> 1) if a line meets the segments AB, AC and bisects the area
> of ABC then the envelope of this line is the hyperbola
> mentioned above.
> 2) if a line meets the segments AB, AC and bisects the
> perimeter of ABC then the envelope of this line is a conic.
> (I did not examined if parabola or hyperbola)
>
> Hence if a line simultaneously bisects the perimeter and the
> area of the triangle ABC then this line must be tangent to
> the above envelopes and therefor the total number of these
> lines perhaps is only 3.

[RG]:
> I think you're getting closer, and I believe that
> the answer is the incentre. My main question is
> still: is this well known?
[FvL]:
>Problem 1761 with solution from Nieuw Tijdschrift voor Wiskunde 67
>(1969-1970) 104-105 states that the lines l bisecting area and
>circumference simultaniously must pass through I and asks for
>construction of this line.

>This is precised in the article

>E.C. Bruissant des Amorie en W.A. van der Spek, Lijnen die de omtrek en
>oppervlakte van een driehoek halveren, NTvW 67 (1969-1970) 161-165

>where it is proved that, assuming a>=b>=c the number of lines that
>bisect area as well as circumference is 1,2 or 3 according to s^2 - 2ab
> <, = or > 0.

>Your question on the 'envelope' is thus not quite hitting the point.
------------------------------------------------
Related to the lines which bisects perimeter and area
of triangle ABC I have the following result.
Let be b>c>a.
If (a+b+c)^2-8bc<0 then we have only one line A1B1 as
solution with A1(0 : 1 :(a+c-b+sqrt((a+b+c)^2-8ab))/(2b)),
B1((c+b-a+sqrt((a+b+c)^2-8ab))/(2(b-c)): 0 : 1).
If (a+b+c)^2-8bc>0,3c>a+b,3a>b+c then we have three
lines A1B1(as in first case),B2C2 and B3C3 with
B2((a+b-c-sqrt((a+b+c)^2-8bc))/(2c) : 0 : 1),
C2(1 : (a+c-b-sqrt((a+b+c)^2-8bc))/(2(c-a)) : 0),
B3((a+b-c+sqrt((a+b+c)^2-8bc))/(2c) : 0 : 1),
C3(1 : (a+c-b+sqrt((a+b+c)^2-8bc))/(2(c-a)) : 0),
If(a+b+c)^2-8bc=0,b<2c then we have two lines with
B2=B3 and C2=C3.
All lines AkBk pass through incenter I.
When I found this solution I saw Floor's message
related to result of E.C. Bruissant des Amorie en
W.A. van der Spek.
Sorry,if this was unnecessary.

Kind regards,
Sincerely,

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• From R.K.Guy:- ... There are only a finite number of such lines, so they have no envelope. John Conway
Message 2 of 2 , Jul 8, 2002
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From R.K.Guy:-

> Dear Hyacinthists,
> -------------------------------------------------
> [RG]:
> > A `proof without words' in the current Math Mag
> > prompts the following question. Answer well known?
> >
> > What is the envelope of straight lines which
> > simultaneously bisect the perimeter and the
> > area of an arbitrary fixed traingle? R.

There are only a finite number of such lines, so
they have no envelope.

John Conway
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