- Dear Hyacinthists,

-------------------------------------------------

[RG]:> A `proof without words' in the current Math Mag

[ND]:

> 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.

> More precisely

[RG]:

> 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.

> I think you're getting closer, and I believe that

[FvL]:

> the answer is the incentre. My main question is

> still: is this well known?

>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,

Milorad R.Stevanovic

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

[Non-text portions of this message have been removed] - From R.K.Guy:-

> Dear Hyacinthists,

There are only a finite number of such lines, so

> -------------------------------------------------

> [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.

they have no envelope.

John Conway