- Dear Eisso

[Eisso]> I was aware of Wolstenholme's book, but I did not realize it was on-line

I have to leave now for a week; just before, here is a summary of my argument :

> (Google books has it too). I find #716 somewhat intriguing for

> historical reasons. Of course, the proof of the statement is trivial,

> provided you know about the nine-point conic and the property that all

> four circles through the mid-points of any three adjacent sides of a

> complete quadrangle pass through the center of the conic of minimal

> eccentricity (PROP I).

Suppose that ABCD is convex and consider an hyperbola h (center W)through A,B,C,D; then there exists an ellipse through A,B,C,D for which the infinite points of h are conjugate; more over, the center We of the ellipse is the reflection of W in the centroid g of A,B,C,D.

As the infinite points of h are conjugate wrt the ellipse, the eccentricity of the ellipse will be minimal when the hyperbola is rectangular.

This fits with your PROP I : as the 9-points circles of BCD pass through the center W0 of the rectangular hyperbola through A,B,C,D, using the reflection wrt g, we get immediately that the circle through the mid-points of AB,AC,AD pass through the reflection of W0 in g.

Friendly. Jean-Pierre - Dear friends,

I calculated a general formula for the eccentricity of a circumscribed conic expressed in barycentric coordinates.

Thanks to the direction Francois Rideau appointed through Mr Wolstenholmes method by using the real or imaginary asymptotes of the conic.

Let ABC be some triangle with extra points P(p,q,r) and U(u,v,w).

In this situation the general equation for the conic is:

qruwxy - prvwxy - qruvxz + pqvwxz + pruvyz - pquwyz = 0

Let e be the eccentricity of the conic.

Now e^2 = 2 X1/(X1-X2)

where:

X1 = Sqrt[a^4 QV RW + b^4 PU RW + c^4 PU QV

- b^2 c^2 PU (-PU+QV+RW)

- c^2 a^2 QV (+PU-QV+RW)

- a^2 b^2 RW (+PU+QV-RW)]

X2 = SA PU + SB QV + SC RW

and:

PU = pu(qw-rv),

QV = qv(ru-pw),

RW = rw(pv-qu).

I checked in a fixed reference situation.

Maybe it is helpful for someone.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, Francois Rideau <francois.rideau@...> wrote:

>

> Yes, there is a general formula for eccentricity, known for a long time

> (before 1886), for example in:

> Mathematical Problems by Mr Wolstenholme.

> The starting point is to look at the angle V of the (real or imaginary)

> asymtots for we have the formula:

> tan²(V) = 4(e²-1)/(e²-2)².

> And it is very easy to write down the equation of the pair of asymptots from

> the general equation of the conic.

> Now if a pair of lines is given in trilinears by the equation:

> Lx² +My² +Nz² +2Uyz+2Vzx+2Wxy = 0

> the angle V of these lines is given by:

> tan²(V) = -D/(R.E)

> where R is the radius of the circumcircle;

> E = L+M+N -2Ucos(A)-2Vcos(B)-2Wcos(C)

> and D is the following 4x4 determinant with lines:

> L(1) = (a, L, W, V)

> L(2) = (b, W, M, U)

> L(3) = (c, V, U, N)

> L(4) = (0, a, b, c)

> where a, b, c are the lengths of the sides BC, CA, AB

> Friendly

> Francois

>

> - Dear François et alii,

now that I have looked at your formula for the relation between

eccentricity and tan(V), I realize that my formula for finding the

minimal eccentricity of all conics circumscribing a quadrangle ABCD can

also be written in the form.

4*csc^2(V) = (cot(A) + cot(C))*(cot(B) + cot(D)),

where csc(V) is a (generally completely superfluous) Anglo-Saxon

notation for 1/sin(V) and V is your angle between the asymptotes of the

conic of minimal eccentricity. Alternatively, if A*B*C*D* denotes the

quadrilateral formed by the circumcenters of BCD, CDA etc., we have

sin^2(V)=area(ABCD)/area(A*B*C*D*).

Of course, this equation is only meaningful for non-convex, non

self-intersecting ABCD, but the formula is easily verified in Geogebra

for these.

Eisso

On 4/4/11 5:31 AM, Francois Rideau wrote:

> Yes, there is a general formula for eccentricity, known for a long time

> (before 1886), for example in:

> Mathematical Problems by Mr Wolstenholme.

> The starting point is to look at the angle V of the (real or imaginary)

> asymtots for we have the formula:

> tan²(V) = 4(e²-1)/(e²-2)².

> And it is very easy to write down the equation of the pair of asymptots from

> the general equation of the conic.

> Now if a pair of lines is given in trilinears by the equation:

> Lx² +My² +Nz² +2Uyz+2Vzx+2Wxy = 0

> the angle V of these lines is given by:

> tan²(V) = -D/(R.E)

> where R is the radius of the circumcircle;

> E = L+M+N -2Ucos(A)-2Vcos(B)-2Wcos(C)

> and D is the following 4x4 determinant with lines:

> L(1) = (a, L, W, V)

> L(2) = (b, W, M, U)

> L(3) = (c, V, U, N)

> L(4) = (0, a, b, c)

> where a, b, c are the lengths of the sides BC, CA, AB

> Friendly

> Francois

>

>

>

--

========================================

Eisso J. Atzema, Ph.D.

Department of Mathematics& Statistics

University of Maine

Orono, ME 04469

Tel.: (207) 581-3928 (office)

(207) 866-3871 (home)

Fax.: (207) 581-3902

E-mail: atzema@...

========================================