- Dear François,

I am not sure the following solution is more elementary than yours, but let me try.

Let O be a point outside the plane ABC. Then the oriented areas of triangles XYZ in

the plane are proportional to the oriented volumes of tetrahedrons OXYZ, or to the

mixed products of vectors <OX,OY,OZ>. Denote vector OX simply by X. Then we have to

prove that

<A',B',C'> (=(1/27)<A+2a,B+2b,C+2c>) = 0.

For the 27*parenthesis, we have:

<A,B,C>+2(<A,B,c>+...)+4(<A,b,c>+<a,B,c>+<a,b,C>)+8<a,b,c>= (as is seen directly from

the figure)

=<A,B,C>+2*(0+0+0)+4(<a,b,c> - <A,B,C>)+8<a,b,c>=

= 3*(4<a,b,c> - <A,B,C>)

and we are done.

Actually, we see that, in general, area A'B'C' = (4 area abc - area ABC)/9.

This is readily generalized: if A', B', C' are defined by AA'=kA'a, ..., then

area A'B'C' = (k^2*area abc - (k-1)*area ABC)/(k+1)^2 and

A',B',C' are collinear iff area abc = ((k-1)/k^2)*area ABC.

Thanks for a nice problem,

Vladimir

----- Original Message -----

From: "Francois Rideau" <francois.rideau@...>

To: <Hyacinthos@yahoogroups.com>

Sent: Sunday, January 08, 2006 2:27 AM

Subject: [EMHL] Re: Affine triangles

Sorry, I correct the little typo in the last sentence:

Given a triangle ABC in the plane and abc an inscribed triangle. Till here

that sounds OK.

We suppose we have the following relation between signed areas:

S(a,b,c) = (1/4) S(A,B,C)

We cut the 3 segments Aa, Bb, Cc in three equal parts so that:

aA' = A'A" = A"A; bB' = B'B" = B"B; cC' = C'C" = C" C

Then the points A', B', C' are on a same line.

Given point "a" on line BC and point "b" on line CA, we have then an affine

construction of the point "c" on line AB so that:

S(a,b,c) = (1/4) S(A,B,C)

Funny this frenzied way of cutting segments in three to cut areas in four.

Friendly

François

Here my solution:

1° I look at he affine map f sending A on a, B on b and C on c.

2° I compute the 3x3 matrix of f in the affine frame {A,B,C}.

3°I compute the characteristic polynomial of Vec(f), the associated linear

map of f.

4°This polynomial is: X^2 + X +1/4 = (X + 1/2)^2

5°I look at the affine map: g = (1/3)(2.f + id). Then Vec(g), the linear

associated map of g, has {0,0} for spectrum and rank 1 and we are done.

May be somebody can give me a more elementary solution? - Dear Vladimir

Thanks with your beautiful solution.

But why using an extra-space?

Keeping your notations and calling <M,N,P> the signed area of the triple

(M,N,P), it's just enough to see that the map:

(M,N,P) --> <M,N,P> is tri-affine alternate just as a determinant function

is multilinear alternate.

The main problem is how to define the signed area without leaving the plane!

Here the way I prefer:

We suppose we have an orientation of the plane, in fact an orientation in

the euclidian vector space associated to the euclidian affine space.

If h and k are two vectors, I note [h,k] the mixed product of these 2

vectors, that is the 2x2 determinant of their components in a direct

orthonormal basis, it's easy to see that this definition don't depend on the

choice of the direct orthonormal basis.

Then for any point O in the plane, I define <M,N,P> by the formula:

<M,N,P> = (1/2)( [Vec(OM),Vec(ON)] + [Vec(ON),Vec(OP)] + [Vec(OP), Vec(OM)]

)

First, it's easy to see that the expression in the second member don't

depend on point O and next that the map (M,N,P) --> <M,N,P> is tri-affine

alternate.

That's the way Hadamard defines signed area in his famous book of

Geometry.In fact he don't speak about vector space and mixed product but the

idea was behind.

If f is the affine map sending M on M', N on N', P on P' and calling g =

Vec(f) the linear map associated to the affine map f, we have:

1°<M',N',P'> = det(g) <M,N,P>

2) <M',N,P> + <M,N',P> + <M,N,P'> = Trace(g) <M,N,P>

It's also easy to define the signed area of an n-gon by:

<A_1, ...,A_n> = (1/2)( [Vec(OA_1),Vec(OA_2] + [Vec(OA_2,Vec(OA_3)] + ... +

[Vec(OA_n),Vec(OA_1)])

and so one.

Reading anglo-saxon books on Geometry, I note they don't like using signed

objects like signed angle of lines or vectors or signed area. I don't

understand why!

In France we do but in fact the situation is more tragic for our teachers

"in general" know nothing in Geometry. In their cursus, emphasis is put on

linear and polynomial algebras so here we get teachers computing very well

the Jordan decomposition of a matrix and for the best the Galois group of a

polynomial but unable to use signed angles not to speak of the signed area

unknown by the crowd (inconnu au bataillon) or to compose two direct

similarity maps in the plane. That's why my fellow countrymen are just able

reading road-signposts that they come near a roundabout when they are not

too drunk!

What is the situation near the Neva river?

Going back to my affine congiguration, we note that 1 is not a root of the

characteristic polynomial: X^2 + X + 1/4 of Vec(f) so we know that f has an

unique fixed point O and as Vec(f) is not diagonalizable then f has an

unique invariant line. It's easy to see that this invariant line is just the

A'B'C' line and O is somewhere on it!

I leave you find a way to construct O.

For those who are interested, I can send a Cabri picture or a scanned image

of this picture as well.

Friendly

François

[Non-text portions of this message have been removed] - On 1/8/06, Francois Rideau <francois.rideau@...> wrote:
>

Dear Vladimir

>

> 1°<M',N',P'> = det(g) <M,N,P>

> 2) <M',N,P> + <M,N',P> + <M,N,P'> = Trace(g) <M,N,P>

> It's also easy to define the signed area of an n-gon by:

> <A_1, ...,A_n> = (1/2)( [Vec(OA_1),Vec(OA_2] + [Vec(OA_2,Vec(OA_3)] + ...

> + [Vec(OA_n),Vec(OA_1)])

> and so one.

>

Again I saw an interesting little typo in my last mail.

Of course , we must read:

2°<M',N,P> + <M,N',P> + <M,N,P'> = (Trace(g) +1) <M,N,P>

I just mixed up with the same trace formula in vector space. Why that extra

" + 1 ", you would tell me? It's just the "hidden" extra-space! Here we call

it the vector extension of the affine space (defined up to isomorphism).

Your solution uses a dropped version of that extra-space. But no matter the

way we define it, the result is the same!

If f is an affine map with g as associated vector map, f has an unique

vector extension h in the vector extension of the affine space and calling

P_h and P_g the characteristic polynomials of g and h, we have the formula:

P_h = (X-1)P_g.

In particular, we have:

1°Det(h) = Det(g)

2°Trace(h) = Trace(g) +1

That explains my formulas in the affine plane. The relation between minimum

polynomials of g and h is more elaborate!

Maybe that explains too why there is no longer geometry teaching in France!

The Jordan decomposition of the vector map h is translated in an "affine

decomposition" of the affine map f. But here we are so "bourbakist", we

prefer computing Jordan decomposition in abstracto just to keep us happy,

forgetting the geometrical applications and they begin in dimension 2 as we

saw it in this little problem.

Friendly

François

[Non-text portions of this message have been removed]