## Re: [EMHL] Re: Affine triangles

Expand Messages
• 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
Message 1 of 3 , Jan 7, 2006
• 0 Attachment
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,

----- 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 the signed area of the triple
Message 2 of 3 , Jan 7, 2006
• 0 Attachment
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
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]
• ... Dear Vladimir Again I saw an interesting little typo in my last mail. Of course , we must read: 2° +
Message 3 of 3 , Jan 8, 2006
• 0 Attachment
On 1/8/06, Francois Rideau <francois.rideau@...> wrote:
>
>
> 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]
Your message has been successfully submitted and would be delivered to recipients shortly.