Sorry my chuff-chuffs are causing all this fuss overseas!

As for me, I try it but I can't remember one single french soldier and

though it was 62 years ago, I still have a vivid memory of tall and strong

GIs giving me a lot of candies and incidentaly chasing the Nazis out off the

landscape!

To come back with Triangle Geometry, I give here a sketch of the proof of

the problem:

> As surprising as it is, even Hamilton Cayley theorem has numerous

Barycentrics of point "a" wrt ABC are: (0, 1-u , u)

> applications in Triangle Geometry.

> Here a problem in Triangle Geometry for Hyacinthos rookies proving what I

> told:

> In the real affine plane a triangle ABC is given.

> We look at an inscribed triangle abc, point "a" on line BC, point "b" on

> line CA, point "c" on line AB, such that signed areas S(A,B,C) and S(a,b,c)

> are equal.

> Let f be the affine map defined by f(A) = a; f(B) = b; f(C) = c.

> 1° Prove that f is of order 3, that is: f^3 = id, has a unique fixed

> point but no invariant lines.

>

Barycentrics of point "b" wrt ABC are: (v, 0 , 1-v)

Barycentrics of point "a" wrt ABC are: (1-w , w , 0)

They are columns of a matrix M called matrix of the affine map f in the

affine basis {A, B, C}.

Let g be the associated linear map of the affine map f.

Char(M) and Char(g) the characteristic polynomials of M and g

Then we have:

Char(M) = (X-1).Char(g)

As consequence, we have:

Trace(M) = 1 + Trace(g) and Det(M) = Det(g).

But as matrix M as only 0 on its main diagonal, we have Trace(g) = -1

and from S(A,B,C) = S(a,b,c), we get Det(g) = 1.

So Char(g) = X^2 + X + 1 and Char(M) = X^3 - 1.

From Hamilton-Cayley theorem, we get M^3 = I (3x3 identity matrix) and so

f^3 = id.

As 1 is not an eigenvalue of g, f has a single fixed point O and as Char(g)

has no real roots(i.e g is not diagonalizable), f has no invariant lines.

Note that f generates a cyclic group G(f) of order 3, hence O is the

centroid of the orbit of any point under action of G(f).

2° Given point "a" on line BC and point "b" on line CA, find an

affineconstruction of point "c" on line AB such that:> S(A,B,C) = S(a,b,c).

I only give here an affine construction of point "c" without proof to let

>

you folks thinking about it.

I would be happy to read some proofs of yours!

We complete parallelograms aCAa' and bCBb'.

Line a'b cuts line BC on U and line b'a cuts line CA on V. Then line UV cuts

line AB in the sought point "c".

Friendly

François

PS

I give me a 3-week break to visit cherries blossoming in Tokyo and look at

some sungakus around here!

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