### Hubert Shutrick

For a triangle ABC, points C_{a} and C_{b} on AB, points A_{b} and A_{c} on BC and points B_{c} and B_{a} on CA lie on a conic if and only if

A proof of this is given in the article which was posted after I had produced an alternative criterion based on the theory of involutions since I was unaware of Carnot's result. Here is the proof that the two are equivalent.

The involution condition is that, if C_{s} on AB, A_{s} on BC and B_{s} on CA are the double points of the involutions given by the pairs (A,B) and (C_{a},C_{b}) on (A,B), (B,C) and (A_{b},A_{c}) on BC and (C,A) and (B_{c},B_{a}) on CA respectively, then the cevians AA_{s}, BB_{s} and CC_{s} should be concurrent thus

It will be enough to find C_{s} so let AC_{a}= a, AC_{s}= s, AC_{b}= b and AB= c for simplicity and even a' = c-a etc.

Then, s satisfies the quadratic equation x^{2} – 2sx + s^{2} = 0 while the pair (A,B) satisfy x^{2} – cx =0 and (C_{a},C_{b}) satisfy x^{2} – (a+b)x + ab . They are in involution if the equations are linearly dependent so the determinant of the coefficients should be zero.

which is a quadratic equation for s

whose solutions are

and therefore

and finally,

after simplification and similarly for the other two sides of the triangle. Hence

Gordon Walsh noticed that there was a simple proof of Carnot's theorem in the book Analytic Conics by D.M.Y.Sommerville using aerial coordinates so let's prove it that way.

With the above notation and the triangle as reference, the coordinates of C_{a} are (a,a',0) and those of C_{b} are (b,b',0) so, if we assume that a conic through them has equation

then p,q and s must satisfy

and, eliminating s, we obtain

Therefore,

In the same way

Stringing them together and cancelling p, gives Carnot's formula.

**Reference** D.M.Y.Sommerville, *Analytic Conics*, G.Bell and Sons Ltd., London 1946