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

{AC_{a}\cdot AC_{b} \over BC_{a}\cdot BC_{b}} \cdot {BA_{b}\cdot BA_{c} \over CA_{b}\cdot CA_{c}}\cdot {CB_{c}\cdot CB_{a }\over AB_{c}\cdot AB_{a}} = 1.

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

{AC_{s} \over BC_{s}} \cdot {BA_{s} \over CA_{s}} \cdot {CB_{s}\over AB_{s}} = 1

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.

\left | \matrix{1& -2s & s^{2}\\ 1& -c & 0 \\ 1 & -(a+b) & ab} \right | = 0

which is a quadratic equation for s

(a + b c)s^{2} -2abs + abc = 0

whose solutions are

s = {ab\pm\sqrt{aba'b'} \over (a + b c)}

and therefore

s' = {a'b'\mp \sqrt{aba'b'} \over (a + b c)}

and finally,

{s \over s'} = \pm \sqrt{ab \over a'b' }

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

{AC_{s}^{2} \over BC_{s}^{2}} = {AC_{a}\cdot AC_{b} \over BC_{a}\cdot BC_{b}}

Direct Proof

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

px^{2} + qy^{2} + rz^{2} + sxy + ... =0,

then p,q and s must satisfy

pa^{2} + qa'^{2} + saa'=0
pxb^{2} + qyb^{2} + sbb' =0

and, eliminating s, we obtain

ab(ab' -a'b)p + a'b'(a'b - ab')q = 0.


q = {ab \over a'b'}p = {AC_{a}\cdot AC_{b} \over BC_{a}\cdot BC_{b}}p

In the same way

r = {BA_{b}\cdot BA_{c} \over CA_{b}\cdot CA_{c}}q
p = {CB_{c}\cdot CB_{a} \over AB_{a}\cdot AB_{c}}q

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

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