Dear Bernard

I just check the MacKay drawing with Cabri.

Let S be any point on this cubic and let A'B'C' be the circumcevian triangle

of S wrt ABC. Then we know that ABC and A'B'C' are orthologic with orthology

centers U and U'.

Then U is conjugate isogonal with S as I said it and so also on the MacKay

cubic and U' is on the butterfly. I use the tool Trace to get the locus of

U' and I saw a beautiful butterfly flying high on the big blue screen.

Notice that the perspectrix of ABC and A'B'C' is the polar of S wrt the

circumcircle, hence orthogonal to OS , (where O is the circumcenter) as we

already know it , courtesy of Sondat for S, O, U, U' are on a same line.

Friendly

François

PS

For Cabri users, here the way to get the MacKay cubic.

1° Choose any point M on the circumcircle.

2° Draw line OM

3° Draw the rectangular hyperbola isogonal of line OM. To get it, it

suffices to construct its intersection N with the circumcircle as the

isogonal conjugate of the point at infinity of line OM. Then use the conic

tool to get the hyperbola through A, B, C, N, H where H is the orthocenter.

4° Draw the 2 intersections of the line OM with the hyperbola.

5° Draw their locus when M moves on the circumcircle with the locus tool.

After that, it is easy to check what I have said after choosing any point S

on the MacKay cubic and so one...

I wonder if GSP can do the same task? Please, overseas, let me know about

it!

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