Install the new FirefoxDear Geometers,
Let ABC be a triangle.
Let B and side AC be the focus and directrix of a parabola.
Let C and side AB be the focus and directrix of a parabola.
Let A and side BC be the focus and directrix of a parabola.
Then, the six points where the three parabola intersect the
sides of the triangle lie on a conic.
Best regards,
Emmanuel Antonio José García
- Dear Emmanuel,
[EAJG]: Let ABC be a triangle.
Let B and side AC be the focus and directrix of a parabola.
Let C and side AB be the focus and directrix of a parabola.
Let A and side BC be the focus and directrix of a parabola.
Then, the six points where the three parabola intersect the sides of the triangle lie on a conic.
*** Actually there are two conics.
Let S denote TWICE the area of ABC.
The three parabolas intersect the sides of ABC at
(0:ab:S), (0:S:ca), (S:0:bc), (ab:0:S), (ca:S:0), and (S:bc:0)
and their harmonic conjugates on the respective sides.
These 6 points lie on the conic
(ca+S)(ab+S)yz + (ab+S)(bc+S)zx + (bc+S)(ca+S)xy - S(x+y+z)(bcx+cay+abz)=0.
The 6 harmonic conjugates lie on another conic
(ca-S)(ab-S)yz + (ab-S)(bc-S)zx + (bc-S)(ca-S)xy + S(x+y+z)(bcx+cay+abz)=0.
Best regards
Sincerely
Paul Yiu
Best regards,
Antonio José García
I found some generalizatuiins.
You don't need parabolas, since you can start with three conics of any three eccentricities. The points (ab:0:eS) and (ca:eS:0) lie on the conic with focus A, directrix BC, and eccentricity e.
Instead of switching all six of the points to their respective harmonic conjugates, you can switch any two (or any four) of those first six points, giving six points that also lie on a conic. So there are 32 different conics here.
--
Barry Wolk
---In advancedplanegeometry@yahoogroups.com, <yiu@...> wrote:
Dear Emmanuel,[EAJG]: Let ABC be a triangle.
Let B and side AC be the focus and directrix of a parabola.
Let C and side AB be the focus and directrix of a parabola.
Let A and side BC be the focus and directrix of a parabola.
Then, the six points where the three parabola intersect the sides of the triangle lie on a conic.
*** Actually there are two conics.
Let S denote TWICE the area of ABC.
The three parabolas intersect the sides of ABC at
(0:ab:S), (0:S:ca), (S:0:bc), (ab:0:S), (ca:S:0), and (S:bc:0)
and their harmonic conjugates on the respective sides.
These 6 points lie on the conic
(ca+S)(ab+S)yz + (ab+S)(bc+S)zx + (bc+S)(ca+S)xy - S(x+y+z)(bcx+cay+abz)=0.
The 6 harmonic conjugates lie on another conic
(ca-S)(ab-S)yz + (ab-S)(bc-S)zx + (bc-S)(ca-S)xy + S(x+y+z)(bcx+cay+abz)=0.
Best regards
Sincerely
Paul Yiu
Best regards,
Antonio José García