SEGOVIA POINT Continued

The antipodal triangle A'B'C' of O (= circumcevian of O) is the reflection of ABC in O. The triangle A2B2C2 is the reflection of the orthic A1B1C1 in N. Consider now the triangle A'2B'2C'2 = the reflection of A2B2C2 in O.

The triangles A'B'C', A2B2C2 are perspective at SEGOVIA Point of ABC.

Similarly the triangles ABC, A'2B'2C'2 are perspective at the SEGOVIA point of A'B'C'.

Denote O1O2O3 = the pedal triangle of O (=medial triangle).

We will work in an acute triangle ABC (similarly if ABC is not acute. Simply we have to change some signs).

ABC, A'2B'2C'2 are perspective <==>

[cotB + cot(O1BA'2)] / [cotC + cot(O1CA'2] * Cyclically = 1

We have:

cot(O1BA'2) = O1B / O1A'2 = (BC/2)/[OA'2 - OO1] = (BC/2)/[HA1 - OO1] =

= sinA /(2cosBcosC - cosA)

Therefore:

[cotB + cot(O1BA'2)] = (cosB/sinB) - [sinA /(2cosBcosC - cosA)] =

cosC[1 + 2cos^2B] / sinB(2cosBcosC - cosA)

and

[cotB + cot(O1BA'2)] / [cotC + cot(O1CA'2] = (cosC/cosB)*(sinC/sinB)*[(1+2cos^2B)/(1+2cos^2C)] (*)

and similarly the other two ratia.

Multiplicating them we get 1, therefore the triangles are perspective.

As for the coordinates of the perspector:

We get it in barycentrics from (*) and are:

(cosA * sinA * (1/(1+2cos^2A)) ::)

These are the barycentrics of the Segovia Point of the circumcevian triangle of O (if I did not make some computational error!)