## Re: [EMHL] Generalized Prasolov point again

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• Dear Darij, and all Hyacinthists [DG] ... exciting ... but I hope ... If P* is the isogonal conjugate of P then it is known that the perpendicular from A to
Message 1 of 1 , Feb 1, 2003
Dear Darij, and all Hyacinthists

[DG]
>In message #6325, Nikolaos Dergiades gave the >following
exciting
>generalization of the Prasolov point:

>> By the way, the above property can be generalized:
>> If P[x : y : z] in barycentrics is a point and A1B1C1
>> is its pedal triangle, A2B2C2 is the reflection of
>> A1B1C1 in the circumcenter of A1B1C1 then the
>> triangles ABC, A2B2C2 are perspective.
>> Perspector is the point P' with barycentrics
>> [1/(x²(A+B)(A+C)+xyC(A+B)+xzB(A+C)-yzA(B+C)): cyclic..]
>> where A = Sa = (b²+c²-a²)/2, B = Sb C = Sc
>> If P=H then P'= X68
>> If P=O then P' = H
>> If P = G then P' I think is not in ETC.

>This is a very nice theorem (thanks very much to >Nikolaos,
but I hope
>somebody can prove it synthetically).

If P* is the isogonal conjugate of P then it is
known that the perpendicular from A to B1C1
passes through P* and the circumcenter of A1B1C1
is the mid-point of PP*.
If x, y, z are the distances of P and
x', y', z' are the distances of P* from BC, CA, AB
respectively then x x' = y y' = z z'.
If PPa is the distance of P from B1C1 then from
the quadrilateral AC1PB1 where AP is the diameter
of its circumircle we have yz = PPa * AP. (1)
If P1 is the reflection of Pa in O then P1 is on the
line AP* (P*P1 = PPa) and on the line B2C2.
Since AP/y = AP*/z' or z' *AP/y = AP*
we get from (1) PPa * AP* = zz' and similarly
(AP*)(P*P1) = (BP*)(P*P2) = (CP*)(P*P3)
This means from the orthogonal collinearity theorem
of the previous Hyacinthos message,
since B2C2 is perpendicular to AP* at P1 etc
that the intersections of the lines
(BC, B2C2), (CA, C2A2), (AB, A2B2) are collinear
and that the triangles ABC, A2B2C2 are perspective.

Best regards