Dear Floor,

very good problem.

I had not noticed the generalization.

Let the projections of P on AB, AC be D, E

AF be the altitude of ABC and

AF' be the altitude of AED.

then the line AP is the diameter of

AEPD and is isogonal conjugate with AF'

and AO is isogonal conjugate with AF.

Hence for angles

(AO,AF') = (AP, AF) or

(AO,ED) = (AP,BC) = f

Hence for areas

(ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =

= R.ED.sinf = 2.(1/2).OA.ED.sinf = (AEOD)

Best regards

Nikos Dergiades

[FvL]

> I wrote:

> > Here is a little anniversary-observation:

> > H_A be the perpendicular foot of A on BC. Let X_B and

> X_C be the perpendicular feet of H_A on AC and AB (yes, well

> known for the Taylor circle). The polygons A X_B O X_C and B

> X_C O X_B C have equal area (O: circumcenter).

>

> I hope that everyone noted that H_A can be replaced by any

> point on BC. So for any point P on BC the quad formed by O,

> A and the perpendicular feet of P on AC and AB has area half

> of triangle ABC.

>

> Kind regards,

> Floor.

>

>

>

> ------------------------------------

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>

>

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