Dear Antreas and Bernard,

Thanks for another surprising perspector!

In Hyacinthos message #10184, Bernard Gibert

wrote:

>> > [APH] Let Ra, Rb, Rc be the orthogonal

>> > projections of H on the cevians of I, [ie

>> > on the angle bisectors of A,B,C, resp.]

>> > and SaSbSc the pedal triangle of I (aka

>> > intouch triangle of ABC).

>> >

>> > Are the triangles RaRbRc, SaSbSc perspective?

>>

>> They are indeed at : [...]

I think that the perspector of the triangles

RaRbRc and SaSbSc is the point X(847) of the

triangle SaSbSc (i. e. of the intouch triangle).

In fact, if we rewrite the problem from the

viewpoint of triangle SaSbSc, we get the

following:

Let ABC be a triangle. Further, let H' be the

orthocenter of the tangential triangle of

triangle ABC, and let D, E, F be the

orthogonal projections of the point H' on the

perpendicular bisectors of the segments BC,

CA, AB. Then show that the lines AD, BE, CF

are concurrent at the point X(847) of

triangle ABC.

In fact, by the definition of X(847), the

line AX(847) passes through the point

BAb /\ CAc, where the point Ab is the point

of intersection of the A-altitude of

triangle ABC with the perpendicular

bisector of the side CA, and the point Ac

is is the point of intersection of the

A-altitude with the perpendicular bisector

of the side AB. Thus, the lines

AD = AX(847), BAb and CAc are concurrent.

In fact, it turns out (conjecture!) that the

following four lines are concurrent:

AD, BAb, CAc and X(5)X(6)X(68)X(155).

I would be glad to see a proof of this (BTW,

I know the synthetic proof of the

collinearity of the points X(5), X(6), X(68),

X(155), so you can define the line

X(5)X(6)X(68)X(155) by any two of them).

Sincerely,

Darij Grinberg