- jean-pierre.ehrmann wrote:
> Dear Antreas, John and other Hyacinthists

ABCP.

> [Antreas]

> > > Let ABC be a triangle and P a point on its circumcircle.

> > > It is well known

> > > that the four orthocenters Hi, i = 1,2,3,4, of the triangles

> > > ABP, PAC, CPB, CAB, resp.,form a quadrilateral congruent to

[snip]> If Ha is the orthocenter of PBC,... the segments AHa, BHb, CHc, PH

That's the locus Antreas asked for. The locus of the center of this

> have the same midpoint.

>

> Hence the congruence is the reflection w.r.t. this point and the

> locus of the the circumcenter of (Ha,Hb,Hc,H) is the circle with

> center H, radius R.

> Best regards. Jean-Pierre

reflection (as P varies on the circumcircle) is also interesting,

because it is a trivial result that the midpoint of PH lies on the

nine-point circle.

Barry Wolk <wolkb AT ccu.umanitoba.ca> [APH]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular

to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

[César Lozada]:

For A,B,C the required loci are the correspondent cevian of N, together with the line at infinity.

For P=N the Euler lines La, Lb, Lc concur at:

Z= (cos(2*A)-2)*cos(B-C)+cos(A) :: (trilinears)

[...]Variation:Let ABC be a triangle, P a point and B', C' the orthogonal projections of P on AC, AB, resp.The locus of P such that the Euler line of AB'C' is perpendicular to BC is the the a-cevian of the isogonal conjugate of N = X54Application:Let ABC be a triangle and A'B'C' the pedal triangle of N.Denote:A", B", C" = the reflections of A in B'C', C'A', A'B', resp.(AA", BB", CC" concur at X54 = isog. conjugate of N)Ab, Ac = the orthogonal projections of A" on AC, AB, resp.L1 = the Euler line of AAbAc. Similarly L2, L3.A*B*C* = the triangle bounded by L1, L2, L3.ABC, A*B*C* are parallelogic with parallelogic center (ABC, A*B*C*) = H of ABC (since L1, L2, L3 are perpendiculars to BC, CA, AB, resp.)APH