- Dear Antreas,

[APH](19845): Given triangle ABC and its circumcircle (O). Construct the circle (Oa) which is tangent to [the segment] BC and tangent to (O) at (A). Let A' be the intersection of the tangents of (Oa) through B,C.

*** Similarly define B' and C'.

The triangle A'B'C' is perspective with ABC at X(31) = (a^3 : b^3 : c^3) in homogeneous barycentric coordinates.

Best regards

Sincerely

Paul - Dear Paul

We can construct triads of circles (Oa),(Ob),(Oc),

each circle of whose is tangent to one (only) side of triangle,

and construct similarly the intersections of tangens A',B',C'

and ask if ABC and A'B'C' are perspective.

Here is another example.

Let HaHbHc be the orthic triangle, and DaDbDc the circumcevian

tiangle of H. The circle (Da,DaHa) is tangent to BC at Ha.

From B,C we draw tangents to the circle (Da,DaHa) intersecting

at A'. Similarly B',C'.

Are the triangles ABC, A'B'C' perspective?

Antreas

PS1: We can take the circles (A,AHa), (B,BHb), (C,CHc)

(tangent to BC,CA,AB at Ha,Hb,Hc, resp.)

PS2: A generalization is this:

Let P be a point, PaPbPc the pedal (or cevian) triangle of P,

(Oa) the circle centered on the circumcircle (O) and tangent

the BC. Similarly (Ob),(Oc). All three circles are centered

on either the positive or the negative sides of BC,CA,AB, resp.

(so we have two cases).

Let A' be the intersection of the tangents to (Oa) from B,C.

Similarly B',C'. Which is the locus of P such that ABC, A'B'C'

are perspective?

APH

--- In Hyacinthos@yahoogroups.com, Paul Y Yiu <yiu@...> wrote:

>

> Dear Antreas,

>

>

> [APH](19845): Given triangle ABC and its circumcircle (O). Construct the circle (Oa) which is tangent to [the segment] BC and tangent to (O) at (A). Let A' be the intersection of the tangents of (Oa) through B,C.

>

> *** Similarly define B' and C'.

>

> The triangle A'B'C' is perspective with ABC at X(31) = (a^3 : b^3 : c^3) in homogeneous barycentric coordinates.

>

> Best regards

> Sincerely

> Paul

> - [APH]
> Let HaHbHc be the orthic triangle, and DaDbDc the circumcevian

I think they are not.

> tiangle of H. The circle (Da,DaHa) is tangent to BC at Ha.

> From B,C we draw tangents to the circle (Da,DaHa) intersecting

> at A'. Similarly B',C'.

>

> Are the triangles ABC, A'B'C' perspective?

If A" is the reflection of A' in BC and A* is the

intersection of AA" and BC, and similarly B*,C*,

then A*,B*,C* are collinear.

And since Da,Db,Dc are the reflections of H in Ha,Hb,Hc, resp.

the problem in simpler form is:

Let HaHbHc be the orthic triangle. The tangents from B,C to the

circle (H,HHa) intersect at Aa. The lines AAa and BC intersect

at A*. Similarly B*,C*.

The points A*,B*,C* are collinear.

[APH]> PS1: We can take the circles (A,AHa), (B,BHb), (C,CHc)

They are perspective at O

> (tangent to BC,CA,AB at Ha,Hb,Hc, resp.)

Antreas