## Re: Triangle construction problem

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• [APH]: [To construct triangle ABC if are given Lb,Lc : two of its sidelines and L : its OI line] ... What do you mean by same procedure ? APH
Message 1 of 17 , Sep 7, 2008
[APH]:
[To construct triangle ABC if are given Lb,Lc : two of
its sidelines and L : its OI line]
>> How about if L = OK line (Brocard axis) ??

[Giovanni Artico]:
> In this case, repeating the same procedure, I get a locus that is not
> a conic.
>
> Am I correct ?

What do you mean by "same procedure"?

APH
• ... I take a variable point on the line L as K, construct the triangle and the corresponding O, and trace the locus of O as K moves on the line L. b.t.w. it
Message 2 of 17 , Sep 7, 2008
--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
>
> What do you mean by "same procedure"?
>

I take a variable point on the line L as K, construct the triangle and
the corresponding O, and trace the locus of O as K moves on the line L.

b.t.w. it seems that also IG, IH, IGe generate a conic: can you check
this?

Best regards

Giovanni
• You can also take the point O (instead of K), draw the circle (O,OA), intersecting Lb,Lc at B,C, and ask for the locus of K as O moves on L = OK. The Genelal
Message 3 of 17 , Sep 8, 2008
You can also take the point O (instead of K), draw the circle
(O,OA), intersecting Lb,Lc at B,C, and ask for
the locus of K as O moves on L = OK.

The Genelal problem is:

Let ABC be a triangle with the sidelines AB = Lb, AC = Lc,
two fixed lines.

Let P be a point lying on a fixed line L.

As B,C move on Lb, Lc, and P move on L so that P = (x:y:z) with
respect ABC, which is the locus of a point Q = (x':y':z') with
respect ABC?

So we can assume that P = O,H,G,N,I, ..., L = PX (with X
a fixed triangle center other than P : O,H,G,N,I, ... )
and ask for the locus of Q = O,H,G,N,I, ...
(but other than P)

APH

--- In Hyacinthos@yahoogroups.com, "Giovanni Artico" <zeroprof@...> wrote:
>
> --- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@> wrote:
> >
> >
> > What do you mean by "same procedure"?
> >
>
> I take a variable point on the line L as K, construct the triangle and
> the corresponding O, and trace the locus of O as K moves on the line L.
>
> b.t.w. it seems that also IG, IH, IGe generate a conic: can you check
> this?
>
> Best regards
>
> Giovanni
>
• Dear Antreas ... I ll use the following : Suppose that the perpendicular at I to AI intersects the sidelines AB and AC at E and E , then the circle touching AB
Message 4 of 17 , Sep 8, 2008
Dear Antreas
> Let L,Lb,Lc be three lines.
> To construct triangle ABC such that Lb,Lc be two of its sidelines
> and L its OI line.

I'll use the following :
Suppose that the perpendicular at I to AI intersects the sidelines AB
and AC at E and E', then the circle touching AB at E and AC at E'
touches the circumcircle of ABC.

Your construction follows : draw the circle c_0 above; it remains to
find the circles touching c_0 and going through A and through the
reflection of A in L - is a very classical problem -; the circumcircle
of ABC is neccesarily such a circle (note that it is possible that the
circle centered at I is an excircle of ABC)
Friendly. Jean-Pierre
• ... This is a bit difficult for me. But playing with your problems I faced with another one: Given the sidelines Lb and Lc and a triangle center X(k), draw the
Message 5 of 17 , Sep 8, 2008
--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
> You can also take the point O (instead of K), draw the circle
> (O,OA), intersecting Lb,Lc at B,C, and ask for
> the locus of K as O moves on L = OK.
>
> The Genelal problem is:
>
> Let ABC be a triangle with the sidelines AB = Lb, AC = Lc,
> two fixed lines.
>
> Let P be a point lying on a fixed line L.
>
> As B,C move on Lb, Lc, and P move on L so that P = (x:y:z) with
> respect ABC, which is the locus of a point Q = (x':y':z') with
> respect ABC?
>
> So we can assume that P = O,H,G,N,I, ..., L = PX (with X
> a fixed triangle center other than P : O,H,G,N,I, ... )
> and ask for the locus of Q = O,H,G,N,I, ...
> (but other than P)
>
> APH

This is a bit difficult for me. But playing with your problems I
faced with another one:

Given the sidelines Lb and Lc and a triangle center X(k), draw the
triangle.

I was able to solve it for various k, but for k=7 (Gergonne point) I
have no solution: can anyone help ?

Best regards

Giovanni

P.S. I'm posting the problems of this type (those I succeeded solving)
in Geometriagon ( www.polarprof.org/geometriagon )
• Dear Giovanni ... A nice problem. We start from a point P and two lines Lb, Lc going through A. L1, L2 are the bisectors of (Lb,Lc) - L1 in the same sector
Message 6 of 17 , Sep 9, 2008
Dear Giovanni

> Given the sidelines Lb and Lc and a triangle center X(k), draw the
> triangle.
>
> I was able to solve it for various k, but for k=7 (Gergonne point) I
> have no solution: can anyone help ?

A nice problem.
We start from a point P and two lines Lb, Lc going through A.
L1, L2 are the bisectors of (Lb,Lc) - L1 in the same sector than P.
L' = harmonic conjugate of AP wrt (Lb,Lc)
L'' = harmonic conjugate of L2 wrt (AP, L')
The perpendicular at P to L1 intersects L'' at Q (this point must lie
on the line BC)
P1 = projection of P upon L1
The perpendicular at Q to AP intersects the perpendicular bisector of
PP1 at F.
For any common point M between the line AP and the circle with diameter
FQ, the line QM intersects Lb, Lc at B,C respectively.
If P is inside ABC, then P is the Gergonne point of ABC.
(This is due to the fact that the line BC must go through Q and touch
the parabola with focus F and with AP with tangent at the vertex)
Friendly. Jean-Pierre
• Dear Jean Pierre, many thanks for your solution. It seems to me like a miracle, as I haven t any idea about why it works so fine. I would be curious to know
Message 7 of 17 , Sep 9, 2008
Dear Jean Pierre, many thanks for your solution. It seems to me like a
miracle, as I haven't any idea about why it works so fine. I would be
curious to know the reasoning you followed, but I would not want to
annoy you with too many questions.
Now I'm blocked with the case k=9 (Mittenpukt), that has some relation
with Gergonne point: so I suspect that you have a solution also for
this case.

I think it would be nice if you could post an article in Forum
Geometricorum with the commented solutions to some of these problems.

Best wishes

Giovanni Artico
• Dear Giovanni! Maybe I m missing something, but your problem ... seems to have a very simple solution by the similarity method: Let rb and rc be the rays on
Message 8 of 17 , Sep 10, 2008
Dear Giovanni!

Maybe I'm missing something, but your problem

> Given the sidelines Lb and Lc and a triangle center X(k), draw the
> triangle...
> for k=7 (Gergonne point)

seems to have a very simple solution by the similarity method:

Let rb and rc be the rays on the given lines that form an angle
containing the given point P. We inscribe a circle in this angle, take
the intersection point T of the circle and AP farthest from A, and
draw a tangent to the circle through T. If the tangent meets both rays
at B' and C' and if P' is the Gergonne point of the triangle AB'C',
then the desired triangle is obtained from AB'C' by the dilation with
center A and ratio AP/AP'.

However, I'd also be very interested in Jean Pierre's article talking
about the ideas behind his solution.

Best regards,
• Dear Vladimir and Giovanni [Vladimir] ... take ... rays ... with ... This is actually very nice and very simple. I think that it is urgent to forget my
Message 9 of 17 , Sep 10, 2008
> Maybe I'm missing something, but your problem
>
> > Given the sidelines Lb and Lc and a triangle center X(k), draw the
> > triangle...
> > for k=7 (Gergonne point)
>
> seems to have a very simple solution by the similarity method:
>
> Let rb and rc be the rays on the given lines that form an angle
> containing the given point P. We inscribe a circle in this angle,
take
> the intersection point T of the circle and AP farthest from A, and
> draw a tangent to the circle through T. If the tangent meets both
rays
> at B' and C' and if P' is the Gergonne point of the triangle AB'C',
> then the desired triangle is obtained from AB'C' by the dilation
with
> center A and ratio AP/AP'.

This is actually very nice and very simple.
problem.
Friendly. Jean-Pierre
• Dear Vladimir, I ll never finish appreciate this list for the amount of surprising solutions it hosts. Your solution is very simple, and it works also for
Message 10 of 17 , Sep 10, 2008