- Dear Martin and all Hyacinthians,

> Construct the triangle ABC if given in position, the

An idea :

> line on which the base BC is located,and the feet of

> the Cevians of O to the two lateral sides.

Let d = BC.

Let Ob and Oc be the feet of the Cevians of O to AC and AB.

Let Mb be the orthogonal projection of Ob on d.

Let Mc be the orthogonal projection of Oc on d.

For a point P, let d_P be the line passing through P and parallel to d.

M = MbOc /\ McOc lies on the A-median of ABC.

Thus AM and d_A are harmonic conjugate wrt AB = AOc and AC = AOb.

The locus of such points A is an hyperbola H.

Let M' be the orthogonal projection of M on d.

Let O' = ObOc /\ d and M" on MO' such that MM" is parallel to d.

Ob, Oc, M, M', M" lies on H.

Thus we can construct H (not necessary).

For A on H, let B = AOc /\ d and C = AOb /\ d, O1 = BOb /\ COc, O2 the

circumcenter of ABC.

The locus of O1 is the rectangular hyperbola H1 passing through Ob, Oc and M',

with center the midpoint between Ob and Oc,

with asymptots respectively parallel and perpendicular to d.

Thus we can construct H1.

The locus of O2 is a cubic K passing through M'.

Now the problem is : constuct K and its intersections with H1

So we obtain 6 possible points O (only 5 : for O = M' the triangle ABC

degenerates to M')

Friendly

Best regards

Gilles - Dear Martin and all Hyacinthians

> [ML]

Let d1 be the asymptot of H which is perpendicular to d.

>

> > Construct the triangle ABC if given in position, the

> > line on which the base BC is located,and the feet of

> > the Cevians of O to the two lateral sides.

>

> [GB]

> Let d = BC.

> Let Ob and Oc be the feet of the Cevians of O to AC and AB.

>

> Let Mb be the orthogonal projection of Ob on d.

> Let Mc be the orthogonal projection of Oc on d.

>

> For a point P, let d_P be the line passing through P and parallel to d.

>

> M = MbOc /\ McOc lies on the A-median of ABC.

>

> Thus AM and d_A are harmonic conjugate wrt AB = AOc and AC = AOb.

>

> The locus of such points A is an hyperbola H.

>

> Let M' be the orthogonal projection of M on d.

> Let O' = ObOc /\ d and M" on MO' such that MM" is parallel to d.

>

> Ob, Oc, M, M', M" lies on H.

>

> Thus we can construct H (not necessary).

>

> For A on H, let B = AOc /\ d and C = AOb /\ d, O1 = BOb /\ COc, O2 the

> circumcenter of ABC.

>

> The locus of O1 is the rectangular hyperbola H1 passing through Ob, Oc and M',

> with center the midpoint between Ob and Oc,

> with asymptots respectively parallel and perpendicular to d.

>

> Thus we can construct H1.

> The locus of O2 is a cubic K passing through M'.

>

> Now the problem is : constuct K and its intersections with H1

> So we obtain 6 possible points O (only 5 : for O = M' the triangle ABC

> degenerates to M')

Let P be the point at infinity on d1.

Then P is a node on K and the line of infinity is tangent in P at K.

So we know 3 common points of H1 and K : M' and P double.

They have 3 other commons points, 2 of those points can be imaginary.

The problem has one or three solutions.

Friendly

Gilles >

[GB]

> > [ML]

> >

> > > Construct the triangle ABC if given in position, the

> > > line on which the base BC is located,and the feet of

> > > the Cevians of O to the two lateral sides.

> > Let d = BC.

to d.

> > Let Ob and Oc be the feet of the Cevians of O to AC and AB.

> >

> > Let Mb be the orthogonal projection of Ob on d.

> > Let Mc be the orthogonal projection of Oc on d.

> >

> > For a point P, let d_P be the line passing through P and parallel

> >

O2 the

> > M = MbOc /\ McOc lies on the A-median of ABC.

> >

> > Thus AM and d_A are harmonic conjugate wrt AB = AOc and AC = AOb.

> >

> > The locus of such points A is an hyperbola H.

> >

> > Let M' be the orthogonal projection of M on d.

> > Let O' = ObOc /\ d and M" on MO' such that MM" is parallel to d.

> >

> > Ob, Oc, M, M', M" lies on H.

> >

> > Thus we can construct H (not necessary).

> >

> > For A on H, let B = AOc /\ d and C = AOb /\ d, O1 = BOb /\ COc,

> > circumcenter of ABC.

Ob, Oc and M',

> >

> > The locus of O1 is the rectangular hyperbola H1 passing through

> > with center the midpoint between Ob and Oc,

triangle ABC

> > with asymptots respectively parallel and perpendicular to d.

> >

> > Thus we can construct H1.

> > The locus of O2 is a cubic K passing through M'.

> >

> > Now the problem is : constuct K and its intersections with H1

> > So we obtain 6 possible points O (only 5 : for O = M' the

> > degenerates to M')

imaginary.

>

> Let d1 be the asymptot of H which is perpendicular to d.

> Let P be the point at infinity on d1.

> Then P is a node on K and the line of infinity is tangent in P at K.

>

> So we know 3 common points of H1 and K : M' and P double.

>

> They have 3 other commons points, 2 of those points can be

>

Thank you,Gilles.

> The problem has one or three solutions.

First I solved similar problem with G and H instead of O,but had no

idea that this case is so difficult.

Best regards,

Martin- An other solution :

A1B1C1is the given first Brocard triangle of the unknown triangle ABC.

Construct first the the circumcircle (O1) and the centroid G of A1B1C1,

it is the centroid of ABC.

The line GA1 intersects the circle (O1) at A2. Construct similarly B2

and C2.

So A2B2C2 is the second Brocard triangle of ABC.

Let Bc be the reflection of B2 wrt the tangent to (O1) at C1.

Let Cb be the reflection of C2 wrt the tangent to (O1) at B1.

The circumcircles of A2BcC1 and A2B1Cb are intersecting at A2 and at the

vertex A of the triangle, we have to construct.

Best regards,

Gilles Boutte

>[DK]

>

>Johnson, page 280:

>

>===

>The solution depends onthe fact that any triangle is inversively similar to

>its first Brocard triangle.

>We locate the Brocard points of the given triangle, also its first Brocard

>triangle; then by similarity we locate on the circumcircle of the given

>triangle the Brocard points of the required triangle.

>The vertices of the latter can then be found at once.

>===

>

>Is this what you did?

>

>Regards,

>Dick Klingens

>

>[AD]

>

>Dear all

>Given the first Brocard triangle of triangle ABC,how to construct

>the triangle ABC ?

>

>My solution has become unnecessarily long.Will anybody tell me a precise

>solution of it?

>

>Yours faithfully

>Atul.A.Dixit

> - Dear Dick and Gilles

Thank you.Actually my solution is somewhat similar(though not efficient) as

that given by Boutte Gilles.Actually,I didn't thought of using the second

Brocard triangle.

Yours faithfully

Atul.A.Dixit

>From: "Dick Klingens" <dklingens@...>

_________________________________________________________________

>Reply-To: Hyacinthos@yahoogroups.com

>To: <Hyacinthos@yahoogroups.com>,<atul_dixie@...>

>Subject: RE: [EMHL] Construction problem

>Date: Sat, 31 Aug 2002 19:00:02 +0200

>

>Johnson, page 280:

>

>===

>The solution depends onthe fact that any triangle is inversively similar to

>its first Brocard triangle.

>We locate the Brocard points of the given triangle, also its first Brocard

>triangle; then by similarity we locate on the circumcircle of the given

>triangle the Brocard points of the required triangle.

>The vertices of the latter can then be found at once.

>===

>

>Is this what you did?

>

>Regards,

>Dick Klingens

>

>::-----Original Message-----

>::From: Atul Dixit [mailto:atul_dixie@...]

>::Sent: Saturday, August 31, 2002 6:42 PM

>::To: Hyacinthos@yahoogroups.com

>::Subject: [EMHL] Construction problem

>::

>::

>::Dear all

>:: Given the first Brocard triangle of triangle ABC,how to construct

>::the triangle ABC ?

>::

>::My solution has become unnecessarily long.Will anybody tell me a precise

>::solution of it?

>::

>::Yours faithfully

>::Atul.A.Dixit

>::

>::

>::

>::_________________________________________________________________

>::Send and receive Hotmail on your mobile device: http://mobile.msn.com

>::

>::

>::

>::

>::

>::Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

>::

>::

>

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