- Dear Luis,

A construction for the first problem is the following:

Construct an angle xAy as the given and on the bisector of this

angle the point I such that the distance from Ay is IK = r.

The circle (I, IK) meets the segment AI at L and the parallel from

L to Ay meets IK at N.

On Ay we take the point D such that ID = m_a

and on the extention of AI we take the segment IE = KD.

If F is the orthogonal projection of E on the line LN

the line IF meets the circle (I, IN) at two points X, X'

Take on AE the point H such that AH = FX (FX > FX')

The perpendicular to AH at H meets the circle (A, m_a)

at M the mid point of BC and hence BC can be constructed

as tangent to the incircle (I, IK).

I think an analogus construction we can have for the other problem

but I have no time to think about it.

Best regards

Nikolaos Dergiades

> Dear Hyacinthists,

>

> Dear Eric,

>

> I have been looking for a "true" geometric construction of

> both problems A(\alpha) , m_a, r and A(\alpha), m_a, r_a,

> where m_a is the median from A.

>

> One can then construct A, I and draw the incircle. And

> similarly, A, I_a and draw the excircle. Now would it be

> possible to construct the point G ?

>

> Best regards,

> Luis

> - There another riddles, affine ones just to forget Euclid

during some time!

Let A_{0}, A_[1}, A_{2}, 3 points in an affine ( possibly euclidian

for Euclid lovers) real plane not in the same line.

Let A_{3} another point acting as a parameter.

Let f the affine map defined by:

f(A_{0}) = A_{1}; f(A_{1}) = A_{2}; f(A_{2}) = A_{3}

We also define recursively the sequence { n --> A_{n} } by:

A_{n+1} = f(A_{n}).

Answer the few following questions and you can imagine

scores of other ones:

1° What are the sets of A_{3} points so that

f has just one fixed point O and examine the map

A_{3} --> O.

2° What is the locus of A_{3} so that f has no fixed points?

3° What is the set of A_{3} so that f has a line of fixed points?

4° What is the locus of A_{3} so that f is parallelogic?

5° If the plane is euclidian, (cheer Euclid!), what is the locus of

A_3 so that f is orthologic and in that case what is the locus of the

fixed point of f?

6°What is the set of A_{3} so that

the sequence { n --> A{n} } is bounded?

7° What is the set of A_{3} so that

the sequence { n --> A_{n} } converges (to the fixed point of f) ?

8° What is the set of A_{3} so that

the sequence { n --> A_{n} } is periodic.

9° What is the set of A_{3} so that

the sequence { n --> A_{n} } has accumulation points?

10° What is the locus of A_{3} so that the range

of the sequence { n --> A_{n} } is on a conic?

Separate on that locus the elliptic, hyperbolic and

parabolic cases.

And so one....

I hope you will find these riddles interesting.

I have all the answers if you want!

Friendly yours

François - Dear Hyacinthists,

Dear Nikolaos Dergiades,

Thank you very much for this solution. It is exactly this

kind of solution I was looking for.

Unfortunately I wasn't able to justify it. Could you please

provide the general ideas leading to this elegant solution?

Maybe afterwards I would be able to understand it and mimic

it to solve the second (with r_a) problem.

Best regards,

Luis

>From: "Nikolaos Dergiades" <ndergiades@...>

>Reply-To: Hyacinthos@yahoogroups.com

>To: <Hyacinthos@yahoogroups.com>

>Subject: RE: [EMHL] Re: AIG construction

>Date: Thu, 24 Mar 2005 21:38:57 +0200

>

>Dear Luis,

>A construction for the first problem is the following:

>Construct an angle xAy as the given and on the bisector of this

>angle the point I such that the distance from Ay is IK = r.

>The circle (I, IK) meets the segment AI at L and the parallel from

>L to Ay meets IK at N.

>On Ay we take the point D such that ID = m_a

>and on the extention of AI we take the segment IE = KD.

>If F is the orthogonal projection of E on the line LN

>the line IF meets the circle (I, IN) at two points X, X'

>Take on AE the point H such that AH = FX (FX > FX')

>The perpendicular to AH at H meets the circle (A, m_a)

>at M the mid point of BC and hence BC can be constructed

>as tangent to the incircle (I, IK).

>

>I think an analogus construction we can have for the other problem

>but I have no time to think about it.

>

>Best regards

>Nikolaos Dergiades

> - Dear Luis,

[LL]>I have been looking for a "true" geometric construction of

[ND]

>both problems A(\alpha) , m_a, r and A(\alpha), m_a, r_a,

>where m_a is the median from A.

> >A construction for the first problem is the following:

[LL]

> >Construct an angle xAy as the given and on the bisector of this

> >angle the point I such that the distance from Ay is IK = r.

> >The circle (I, IK) meets the segment AI at L and the parallel from

> >L to Ay meets IK at N.

> >On Ay we take the point D such that ID = m_a

> >and on the extention of AI we take the segment IE = KD.

> >If F is the orthogonal projection of E on the line LN

> >the line IF meets the circle (I, IN) at two points X, X'

> >Take on AE the point H such that AH = FX (FX > FX')

> >The perpendicular to AH at H meets the circle (A, m_a)

> >at M the mid point of BC and hence BC can be constructed

> >as tangent to the incircle (I, IK).

> Thank you very much for this solution. It is exactly this

I worked as follows:

> kind of solution I was looking for.

>

> Unfortunately I wasn't able to justify it. Could you please

> provide the general ideas leading to this elegant solution?

>

> Maybe afterwards I would be able to understand it and mimic

> it to solve the second (with r_a) problem.

In Cartesian coordinates we consider the points

A(0,0), I(d,0), M(g,h).

Let S(2g,2h) be the reflection of A in M.

The tangent of the angle IAB is t = r/sqrt(d^2-r^2).

the equation of the line AB is y= tx

the equation of the line AC is y= -tx

The line SC: y-2h = t(x-2g) is parallel to AB

Hence solving the system of SC, AC we find

the point C(g-h/t,h-tg). Similarly we find the point

B(g+h/t,h+tg) and then the equation of the line

BC: t^2gx-hy+h^2-t^2g^2=0

From the formula of the distance r of I from BC

substituting t and h from h^2=m^2-g^2 where m = AM

we get the equation

(dg)^2=m^2(d^2-r^2) which is not true

or the equation

dg(dg-2r^2)=(d^2-r^2)(m^2-r^2) or

g(g-2r^2/d)=(1-(r/d)^2)(m^2-r^2) or

g(g-2IN)=cos^2(A/2)(m^2-r^2) or

g(g-XX')=(cos(A/2)IE)^2 or

g(g-XX')=NF^2 or g = FX = AH

Since g is the first coordinate of M the

perpendicular to AH at H meets the circle (A, m)

at M the mid point of BC and hence BC can be constructed

as tangent to the circle (I, IK).

This proof and construction is the same

if r is the inradius but we must have d<m or

if r is the a_exradius but we must have d>m.

Best regards

Nikolaos Dergiades - Dear Hyacinthists,

Dear Nikolaos Dergiades,

Thank you for your reply. I will work through your

justification and the solution that followed.

Now an easier and nice construction problem.

Or rather two.

Construct ABC given (a, h_a, m_b\pm m_c) where

\pm = + -

Taken from Court.

Best regards,

Luis

>From: "Nikolaos Dergiades" <ndergiades@...>

[....]

>Reply-To: Hyacinthos@yahoogroups.com

>To: <Hyacinthos@yahoogroups.com>

>Subject: RE: [EMHL] alpha, m_a, r (r_a) construction [was AIG construction]

>Date: Wed, 30 Mar 2005 17:07:28 +0300

>

>

>Dear Luis,

>

> > Unfortunately I wasn't able to justify it. Could you please

> > provide the general ideas leading to this elegant solution?

> >

>

[....]

>I worked as follows:

>In Cartesian coordinates we consider the points

>A(0,0), I(d,0), M(g,h).

>Let S(2g,2h) be the reflection of A in M.

>Best regards

>Nikolaos Dergiades