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• 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
Message 1 of 8 , Mar 24, 2005
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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

> 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
Message 1 of 8 , Mar 25, 2005
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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
Message 1 of 8 , Mar 28, 2005
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Dear Hyacinthists,

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

>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
>
• Dear Luis, [LL] ... [ND] ... [LL] ... 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
Message 1 of 8 , Mar 30, 2005
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Dear Luis,

[LL]
>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.

[ND]
> >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).

[LL]
> 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.

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.

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
• 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
Message 1 of 8 , Mar 30, 2005
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Dear Hyacinthists,

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

>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