[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
[obm-l] "german" triangle construction
Sauda,c~oes,
Alguns membros da lista sabem que escrevi
(e inclusive têm um exemplar) um livro em
francês sobre construção de triângulos.
Desde então fiz muitas descobertas que espero
colocar numa possível edição em português.
No momento, gostaria de mostrar a solução abaixo,
que não aparece no livro. Infelizmente o arquivo .jpg
que poderia acompanhá-la é muito grande. Posso
enviá-lo para o email pessoal dos interessados
O ponto importante da construção é o lugar geom (lg).
dado pela parábola.
Apreciaria se alguém pudesse dar uma outra prova
para o lg ( seria possível uma prova sintética?) e
também uma análise do lg e a figura feita (parábola)
por um programa de geometria.
Em palavras, o problema é:
Qual o lg de O quando B se move na reta r, tal que
BM_b = m_b, m_b (mediana) fixo?
A
-------|------------|--------------------------------------- r
B H_a
Obrigado.
[]'s
Luis
-----Mensagem Original-----
De: Barukh Ziv <zbaruh@yahoo.co.uk>
Para: Luis Lopes <llopes@ensrbr.com.br>
Enviada em: sexta-feira, 17 de maio de 2002 15:23
Assunto: Re: En: [EMHL] "german" triangle construction
> Dear Luis,
>
> Thank you very much for your last messages, the
> solution
> in German book and Sketchpad figure reprodicing
> Gilles's
> construction. Your notion about the Portugese book
> with
> construction of conics is very intriguing; may I ask
> you
> to scan a page or two for me (if time permits)?
>
> The solution in German book is along the same lines as
> two other solutions we discussed (a, m_a, s_a and m_a,
> s_a, R). In what follows I present my solution of the
> problem h_a, s_a, m_b. Please see the attached file,
> that contains four figures: two upper are the
> construction
> steps; two lower are for the justification. My aim is
> to
> explain it as thoroughly as my skills permit. Please
> remember that in my figures I use 'ta' for what you
> use s_a.
>
> The basic idea is to construct the circumcemter O of
> the sought triangle ABC. I use the method of intersection
> of loci. So, this approach is essentially the same as
> Gilles's, but the element constructed is different.
>
> The analysis goes as follows:
>
> 1. First, we know that O lies on a line which is
> reflection of AHa in ATa (since angles HaATa and TaAO are
> equal).
>
> 2. Second, assume the segment AHa is fixed; then the
> vertex B may move (on a line BC, obviously), so that the
> length of the median BMb is constant and equal m_b; what
> is the locus of circumcenters O of triangles ABC (one of
> such triangles is shown at the left lower figure
> attached)?
> It turns out that the locus is a parabola with the
> following parameters:
>
> 2a. Its axis is parallel to AHa.
> 2b. Its directrix d (perpendicular to AHa,
> according to 2a), is m_b^2/(4*h_a) away from Ha.
> 2c. Its focus F is 9/8*h_a away from d, and m/4
> away from AHa, where m is defined as follows: in
> a right triangle with hypotenuse m_b and
> one side h_a/2, the other side is m (see the upper
> left figure, triangle P1HaM).
>
> The proof of this assertion is given below, at the
> end of explanation.
>
> The important outcomes of this analysis are:
>
> (*) The triangle ABC is constructible with ruler and
> compass, as its circumcenter is an intersection of a
> straight line and a parabola (cf. discussion of this in J.
> Petersen's book on page 98, theorem 1).
>
> (**) the essential elements of the parabola - its
> directix and focus - may be easily constructed from given
> data.
>
> Actually, the upper left figure depicts the
> construction of F and d. After the above explanations,
> I hope the steps are clear.
>
> The second step - after F, d and also the line AE (the
> reflection
> of AHa in ATa) are constructed - is to find the
> intersection of
> the line with the parabola. Fortunately, this has a
> very elegant
> solution given in an excellent book "Ruler and
> Compass" by H. P.
> Hudson (page 85-86). It provides a construction for
> ANY conic, but the case of parabola is particularly
> simple. Let me guide you through the
> construction (upper right figure):
>
> C.1. Construct AE, the reflection of AHa in ATa.
> C.2. Draw line FE.
> C.3. Take any point on a line AE, eg. D.
> C.4. Draw a circle centered at D and tangent to
> directrix d.
> C.5. Let G, G' be points of intersect of FE and the
> circle.
> C.6. Draw FO||DG, FO'||DG' (only the first is
> shown). O, O'
> are the circumcenters of the 2 solutions.
>
> To prove the construction, notice that (see lower
> right figure)
> triangles OFE and DGE, ORE and DQE are similar,
> therefore:
>
> OE OF OR OF DG
> ---- = ---- = ----, or ----- = ---- = 1,
> DE DG DQ OR DQ
>
> or OF = OR. Thus, O is equidistant from the focus F
> and the directrix of the parabola, that is, lies on it
> (see T4 below)!
> Beautiful, isn't it?!
>
> Once we construct O, two vertices B and C are easily
> obtained.
>
> As promised, I conclude the explanation by proving the
> assertion
> 2 above. I will use the lower left figure for
> reference. For the
> convinience, the following notations are used:
>
> h = h_a/2, m^2 = m_b^2 - h^2 (compare with 2c above).
>
>
> In what follows, I assume the cartesian coordinates
> are used, and
> will use the following basic statements about straight
> lines and
> parabolas:
>
> T1. The line passing through points (x0, y0) and (x1,
> y1), has a
> slope L = (y1-y0)/(x1-x0).
>
> T2. If a line has a slope L, its perpendicular has a
> slope -1/L.
>
> T3. The line with a slope L and passing through point
> (x0, y0),
> has an equation y-y0 = L*(x-x0).
>
> T4. Parabola is a locus of points equidistant from a
> given point F
> (focus) and a given line d (directrix).
>
> T5. If the parabola is given by an equation y-y0 =
> p*(x-x0)^2, its
> directrix d is parallel to x-axis; focus F is (x0,
> y0+1/4p),
> and the distance from F to d is 1/2p.
>
> Let the cartesian coordinates be chosen so that Ha is
> at the origin,
> and A is on y-axis. Then, BC lies on x-axis. Assume B
> at (2*b, 0).
> Then, it is straightforward to see that the important
> points have
> the following coordinates:
>
> Ha(0, 0); A(0, 2*h); B(2*b, 0); Mb(2*b+m, h);
> Mc(b, h).
>
> The circumcenter O is an intersection of perpendicular
> bisectors
> of AB and AC. Designate them as lb and lc accordingly;
> they pass
> through the points Mb and Mc (see drawing). Applying
> T1-T3, we
> get the following equations:
>
> lb: y-h = (2*b+m)/h * (x - 2*b-m)
> (1)
> lc: y-h = b/h * (x-b)
> (2)
>
> Solving linear system (1)-(2), we get the following
> formulas for
> coordinates xO, yO of circumcenter O for a specific b:
>
> xO = 3*b + m
> yO = b/h * (2*b + m)
> (3)
>
> The above expressions suggest that xO, yO satisfy the
> following
> relation:
> y = p*x^2 + q*x + r,
> (4)
>
> where p, q, r do not depend on b. Indeed, plugging (3)
> into (4),
> we get:
>
> p = 2/(9*h), q = -m/(9*h), r = h -
> m^2/(9*h). (5)
>
> Rearranging the terms in (4) with (5), we get:
>
> y-y0 = p*(x-x0)^2,
>
> where x0 = -p/2q = m/4, y0 = r - q^2/4p = h -
> m^2/(8*h).
>
> Applying T5, we conclude that the locus of O is indeed
> parabola,
> with directrix d parallel to line BC, and distant from
> Ha:
>
> y0 - 1/4p = - (m^2 + h^2)/(8*h) = -
> m_b^2/(4*h_a).
>
> The focus F is 1/2p = 9/8*h_a far away from d. Q.E.D.
>
> Let me stop here. I will write my comments about the
> solution in
> a separate mail. And yours are always welcome...
>
> Best regards,
> Barukh.
>
=========================================================================
Instruções para entrar na lista, sair da lista e usar a lista em
http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html
O administrador desta lista é <nicolau@mat.puc-rio.br>
=========================================================================