[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 t�m um exemplar) um livro em
franc�s sobre constru��o de tri�ngulos.
Desde ent�o fiz muitas descobertas que espero
colocar numa poss�vel edi��o em portugu�s.
No momento, gostaria de mostrar a solu��o abaixo,
que n�o aparece no livro. Infelizmente o arquivo .jpg
que poderia acompanh�-la � muito grande. Posso
envi�-lo para o email pessoal dos interessados
O ponto importante da constru��o � o lugar geom (lg).
dado pela par�bola.
Apreciaria se algu�m pudesse dar uma outra prova
para o lg ( seria poss�vel uma prova sint�tica?) e
tamb�m uma an�lise do lg e a figura feita (par�bola)
por um programa de geometria.
Em palavras, o problema �:
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.
>
=========================================================================
Instru��es 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 � <nicolau@mat.puc-rio.br>
=========================================================================