Re: [help-3dldf] Honza's puzzler

[L. Nobre G.]

On Mon, 17 Jan 2005, Larry Siebenmann wrote:

> On Mon, 17 Jan 2005 14:06:11, on the address@hidden list, Honza
> Prachar <address@hidden> wrote :
>
>  > Hello!
>  >
>  > Do you have any idea how to do a tangent to path p from point z.
>  >
>  > `Honza` Prachar

>      The condition of tangency is that the two vectors in R^2 that
> are repectively the t-derivative  p'(t) and the difference
>  p(t) - z be linearly dependent. But linear dependency occurs if
> and only if the following determinant condition holds:
>
>       D(t) := det( p'(t) , p(t) - z ) = 0
>
>      This determinant is clearly of degree 5 in t, at most. But
> inspection shows that the term of degree 5 in t vanishes. So the
> degree is 4 or less.  D(t)  is, in fact, generically of degree 4 in
> t.

>      But there may still exist an elegant geometrical solution
> within MP ...

Some elegance remains if path p is a circle.
Being c the center of the circle and r its radius, the tangent vector v is

numeric coss, sinn;
coss = r/abs(c-z);
sinn = sqrt(1-coss**2);
pair axis;
axis = unitvector(c-z);
v := coss*axis + sinn*(ypart axis, -xpart axis);
%%%% the above ^ plus sign may be a minus sign %%%%

z must not be inside the circle.

Lu\'{\i}s Nobre Gon\c{c}alves - http://matagalatlante.org