[metapost] Fwd: Honza's puzzler
Laurence Finston
lfinsto1 at gwdg.de
Mon Jan 17 22:08:05 CET 2005
------ Forwarded message -------
From: Larry Siebenmann <laurent at math.toronto.edu>
To: help-3dldf at gnu.org, laurent at math.toronto.edu, lfinsto1 at gwdg.de,
metafont at ens.fr, pracj3am at mbox.troja.mff.cuni.cz
Date: Mon, 17 Jan 2005 15:41:56 -0500
On Mon, 17 Jan 2005 14:06:11, on the metapost at tug.org list, Honza
Prachar <pracj3am at mbox.troja.mff.cuni.cz> wrote :
> Hello!
>
> Do you have any idea how to do a tangent to path p from point z.
>
> `Honza` Prachar
Hi 'Honza'!
That is a cherry-bomb in a tearoom ;-) We like to think that
MP/MF does simple things with gracefully!
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. One can see this geometrically by considering any bezier cubic
with a double point. Taking arbitrary z in the region 'just below'
the double point there are obviously four distinct lines on point z
tangent to path p.
xxx
xxxx xxxx
xxx xxx
xx xx
xxx xxx
xxxx xxxx
xxxxx
xxxxx Z xxxxx
xxxxxx xxxxxx
xxxxxx xxxxxxx
Now if MP could intersect quartic bezier curves is would be a
simple matter to find the roots:- just intersect
the line x=0 with the curve q(t)=(t,D(t)).
Until then, since quartics are solvable by extraction of
square roots, one can squeeze an (excruciatingly boring?) solution
out of metafont -- using sqrt.
But there may still exist an elegant geometrical solution
within MP ...
Cheers
Laurent S.
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