Cinderella Blog - Geometry

SchwerpunktVektoriell.cdy

anonymous@cinderella.de (Anonymous Cinderella Submission) — Wed, 13 May 2009 16:02:53 +0200

Bitte schalten Sie Java ein, um eine Cinderella-Konstruktion zu sehen.

Triangle Centers: The Apollonius Point.cdy

anonymous@cinderella.de (Anonymous Cinderella Submission) — Tue, 20 Jan 2009 16:06:38 +0100

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Construction of the Apollonius Point. See http://faculty.evansville.edu/ck6/tcenters/recent/apollon.html
for the definition, and http://whistleralley.com/tangents/tangents.htm and
http://whistleralley.com/inversion/inversion.htm for the methods of construction.

The triangle is defined by free points A, B, and C. Construction of the three excircles
Ea, Eb, and Ec was pretty simple given the angle-bisection tool. Construction of the
circle E tangent to Ea, Eb, and Ec and encompassing them was not so simple.
Cinderella's tool for inversions (mirroring defined by a circle) was invaluable in this
complicated construction. No CindyScript was used.

This construction: Just for fun. (To see how it was done, download the construction.
Use the bookmarklet you can find at the Cinderella Support Forum, topic
"Learning from blog posts".)

For more information on different triangle centers, see
http://faculty.evansville.edu/ck6/tcenters/index.html.

-- David Bakin

Construction of a Conic from 5 Points via Pascal's Theorem (projective).cdy

anonymous@cinderella.de (Anonymous Cinderella Submission) — Fri, 02 Jan 2009 22:51:13 +0100

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Construction of a Conic from the 5 points A,B',C,A',B via (the converse of)
Pascal's Theorem.

Let x be an arbitrary line through B, then we're looking for the locus of the
points C', where C' is a point on the conic, and is the 6th point of the hexagon
inscribed in the conic. The line x is thus the line BC'.

Find A'' as the intersection of BC' (x) and B'C. FInd C'' as the intersection of
AB' and B'A. The pascal line then goes through A'' and C''. B'' is the
intersection of A'C and the pascal line. Finally, C' is the intersection of
BC' (x) and AB''.

In this Cinderella construction, the points A,B',C,A',B are free. The red
conic is the locus of point C' as the orange line x rotates.

This construction is described in "Introduction to Projective Geometry" by
C. R. Wylie, Jr. (Dover, 2008).

-- David Bakin

Maclaurin Construction of a Conic from 5 Points (projective).cdy

anonymous@cinderella.de (Anonymous Cinderella Submission) — Fri, 02 Jan 2009 21:46:11 +0100

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Construction of a conic from the 5 points P0 ... P4.

Construct Q3, Q'3, Q4, Q'4, V.

Now let there be an arbitrary line, p, through P1. Q is the point where the line p
meets p2, then Q' is the point where VQ meets p1. Now, P is the point where P2Q'
meets P1Q. The locus of points P as p rotates is the conic through the 5 given
points.

In this Cinderella construction, the points P0 ... P4 are free. The light blue conic
is the locus of point P as the orange line l rotates.

This is the construction of Colin Maclaurin (1698-1746), as described in "Introduction
to Projective Geometry" by C. R. Wylie, Jr. (Dover, 2008).

-- David Bakin

Basic Monodromy

richter@cinderella.de (Jürgen Richter-Gebert) — Tue, 18 Sep 2007 10:33:38 +0200

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Doppelverhaeltnis

anonymous@cinderella.de (Anonymous Cinderella Submission) — Wed, 04 Jul 2007 09:35:58 +0200

Bitte schalten Sie Java ein, um eine Cinderella-Konstruktion zu sehen.

Doppelverhaeltnis

anonymous@cinderella.de (Anonymous Cinderella Submission) — Wed, 04 Jul 2007 09:35:20 +0200

Bitte schalten Sie Java ein, um eine Cinderella-Konstruktion zu sehen.
Bitte schalten Sie Java ein, um eine Cinderella-Konstruktion zu sehen.

Thales

anonymous@cinderella.de (Anonymous Cinderella Submission) — Tue, 06 Mar 2007 19:21:33 +0100

Bitte schalten Sie Java ein, um eine Cinderella-Konstruktion zu sehen.
Ein rechtwinkliges Dreieck

Rechtwinkling, Gleichschenklig, Gleichseitig, Beliebig

anonymous@cinderella.de (Anonymous Cinderella Submission) — Thu, 01 Mar 2007 13:03:14 +0100

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Klicken Sie auf die SchaltflŠchen!

ArcOnConic2.cdy

anonymous@cinderella.de (Anonymous Cinderella Submission) — Tue, 20 Feb 2007 21:04:54 +0100

Bitte schalten Sie Java ein, um eine Cinderella-Konstruktion zu sehen.
A Construction for Arcs on a conic:

Here comes a non-trivial Solution to a non-trivial problem.
How to get an arc on a conic.

(Perhaps I will implement a method ArcOnConic based on this if I have more time).
In fact it is a nice exercise in projective geometry.

So assume you want to have an arc on a conic C through three points a,b,c on C.

You need a circle K and three points a' b' c' on the circle and furthermore (and this is the Problem)
a projective transformation that maps

a' --> a
b' --> b
c' --> c
and
K --> C.

You can specify such a Transformation by the image of four points. So you need one more point pair d, d' with
d' --> d

such that also K --> C.

Now you need a property of conics and crossratios.
If you have four points fixed a,b,c,d on a fixed conic and a fifth point e on the conic the the
crossratio of the four lines|e,a|,|e,b|,|e,c|,|e,d| is independent on the choice of e. Thus we can call it the
crossratio of a,b,c,d with respect to C.

Thus we can choose an arbitrary point d on the conic.

d' on the circle must be chosen such that the
crossratio of a,b,c,d w.r.t C is the same as the crossratio of
a', b', c', d' w.r.t K.

So how do we get the point d'.

One possibility (also a tricky one) is as follows:

intersect the four lines|e,a|,|e,b|,|e,c|,|e,d| with a line.
your get four intersections a'', b'', c'', d''. Now define a Moebius transformation
that maps

a'' --> a'
b'' --> b'
c'' --> c'

the Moebius transformation maps d'' also to the circle K....
and (since the world is nice and miracles occur sometimes)
maps it to the position with exactly the right cross ratio. Call the mapped point
d', define the projective transformation

a' --> a
b' --> b
c' --> c
d' --> d

draw the arc a', b', c'
and map it through the projective transformation.
Et voila -- your done.

Uff it took me two minutes to do the construction. And 20 minutes to explain it.