Re: Tarski's Axioms for Geometry -- a draft version for DC Proof
From: Rainer Rosenthal (r.rosenthal_at_web.de)
Date: 08/12/04
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Date: Thu, 12 Aug 2004 23:55:50 +0200
"philippe 92"
>
> The practical constructions are just much harder.
>
> Example : Midpoint of segment AB with Sparski.
> ...
>
> Even worse with Poncelet-Steiner, that is using the compass
> only once.
Many thanks for the information and the construction.
For the 60° construction there is indeed a very simple
way. First you make a right angle: two lines intersect
in point X. Make a circle round X. The four corners
form a rectangle. Let C one of the corners, and draw
points A and B such that angle <)ACB = 90° and |AC|=2*|BC|.
Circle around B with radius |AC| cuts AC in P. Then we
have angle <)ABP = 60°. (Solution due to Klaus Nagel.)
When I invented this 60° problem, I had in mind the
following construction, which is presented in alt.math.
recreational also:
: R-----+-----+
: | I| | We have points A,P,Z,M,R,Y
: | ' | | in a square grid as shown.
: M-----Y-----+
: | | | Point I lies on RY
: | | | and we have |ZM|=|ZI|
: +-----Z-----+
: | | | It follows <)API = 60°
: | | |
: A-----+-----P
Sparski allows the square grid, so it allows construction
of point I also. I did not find an elegant algebraic
proof, but I would welcome any such proofs.
Au revoir,
Rainer
-- Rainer Rosenthal, r.rosenthal@web.de _____________________ | _ | | | (_) | Given A, P and a circle. Find B, C on the | | A P | circle with P on BC and area(ABC)=maximum. | |__________|_(http://chephip.free.fr/ie/sol117d_en.html)___|
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