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hyperbolic geometry

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Karolina Barone

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Nov 15, 2010, 8:23:18 AM11/15/10
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I'm Brazilian teacher and I have just entered this forum. My English is very poor, but I'll try to explain my doubt.

I study hyperbolic geometry and I have the book "Introduction to Non-Euclidean Geometry", by Harold E. Wolfe. I love this book, but some exercises are making me crazy. In the chapter about Saccheri Quadrilateral, there is a exercise that says:

Prove that, if perpendiculars are drawn from the extremities of one side of a triangule to the line passing through the midpoints of the other two sides, a Saccheri Quadrilateral is formed. (page 81)

How can I prove that? In the NonEuclid software is easy to see this, but prove is hard.

Thank you very much,

Karolina.

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gudi

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Nov 16, 2010, 10:57:45 PM11/16/10
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On Nov 15, 6:23 pm, Karolina Barone <karolinabar...@yahoo.com.br>
wrote:

I do not have access to Wolfe's book, but guess it is a misprint. I
imagine a possible situation to be :-

Prove that, if perpendiculars are drawn from the extremities of one

side of an _ irregular triangular prism_ to the line passing through


the midpoints of the other two sides, a Saccheri Quadrilateral is
formed.

It is easy to find the two right angles where perpendiculars are
drawn,and check that the sum of four angle is less than 360 degrees.

Let ABC be a triangle in X-Y plane, A is on origin and side BC is a
rubber elastic band. Now split A into A1 and A2, letting
A2 travel along Z axis. You get all Saccheri Quadrilaterals as (A1, B,
C, A2)

Examples of general skew quadrilaterals in general not having two
right angles at neighboring two vertices:

A pair of opposite skew sides removed from a tetrahedron.

If a point of any quadrilateral is lifted away from its plane and
rejoined to connecting vertices.

If two adjacent sides of any quadrilateral is rotated about a diagonal
to raise it from its plane.

( earlier errors corrected)

Narasimham

Karolina Barone

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Nov 19, 2010, 7:44:24 AM11/19/10
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Narasimhan:

It isn't a misprint. The construction is possible in the Poincare's disk, in NonEuclid 2007.04. This software is free and you can get it in
http://cs.unm.edu/~joel/NonEuclid/NonEuclid.html
http://cs.unm.edu/~joel/NonEuclid/

Thank you for your contribution, anyway.

Karolina.

gudi

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Nov 21, 2010, 10:02:41 AM11/21/10
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On Nov 19, 5:44 pm, Karolina Barone <karolinabar...@yahoo.com.br>
wrote:
> Narasimhan:
>
> It isn't a misprint. The construction is possible in the Poincare's disk, in NonEuclid 2007.04. This software is free and you can get it inhttp://cs.unm.edu/~joel/NonEuclid/NonEuclid.htmlhttp://cs.unm.edu/~joel/NonEuclid/

>
> Thank you for your contribution, anyway.
>
> Karolina.

Was guessing it when I do not know what a " triangule" is.

An elementary geometrical surface visualization:

A straight line AB rotates about A on z-axis simultaneously
translating along z-axis (from A to C) to a new position CD, anyhow
forming right angles at A and C on the axis. Also right angles are
made by AB and CD to helix line BD at B and D.The sum of two corner
angles of this swept helicoidal surface at B and D is 2 right angles.

If now C is connected to D directly by a straight line instead of the
helical line, then the sum of two corner angles of another hyperbolic
paraboloid surface at B and D containing that line is less than 2
right angles, which is the saccheri quadrilateral.

This is the classical 'acute angle hypothesis' of hyperbolic
geometry . ( E.g., Lectures in Classical Differential geometry by DJ
Struik page 151)

Another example of two asymptotic lines that may be viewed as parallel
to z-axis on a catenoid surface of revolution:
r = c cosh(z/c) , z = c (+ /- th + A) in cylindrical coordinates
( r,th,z).Saccheri quadrilaterals can be drawn between any four points
given this way.

Narasimham

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