Difference between revisions of "Projective geometry (simplest cases)"
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Similarly points <math>K, E, </math> and <math>D'</math> are collinear. | Similarly points <math>K, E, </math> and <math>D'</math> are collinear. | ||
− | We use the symmetry lines <math>DF</math> and <math>D'E</math> with respect <math>\ell</math> and get symmetry <math>K</math> and <math>C</math> with respect <math>\ell, KB'</math> and CB with respect <math>\ell</math> | + | We use the symmetry lines <math>DF</math> and <math>D'E</math> with respect <math>\ell</math> and get in series |
+ | |||
+ | <math>-</math> symmetry <math>K</math> and <math>C</math> with respect <math>\ell,</math> | ||
+ | |||
+ | <math>-</math> symmetry <math>KB'</math> and CB with respect <math>\ell,</math> | ||
+ | |||
+ | <math>-</math> symmetry <math>X</math> and <math>Y</math> with respect <math>\ell.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 14:20, 3 November 2024
Projective geometry contains a number of intuitively obvious statements that can be effectively used to solve some Olympiad mathematical problems.
Useful simplified information
Let two planes and
and a point
not lying in them be defined in space. To each point
of plane
we assign the point
of plane
at which the line
intersects this plane. We want to find a one-to-one mapping of plane
onto plane
using such a projection.
We are faced with the following problem. Let us construct a plane containing a point and parallel to the plane
Let us denote the line along which it intersects the plane
as
No point of the line
has an image in the plane
Such new points are called points at infinity.
To solve it, we turn the ordinary Euclidean plane into a projective plane. We consider that the set of all points at infinity of each plane forms a line. This line is called the line at infinity. The plane supplemented by such line is called the projective plane, and the line for which the central projection is not defined is called (in Russian tradition) the exceptional line of the transformation. We define the central projection as follows.
Let us define two projective planes and
and a point
For each point of plane
we assign either:
- the point of plane
at which line
intersects
- or a point at infinity if line does not intersect plane
We define the inverse transformation similarly.
A mapping of a plane onto a plane is called a projective transformation if it is a composition of central projections and affine transformations.
Properties of a projective transformation
1. A projective transformation is a one-to-one mapping of a set of points of a projective plane, and is also a one-to-one mapping of a set of lines.
2. The inverse of a projective transformation is projective transformation. The composition of projective transformations is a projective transformation.
3. Let two quadruples of points and
be given. In each quadruple no three points lie on the same line: Then there exists a unique projective transformation that maps
to
to
to
to
4. There is a central projection that maps any quadrilateral to a square. A square can be obtained as a central projection of any quadrilateral.
5. There is a central projection that maps a circle to a circle, and a chosen interior point of the first circle to the center of the second circle. This central projection maps the polar of the chosen point to the line at infinity.
6. The relationships of segments belonging to lines parallel to the exceptional line are the same for images and preimages.
Projection of a circle into a circle
Let a circle with diameter
and a point
on this diameter
be given.
Find the prospector of the central projection that maps the circle into the circle
and the point
into point
- the center of
Solution
Let be the center of transformation (perspector) which is located on the perpendicular through the point
to the plane containing
Let
be the diameter of
and plane
is perpendicular to
Spheres with diameter and with diameter
contain a point
, so they intersect along a circle
Therefore the circle is a stereographic projection of the circle
from the point
That is, if the point lies on
, there is a point
on the circle
along which the line
intersects
It means that is projected into
under central projection from the point
is antiparallel
in
is the symmedian.
Corollary
Let
The inverse of a point
with respect to a reference circle
is
The line throught in plane of circle
perpendicular to
is polar of point
The central projection of this line to the plane of circle from point
is the line at infinity.
vladimir.shelomovskii@gmail.com, vvsss
Butterfly theorem
Let be the midpoint of a chord
of a circle
through which two other chords
and
are drawn;
and
intersect chord
at
and
correspondingly.
Prove that is the midpoint of
Proof
Let point be the center of
We make the central projection that maps the circle into the circle
and the point
into the center of
Let's designate the images points with the same letters as the preimages points.
Chords and
maps into diameters, so
maps into rectangle and in this plane
is the midpoint of
The exceptional line of the transformation is perpendicular to so parallel to
The relationships of segments belonging to lines parallel to the exceptional line are the same for images and preimages. We're done! .
vladimir.shelomovskii@gmail.com, vvsss
Sharygin’s Butterfly theorem
Let a circle and a chord
be given. Points
and
lyes on
such that
Chords
and
are drawn through points
and
respectively such that quadrilateral
is convex.
Lines and
intersect the chord
at points
and
Prove that
Proof
Let us perform a projective transformation that maps the midpoint of the chord to the center of the circle
. The image
will become the diameter, the equality
will be preserved.
Let and
be the points symmetrical to the points
and
with respect to line
the bisector
Denote (Sharygin’s idea.)
is cyclic
is cyclic
points
and
are collinear.
Similarly points and
are collinear.
We use the symmetry lines and
with respect
and get in series
symmetry
and
with respect
symmetry
and CB with respect
symmetry
and
with respect
vladimir.shelomovskii@gmail.com, vvsss