Projective geometry (simplest cases)
Projective geometry contains a number of intuitively obvious statements that can be effectively used to solve some Olympiad mathematical problems. We explore only central projection.
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 vertices of any quadrilateral to vertices of the square.
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.
7. The double (anharmonic) ratio for given four points on a line (or on a circle) is a number
It is the only projective invariant of a quadruple of such points.
Contents
- 1 Projection of a circle into a circle
- 2 Butterfly theorem
- 3 Sharygin’s Butterfly theorem
- 4 Semi-inscribed circle
- 5 Fixed point
- 6 Set of lines
- 7 Set of lines in circle and triangle
- 8 Projection of any triangle into regular one
- 9 Some properties of an equilateral triangle
- 10 Invariant for points on circles
- 11 Two generated triangles
- 12 Sphere and two points
- 13 Projecting non-convex quadrilateral into rectangle
- 14 Projecting convex quadrilateral into square
- 15 Two lines and two points
- 16 Crossing lines
- 17 Convex quadrilateral and point
- 18 Theorem on doubly perspective triangles
- 19 Complete quadrilateral theorem
- 20 Medians crosspoint
- 21 Six segments
- 22 Sines of the angles of a quadrilateral
- 23 Triangle and line
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.
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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! .
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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
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Semi-inscribed circle
Let triangle and circle
centered at point
and touches sides
and
at points
and
be given.
Point is located on chord
so that
Prove that points and
(the midpoint
are collinear.
Proof
Denote point on line
such that
Therefore line is the polar of
Let us perform a projective transformation that maps point to the center of
Image is the point at infinity, so images
and
are parallel.
Image is diameter, so image
is midpoint of image
and image
is midpoint of image
so image
is parallel to the line at infinity and the ratio
is the same as ratio of images.
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Fixed point
Let triangle and circle
centered at point
and touches sides
and
at points
and
be given.
The points and
on the side
are such that
The cross points of segments and
with
form a convex quadrilateral
Point lies at
and satisfies the condition
Prove that
Proof
Let us perform a projective transformation that maps point to the center of
Image is the point at infinity, so images
and
are parallel. The plane of images is shown, notation is the same as for preimages.
Image is diameter
image
is parallel to the line at infinity, so in image plane
Denote
is rectangle, so
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Set of lines
Let points at the line
at the line
at the line
be given.
Denote circle
Prove that lines and
are concurrent.
Proof
According the Pascal theorem in case shown in diagram points and
are collinear at Pascal line.
Note that lines
Let point be the pole of Pascal line.
Let us perform a projective transformation that maps point to the center of
Then image of is parallel to image of
image of
is parallel to image of
, image of
is parallel to image of
As shown in ( Set of parallel lines) images of lines and
are concurrent.
Therefore lines and
are concurrent.
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Set of lines in circle and triangle
Let points at the line
at the line
at the line
be given.
Denote circle
Prove that lines and
are concurrent.
Proof
Let us consider the inscribed hexagon shown in diagram. According the Pascal theorem points
and
are collinear at Pascal line. Note that lines
The Pascal line has no common point with so the pole of this line is the inner point of the circle.
Let us perform a projective transformation that maps this pole to the center of
Image of is the point at infinity, so image
is parallel to image
Similarly, images
and
are parallel to images
and
As shown in ( Set of lines in triangle) images of lines and
are concurrent. Therefore lines
and
are concurrent.
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Projection of any triangle into regular one
Find a projective transformation that maps the given triangle into a regular one, and its inscribed circle into a circle.
Solution
Any point of tangency of the circle and line and any crosspoint of the lines are invariants of any projective transformation. Therefore, the Gergonne point of preimage maps into Gergonne point of image.
We make transformation which maps the Gergonne point of given triangle into the center if the incircle. According with Lemma, given triangle maps into regular one.
Lemma
Let the Gergonne point of coincide with the incenter of this triangle. Prove that
is regular.
Proof
The inradius connect the incenter and point of tangency, bisector connect vertex and incenter, Gergonne point belong the line connect vertex and point of tangency, so these objects lie at the same line.
The radius is perpendicular to the side at the points of tangency, which means that the bisector coincide with the altitude of the triangle. The axial symmetry with respect to bisector maps one side of triangle to another, the base side is perpendicular to the bisector so symmetric sides are equal. Applying symmetry with respect to another bisector, we find that all three sides are equal and the triangle is regular.
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Some properties of an equilateral triangle
Let an equilateral triangle be given.
Denote
the incircle of
Points are located on
so that ordered triplets of points
are collinear.
Find and
Prove that quadrilateral is cyclic,
is tangent to
and symmetric to
with respect
Proof
1. The circumradius is twice the inradius
From considerations of symmetry we conclude that
is regular, so
where
is the golden ratio:
3. Let be the angle of rotation of the spiral similarity that maps
into
By applying the Law of Sines on
we get
is A-bisector of
So
is the midpoint
Similarly,
Corollary
Lines and
are concurrent.
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Invariant for points on circles
Let a circle set of points
on it, and a point
outside plane
be given.
Let the central projection
on some plane
from
be the circle
Prove that
Proof
Denote - the diameter
perpendicular to
Projection on
is the diameter
so bisectors
and
lies in the plane perpendicular to
Therefore there is the sphere which contains and
The power of a point with respect sphere is
The plane cross sphere by circle, so quadrilateral
is cyclic.
Corollary
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Two generated triangles
Let triangle be given. Denote
the incircle of
. Points
are located on
so that ordered triplets of points
are collinear.
Prove that:
1. Lines and
are concurrent.
2. Lines and
are concurrent.
Proof
We make transformation which maps the incircle into incircle and the Gergonne point of given triangle
into the center of this incircle.
The image of is a regular triangle shown in ( Some properties of an equilateral triangle ).
All statements are projective invariants, they are true for images, which means they are also true for preimages.
Corollary
To build a configuration, we construct points and point
Line
cross
at points
and
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Sphere and two points
Let a sphere and points
and
be given in space. The line
does not has the common points with the sphere. The sphere is inscribed in tetrahedron
Prove that the sum of the angles of the spatial quadrilateral (i.e. the sum
does not depend on the choice of points
and
Proof
Denote points of tangency
and faces of
(see diagram),
It is known that
Similarly,
The sum not depend on the choice of points
and
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Projecting non-convex quadrilateral into rectangle
Let a non-convex quadrilateral be given. Find a projective transformation of points
into the vertices of rectangle.
Solution
WLOG, point is inside the
Let and
be the rays,
be any point on segment
Planes
and
are perpendicular, planes
and
are parallel, so image
is line at infinity and
is rectangle.
Let's paint the parts of the planes and
that maps into each other with the same color.
maps into
(yellow).
Green infinite triangle between and
maps into
where plane
is parallel to plane
Blue infinite quadrilateral between and
with side
maps into quadrilateral
Therefore inner part of quadrilateral maps into external part of rectangle
For example
maps into
where
is the intersection of planes
and
Note that the lines through pairs of points (for example, ) maps into the corresponding lines, and the intersection point of
and
maps into the center of the rectangle.
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Projecting convex quadrilateral into square
Let be a convex quadrilateral with no parallel sides.
Find the projective transformation of into the square
if the angle between the planes
and
is given. This angle is not equal to
or
Solution
Denote
Let be the point satisfying the conditions
The locus of such points is the intersection circle of spheres with diameters and
Let be the perspector and the image plane be parallel to plane
We use the plane contains
so image
Then image is the line at infinity, point
is point at infinity, so images
(line
) and
(line
) are parallel to
Similarly point is the point at infinity, so images
is the rectangle.
Point is the point at infinity, so
Point
is the point at infinity, so
is the square.
Let be such point that
The angle between and plane
is the angle we can choose. It is equal to the angle between planes
and
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Two lines and two points
Let lines and
intersecting at point
, a point
not lying on any of these lines, and points
and
on line
be given.
Find the locus of points
Solution
Let be the point
such that
be the midpoint
Let us prove that the points
and
are collinear.
The quadrilateral is convex. We make the projective transformation of
into the square.
Then line is the line at infinity,
so image
is the midpoint of image
image
is the center of the square.
Therefore images and
are parallel and points
and
are collinear.
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Crossing lines
Let a convex quadrilateral be given.
Denote
Prove that lines
and
are collinear.
Solution
The quadrilateral is convex.
We make the projective transformation of into the square.
Then image of the line is the line at infinity, image of
is the center of the square.
Images of and
are parallel, so image
is the midpoint of the image
Similarly images of
and
are midpoints of the square sides.
Therefore images and
are parallel, they are crossed at the point in infinity witch lyes at the line at infinity, that is at
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Convex quadrilateral and point
Let a convex quadrilateral and an arbitrary point
be given,
Prove that lines and
are concurrent.
Proof
The quadrilateral is convex. We make the projective transformation of
into the square. Then image of the line
is the line at infinity, images of
and
are parallel. Similarly
We use the Cartesian coordinate system with
Then
So line is
line
is
line
is
These lines contain point
Therefore preimages of and
are concurrent in preimage of the point
.
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Theorem on doubly perspective triangles
Let two triangles and
be given. Let the lines
and
be concurrent at point
and the lines
and
be concurrent at point
Prove that the lines and
are concurrent (the theorem on doubly perspective triangles).
Proof
WLOG, the quadrilateral is convex.
We make the projective transformation of into the square.
Then image of the line contains point is the line at infinity, images of
and
are parallel. Similarly
We use the Cartesian coordinate system with
So the line
is
line
is
line
is
These lines contain point
Therefore preimages of and
are concurrent in point
.
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Complete quadrilateral theorem
Let points no three of which are collinear, be given.
Prove that
Proof
We make the projective transformation of the vertices of into vertices of the square. Then image of the point
is the point at infinity, image of
is the center of the square, images of
and
are parallel, so for images
and
The double ratio is the projective invariant of a quadruple of collinear points
so the equality also holds for the preimages.
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Medians crosspoint
Let a convex quadrilateral and line
in common position be given (points
and
not belong
sides and diagonals are not parallel to
Denote
Denote
and
midpoints of
and
respectively.
Prove that lines and
are collinear.
Proof
Let the angle be fixed and the line
moves in a plane parallel to itself.
Then the line on which the median of the triangle lies is also fixed. Similarly, lines
and
are fixed. Denote
Let moves in a plane parallel to itself to the position where
It is known ( Six segments) that
After some simple transformations one can get
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Six segments
Let a convex quadrilateral and line
in common position be given (points
and
not belong
sides and diagonals are not parallel to
Denote
Prove that
Proof
By applying the law of sines, we get:
(see Sines of the angles of a quadrilateral)
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Sines of the angles of a quadrilateral
Let a convex quadrilateral be given. Prove that
Proof
By applying the law of sines, we get:
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Triangle and line
Let triangle and line
be given.
Denote
Let and
be the points on
such that
Prove that points are collinear.
Proof
Under projective transformation line from point
to
points
and
maps into points
and
so
Similarly,
We multiply these three equations and get:
We use the Menelaus' theorem for
and a transversal line
and get:
We make reduction of fractions, we take into account the given condition
and get:
Therefore in accordance with the Menelaus' theorem for
points
and
are collinear.
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