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
[hide]- 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 vladimir.shelomovskii@gmail.com, vvsss
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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