Difference between revisions of "Projective geometry (simplest cases)"

Revision as of 07:49, 10 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 $\Pi$ and $\Pi'$ and a point $O$ not lying in them be defined in space. To each point $A$ of plane $\Pi$ we assign the point $A'$ of plane $\Pi'$ at which the line $AO$ intersects this plane. We want to find a one-to-one mapping of plane $\Pi$ onto plane $\Pi'$ using such a projection.

We are faced with the following problem. Let us construct a plane containing a point $O$ and parallel to the plane $\Pi'.$ Let us denote the line along which it intersects the plane $\Pi$ as $\ell.$ No point of the line $\ell$ has an image in the plane $\Pi'.$ 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 $\Pi$ and $\Pi'$ and a point $O.$

For each point $A$ of plane $\Pi$ we assign either:

- the point $A'$ of plane $\Pi'$ at which line $AO$ intersects $\Pi',$

- or a point at infinity if line $AO$ does not intersect plane $\Pi'.$

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 $A, B, C, D$ and $A', B', C', D'$ be given. In each quadruple no three points lie on the same line: Then there exists a unique projective transformation that maps $A$ to $A',$ $B$ to $B', C$ to $C', D$ to $D'.$

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 $\omega$ with diameter $PQ$ and a point $A$ on this diameter $(2AP < PQ)$ be given.

Find the prospector of the central projection that maps the circle $\omega$ into the circle $\omega'$ and the point $A$ into point $O'$ - the center of $\omega'.$

Solution

Let $S$ be the center of transformation (perspector) which is located on the perpendicular through the point $P$ to the plane containing $\omega.$ Let $P = \omega \cap \omega', PQ'$ be the diameter of $\omega', PO' = O'Q'$ and plane $\omega'$ is perpendicular to $SQ.$

Spheres with diameter $PQ$ and with diameter $PS$ contain a point $Q'$ , so they intersect along a circle $\omega'.$

Therefore the circle $\omega$ is a stereographic projection of the circle $\omega'$ from the point $S.$

That is, if the point $M$ lies on $\omega$ , there is a point $M'$ on the circle $\omega'$ along which the line $SM$ intersects $\omega'.$

It means that $\omega$ is projected into $\omega'$ under central projection from the point $S.$

$PQ \perp SQ, PQ' \perp SQ \implies PQ'$ is antiparallel $PQ$ in $\triangle SPQ.$

$PO' = O'Q' \implies SA$ is the symmedian. $\frac {AQ}{AP} = \frac {SQ^2}{SP^2} \implies \frac {QP - AP}{AP} = \frac {SP^2 + PQ^2}{SP^2} \implies SP = \frac {QP}{\sqrt{\frac {QP}{AP}-2}}.$

Corollary

Let $Q'P || SA_0 \implies SA_0 \perp SQ \implies SP^2 = QP \cdot PA_0 = \frac {QP^2}{\frac {QP}{AP} - 2} \implies$ $PA_0 = \frac {AP \cdot OP}{OP – AP} \implies OA \cdot OA_0 = OP^2 .$ The inverse of a point $A$ with respect to a reference circle $\omega$ is $A_0.$

The line throught $A_0$ in plane of circle $\omega$ perpendicular to $PQ$ is polar of point $A.$

The central projection of this line to the plane of circle $\omega'$ from point $S$ is the line at infinity.

vladimir.shelomovskii@gmail.com, vvsss

Butterfly theorem

Let $M$ be the midpoint of a chord $PQ$ of a circle $\omega,$ through which two other chords $AB$ and $CD$ are drawn; $AC$ and $BD$ intersect chord $PQ$ at $X$ and $Y$ correspondingly.

Prove that $M$ is the midpoint of $XY.$

Proof

Let point $O$ be the center of $\omega, OM \perp PQ.$

We make the central projection that maps the circle $\omega$ into the circle $\omega'$ and the point $M$ into the center of $\omega'.$

Let's designate the images points with the same letters as the preimages points.

Chords $AB$ and $CD$ maps into diameters, so $ACBD$ maps into rectangle and in this plane $M$ is the midpoint of $XY.$

The exceptional line of the transformation is perpendicular to $OM,$ so parallel to $PQ.$

The relationships of segments belonging to lines parallel to the exceptional line are the same for images and preimages. We're done! $\blacksquare$ .

vladimir.shelomovskii@gmail.com, vvsss

Sharygin’s Butterfly theorem

Let a circle $\Omega$ and a chord $PQ$ be given. Points $E$ and $F$ lyes on $PQ$ such that $PE = QF (2PE > PQ).$ Chords $AB$ and $CD$ are drawn through points $E$ and $F,$ respectively such that quadrilateral $ACBD$ is convex.

Lines $AC$ and $BD$ intersect the chord $PQ$ at points $X$ and $Y.$

Prove that $PX=QY.$

Proof

Let us perform a projective transformation that maps the midpoint of the chord $PQ$ to the center of the circle $\Omega$ . The image $PQ$ will become the diameter, the equality $PE = QF$ will be preserved.

Let $B'$ and $D'$ be the points symmetrical to the points $B$ and $D$ with respect to line $\ell,$ the bisector $PQ.$

Denote $\omega = \odot AXE, K = \omega \cap \Omega.$ (Sharygin’s idea.)

$AKXE$ is cyclic $\implies \angle KXE = 180^\circ - \angle KAE.$

$AKB'B$ is cyclic $\implies \angle KB'B = 180^\circ - \angle KAB = \angle KXE.$

$PQ||B'B \implies$ points $K, X,$ and $B'$ are collinear.

Similarly points $K, E,$ and $D'$ are collinear.

We use the symmetry lines $DF$ and $D'E$ with respect $\ell$ and get in series

$-$ symmetry $K$ and $C$ with respect $\ell,$

$-$ symmetry $KB'$ and CB with respect $\ell,$

$-$ symmetry $X$ and $Y$ with respect $\ell.$

vladimir.shelomovskii@gmail.com, vvsss

Semi-inscribed circle

Let triangle $\triangle ABC$ and circle $\omega$ centered at point $O$ and touches sides $AB$ and $AC$ at points $E$ and $F$ be given.

Point $D$ is located on chord $EF$ so that $DO \perp BC.$

Prove that points $A, D,$ and $M$ (the midpoint $BC)$ are collinear.

Proof

Denote $K = AO \cap EF, r = OE, A' -$ point on line $DO$ such that $AA' || BC.$

$\frac {OK}{OD} = \frac{OA'}{AO}, \frac {OK}{OE} = \frac{OE}{AO} \implies OD \cdot OA' = r^2 .$

Therefore line $AA'$ is the polar of $D.$

Let us perform a projective transformation that maps point $D$ to the center of $\omega.$

Image $A$ is the point at infinity, so images $AB, AD,$ and $AC$ are parallel.

Image $EF$ is diameter, so image $D$ is midpoint of image $EF$ and image $M$ is midpoint of image $BC.$

$BC || AA',$ so image $BC$ is parallel to the line at infinity and the ratio $\frac {BM}{MC}$ is the same as ratio of images.

vladimir.shelomovskii@gmail.com, vvsss

Fixed point

Let triangle $\triangle ABC$ and circle $\omega$ centered at point $O$ and touches sides $AB$ and $AC$ at points $E$ and $F$ be given.

The points $P$ and $Q$ on the side $BC$ are such that $BP = QC.$

The cross points of segments $AP$ and $AQ$ with $\omega$ form a convex quadrilateral $KK'LL'.$

Point $D$ lies at $FE$ and satisfies the condition $DO \perp BC.$

Prove that $D \in KL.$

Proof

Let us perform a projective transformation that maps point $D$ to the center of $\omega.$

Image $A$ is the point at infinity, so images $AB, AP, AQ,$ and $AC$ are parallel. The plane of images is shown, notation is the same as for preimages.

Image $EF$ is diameter $\omega,$ image $BC$ is parallel to the line at infinity, so in image plane $\frac {BP}{BC} = \frac {QC}{BC}, BP = QC.$

Denote $M = EF \cap PK, N = EF \cap QL.$ $\frac {EM}{BP} = \frac {NF}{QC} \implies EM = NF \implies MD = ND,$ $KM \perp EF, LN \perp EF \implies$ $KK'LL'$ is rectangle, so $D \in KL. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Sphere and two points

Let a sphere $\omega$ and points $A$ and $B$ be given in space. The line $AB$ does not has the common points with the sphere. The sphere is inscribed in tetrahedron $ABCD.$

Prove that the sum of the angles of the spatial quadrilateral $ACBD$ (i.e. the sum $\angle ACB + \angle CBD + \angle ADB + \angle DAC)$ does not depend on the choice of points $C$ and $D.$

Proof

Denote $K, L, P, Q$ points of tangency $\omega$ and faces of $ABCD$ (see diagram), $\alpha = \angle AQB.$ $AQ = AK = AL, BP = BL = BQ \implies \angle ALB = \alpha.$

It is known that $\angle QBD = \angle PBD, \angle PBC = \angle LBC, \angle LAC = \angle KAC, \angle KAD = QAD.$ $\angle ACB = 180^\circ - \angle CBA - \angle CAB = 180^\circ - (\angle LBC + \angle LBA) - (\angle LAC + \angle LAB) =$ $= (180^\circ - \angle LBA - \angle LAB) - (\angle PBC + \angle KAC) = \alpha - \angle PBC - \angle KAC.$ Similarly, $\angle ADB = \alpha - \angle PBD + \angle KAD.$ $\angle ACB + \angle CBD + \angle ADB + \angle DAC =$ $= (\alpha - \angle PBC - \angle KAC) +(\angle PBC + \angle PBD) + (\alpha - \angle PBD - \angle KAD) + (\angle QAD + \angle QAC) = 2 \alpha,$ The sum not depend on the choice of points $C$ and $D.\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 176: / Line 176: @@
 <cmath>KM \perp EF, LN \perp EF \implies</cmath>
 <math>KK'LL'</math> is rectangle, so <math>D \in KL. \blacksquare</math>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Sphere and two points==
+[[File:Sphere and 2 points.png|350px|right]]
+Let a sphere <math>\omega</math> and points <math>A</math> and <math>B</math> be given in space. The line <math>AB</math> does not has the common points with the sphere. The sphere is inscribed in tetrahedron <math>ABCD.</math>
+Prove that the sum of the angles of the spatial quadrilateral <math>ACBD</math> (i.e. the sum <math>\angle ACB + \angle CBD + \angle ADB + \angle DAC)</math> does not depend on the choice of points <math>C</math> and <math>D.</math>
+<i><b>Proof</b></i>
+Denote <math>K, L, P, Q </math> points of tangency <math>\omega</math> and faces of <math>ABCD</math> (see diagram), <math>\alpha = \angle AQB.</math>
+<math>AQ = AK = AL, BP = BL = BQ \implies \angle ALB = \alpha.</math>
+It is known that <math>\angle QBD = \angle PBD, \angle PBC = \angle LBC, \angle LAC = \angle KAC, \angle KAD = QAD.</math>
+<cmath>\angle ACB = 180^\circ - \angle CBA - \angle CAB = 180^\circ - (\angle LBC + \angle LBA) - (\angle LAC + \angle LAB) =</cmath>
+<cmath>= (180^\circ - \angle LBA - \angle LAB) - (\angle PBC + \angle KAC) = \alpha - \angle PBC - \angle KAC.</cmath>
+Similarly, <cmath>\angle ADB = \alpha - \angle PBD + \angle KAD.</cmath>
+<cmath>\angle ACB + \angle CBD + \angle ADB + \angle DAC =</cmath>
+<cmath>= (\alpha - \angle PBC - \angle KAC) +(\angle PBC + \angle PBD) + (\alpha - \angle PBD - \angle KAD) + (\angle QAD + \angle QAC) = 2   \alpha,</cmath>
+The sum not depend on the choice of points <math>C</math> and <math>D.\blacksquare</math>
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Projective geometry (simplest cases)"

Revision as of 07:49, 10 November 2024

Contents

Projection of a circle into a circle

Butterfly theorem

Sharygin’s Butterfly theorem

Semi-inscribed circle

Fixed point

Sphere and two points