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  • What is projective geometry and example?

    projective geometry, branch of mathematics that deals with the relationships between geometric figures and the images, or mappings, that result from projecting them onto another surface. Common examples of projections are the shadows cast by opaque objects and motion pictures displayed on a screen.

  • What is projective geometry used for?

    Sava College. By an extension, Descriptive or Projective Geometry, it can be used to transform the Three-Dimensional Space into a Tetra-Dimensional Space and the other, being the only branch of mathematics that can directly describe a four-dimensional space.

  • What are the basics of projective geometry?

    Projective geometries are characterised by the "elliptic parallel" axiom, that any two planes always meet in just one line, or in the plane, any two lines always meet in just one point. In other words, there are no such things as parallel lines or planes in projective geometry.
  • Although very beautiful and elegant, we believe that it is a harder approach than the linear algebraic approach. In the linear algebraic approach, all notions are considered up to a scalar. For example, a projective point is really a line through the origin.
IMO Training 2010 Projective Geometry Alexander Remorov

Projective Geometry

Alexander Remorov

alexanderrem@gmail.com

Harmonic Division

Given four collinear pointsA;B;C;D, we dene theircross-ratioas: (A;B;C;D) =!CA!

CB:!DA!

DB(1) Note thatthe lengths are directed. When the cross-ratio is equal to1, we say that (A;B;C;D) is aharmonic bundle. A particular case which occurs in many problems is whenA;C;B;Dare on a line in this order. LetPbe a point not collinear withA;B;C;D; we dene thepencilP(A;B;C;D) to be made up of 4 linesPA;PB;PC;PD.P(A;B;C;D) is calledharmonicwhen (A;B;C;D) is harmonic. Lemma 1:A pencilP(A;B;C;D) is given. The linesPA;PB;PC;PDintersect a linelatA0;B0; C

0;D0respectively. Then (A0;B0;C0;D0) = (A;B;C;D).

Proof:WologA;C;B;Dare collinear in this order. Using Sine Law in4CPA;4CPB;4DPA,

4DPB, we get (the lengths and angles are directed):

CA!

CB:!DA!

This gives a "trigonometric denition" corresponding to a cross ratio. Lemma 1 follows from (2). Therefore for any pencilP(A;B;C;D), we can dene its cross ratio to be: (PA;PB;PC;PD) = (A;B;C;D). This denition is ok because of lemma 1. Corollary 1:This is extremely useful. Using the same notation as in lemma 1, if (A;B;C;D) is harmonic then so is (A0;B0;C0;D0).A B C DE F

GLemma 2:In4ABC, pointsD;E;Fare on sides

BC;CA;AB. LetFEintersectBCatG. Then

(B;C;D;G) is harmonic iAD;BE;CFare concur- rent.

Proof:Use Ceva and Menelaus.

1 IMO Training 2010 Projective Geometry Alexander Remorov

Lemma 3:Consider pointsA;B;C;Don a circle. Let

Pbe any point on the circle. Then the cross ratio

(PA;PB;PC;PD) does not depend onP.

Proof:From (2) it follows that

:DADB j(3) If (PA;PB;PC;PD) =1, the quadrilateralACBDis calledharmonic. From lemma 3, if

A;C;B;Dare on a circle in this order, andjCACB

j=jDADB j, thenACBDis harmonic. The nice thing about lemma 3 is that it allows you to use harmonic pencils for circles. Lemma 4:A pointPis outside or on a circle!. LetPC;PDbe tangents to!, andlbe a line throughPintersecting!atA;B(so thatP;A;Bare collinear in this order). LetABintersect CDatQ. ThenACBDis a harmonic quadrilateral and (P;Q;A;B) is harmonic.B CD A

PQProof:4PAD 4PDB=)ADDB

=PAPD . Similarly ACCB =PAPC . BecausePC=PD, it follows thatACBD is harmonic.

We can now apply lemma 3.We takeP=C(!)and con-

sider the intersection ofC(A;B;C;D) with linel. Since ACBDis harmonic, the resulting 4 points of intersec- tion form a harmonic bundle, hence (P;Q;A;B) is har- monic. Corollary 2:PointsA;C;B;Dlie on a line in this order, andMis the midpoint ofCD. Then (A;B;C;D) is har- monic iACAD=ABAM. Proof:Whenever you see things likeACADand circles, trying Power of a Point is a good idea. AssumeACAD=ABAM. Consider the circle centred atMpassing throughC;D. Let ATbe a tangent fromAto this circle. ThenACAD=AT2. HenceABAM=AT2and

4ATM 4ABT. Since\ATM= 90it follows that\ABT= 90. By lemma 4 (A;B;C;D) is

harmonic. The converse of the corollary is proved in the same way. B

CDAPLemma 5:PointsA;C;B;Dlie on a line in this

order.Pis a point not on on this line. Then any two of the following conditions imply the third:

1. (A;B;C;D) is harmonic.

2.PBis the angle bisector of\CPD.

3.AP?PB.

Proof:Straightforward application of Sine Law.

2 IMO Training 2010 Projective Geometry Alexander Remorov

Poles and Polars

Given a circle!with centerOand radiusrand any pointA6=O. LetA0be the point on rayOAsuch thatOAOA0=r2. The linelthroughA0perpendicular toOAis called thepolarof Awith respect to!.Ais called thepoleoflwith respect to!. Lemma 6:Consider a circle!and a pointPoutside it. LetPCandPDbe the tangents fromP to!. ThenSTis the polar ofPwith respect to!.

Proof:Straightforward.

Note:Using the same notation as in lemma 4, it follows thatQlies on the polar ofPwith respect to!. La Hire's Theorem:This is extremely useful.A pointXlies on the polar of a pointYwith respect to a circle!. ThenYlies on the polar ofXwith respect to!.

Proof:Straightforward.

B C DA O PQ R E FBrokard's Theorem:The pointsA;B;C;Dlie in this or- der on a circle!with centerO.ACandBDinter- sect atP,ABandDCintersect atQ,ADandBC intersect atR. ThenOis the orthocenter of4PQR.

Furthermore,QRis the polar ofP,PQis the po-

lar ofR, andPRis the polar ofQwith respect to

Proof:LetQPintersectBC;ADatF;E, respec-

tively. From lemma 2 it follows that (R;E;A;D) and (R;F;B;C) is harmonic. From lemma 4 it follows thatEFis the polar ofR. HencePQis the po- lar ofR. SimilarlyPRis the polar ofQandRQ is the polar ofP. The fact thatOis the orthocen- ter of4PQRfollows from properties of poles and po- lars. The congurations in the above lemmas and theorems come up in olympiad problems over and over again. You have to learn to recognize these congurations.Sometimes you need to com- plete the diagram by drawing extra lines and sometimes even circles to arrive at a "standard" conguration. 3 IMO Training 2010 Projective Geometry Alexander Remorov

Problems

Many of the following problems can be done without using projective geometry, however try to use it in your solutions.

0.[Useful]Mis the midpoint of a line segmentAB. LetP1be a point at innity on lineAB.

Prove that (M;P1;A;B) is harmonic.

1.[Useful] PointsA;C;B;Dare on a line in this order, so that (A;B;C;D) is harmonic. LetM

be the midpoint ofAB. Prove thatAM2=MCMD. (There is a purely algebraic way to do this, as well as a way using poles and polars. Try to nd the latter).

2.The tangents to the circumcircle of4ABCatBandCintersect atD. Prove thatADis the

symmedian of4ABC.

3.ADis the altitude of an acute4ABC. LetPbe an arbitrary point onAD.BP;CPmeet

AC;ABatM;N, respectively.MNintersectsADatQ.Fis an arbitrary point on sideAC.FQ intersects lineCNatE. Prove that\FDA=\EDA.

4.PointMlies on diagonalBDof parallelogramABCD. LineAMintersects sideCDand line

BCat pointsKandN, respectively. LetC1be the circle with centerMand radiusMAandC2be the circumcircle of triangleKCN.C1,C2intersect atPandQ. Prove thatMP;MQare tangent toC2.

5.(IMO 1985) A circle with centreOpasses through verticesA;Cof4ABCand intersects its sides

BA;BCat distinct pointsK;N, respectively. The circumcircles of4ABCand4KBNintersect at pointBand another pointM. Prove that\OMB= 90.

6.(Vietnam 2009) LetA;Bbe two xed points andCis a variable point such that\ACB=, a

constant in the range [0 ;180]. The incircle of4ABCwith incentreItouches sidesAB;BC;CA at pointsD;E;F, respectively.AI;BIintersectEFatM;Nrespectively. Prove that the length ofMNis constant and the circumcircle of4DMNpasses through a xed point.

7.(Vietnam 2003) CirclesC1andC2are externally tangent atM, and radius ofC2is greater than

radius ofC1.Ais a point onC2which does not lie on the line joining the centers of the circles. Let BandCbe points onC1such thatABandACare tangent toC1. LinesBMandCMintersect C

2again atEandF, respectively. LetDbe the intersection of the tangent toC2atAand line

EF. Show that the locus ofDasAvaries is a straight line.

8.(SL 2004 G8) In a cyclic quadrilateralABCD, letEbe the intersection ofADandBC(so

thatCis betweenBandE), andFbe the intersection ofACandBD. LetMbe the midpoint of sideCD, andN6=Mbe a point on the circumcircle of4ABMsuch thatAMMB =ANNB . Show that

E;F;Nare collinear.

9.(SL 2006, G6) Circles!1and!2with centresO1andO2are externally tangent at pointDand

internally tangent to a circle!at pointsEandFrespectively. Linelis the common tangent of!1 and!2atD. LetABbe the diameter of!perpendicular tol, so thatA;E;O1are on the same side ofl. Prove thatAO1;BO2;EFandlare concurrent. 4 IMO Training 2010 Projective Geometry Alexander Remorov Hints

0.This is obvious. However, make sure you know this fact! It is used for several other problems

on the handout.

1.It is not hard.

2.It is not hard.

3.There are no circles involved.

4.You almost have a harmonic bundle. Draw the fourth point.

5.Draw tangents fromBto the circumcircle ofAKCN. Now draw another circle with centreB.

Complete the diagram.

6.First proveAN?MN. Do this using the results in the handout.

7.Consider the homothethy carrying one circle to the other.

8.Don't be scared that this is a G8. Complete the diagram.

9.First proveA;D;Fare collinear. Complete the diagram.

5 IMO Training 2010 Projective Geometry Alexander Remorov

References

1Cosmin Pohoata, Harmonic Division and its Applications,

2Milivoje Lukic, Projective Geometry,

3Yufei Zhao, Circles,

4Yufei Zhao, Cyclic Quadrilaterals - The Big Picture,

5Yufei Zhao, Lemmas in Euclidean Geometry,

6Various MathLinks Forum Posts,

6quotesdbs_dbs43.pdfusesText_43
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