Cyclic Quadrilaterals | The Big Picture

1 Winter Camp 2009 Cyclic QuadrilateralsYufei ZhaoCyclic Quadrilaterals The Big PictureYufei important skill of an olympiad geometer is being able to recognize known , many geometry problems are built on a few common themes. In this lecture, we will exploreone such What Do These Problems Have in Common?1. (IMO 1985) A circle with centerOpasses through the verticesAandCof triangleABCand intersects segmentsABandBCagain at distinct pointsKandN, respectively. Thecircumcircles of trianglesABCandKBNintersects at exactly two distinct that OMB= 90 .ACBKNMO2. (Russia 1995; Romanian TST 1996; Iran 1997) Consider a circle with diameterABand centerO, and letCandDbe two points on this circle. The lineCDmeets the lineABat a pointMsatisfyingMB < MAandMD < MC. LetKbe the point of intersection (different fromO) of the circumcircles of trianglesAOCandDOB.

2 Show that MKO= 90 .ABCDOMK3. (USA TST 2007) TriangleABCis inscribed in circle . The tangent lines to atBandCmeet atT. PointSlies on rayBCsuch thatAS AT. PointsB1andC1lies on rayST(withC1in betweenB1andS) such thatB1T=BT=C1T. Prove that trianglesABCandAB1C1are similar to each Camp 2009 Cyclic QuadrilateralsYufei ZhaoABCB1C1 STAlthough these geometric configurations may seem very different at first sight, they are actuallyvery related. In fact, they are all just bits and pieces of one big diagram!2 One Big DiagramABCDPQROMF igure 1: The big this lecture, we will try to understand the features of Figure 1. There are a lot of thingsgoing on in this diagram, and it can be frightening to look at. Don t worry, we will go through it2 Winter Camp 2009 Cyclic QuadrilateralsYufei Zhaobits and pieces at time.

3 In the process, we will discuss some geometric techniques that are usefulin other places as well.(Can you tell where to find each of the problems in Section 1 in Figure 1? You probably can tat this point, but hopefully you will be able to by the end of this lecture.)3 Miquel s Theorem and Miquel PointFact 1(Miquel s Theorem).LetABCbe a triangle, and letX,Y,Zbe points on linesBC,CA,AB,respectively. Assume that the six pointsA,B,C,X,Y,Zare all distinct. Then the circumcircles ofAY Z,BZX,CXYpass through a common 2: Diagram for Fact 1 (Miquel s Theorem).Exercise Fact 1. (This is very easy. Just chase1a few angles.)Fact 2(Miquel point).Let`1,`2,`3,`4be four lines in the plane, no two parallel. LetCijkdenotethe circumcircle of the triangle formed by the lines`i,`j,`k(these circles are calledMiquel circles).

4 ThenC123,C124,C134,C234pass through a common point (called theMiquel point).Exercise Fact 2. (Hint: apply Theorem 1)We want to specialize to the case of a Cyclic 3: Miquel point for a Cyclic you are bothered by configuration and orientation issues (and you should be!), use directed Camp 2009 Cyclic QuadrilateralsYufei ZhaoFact a quadrilateral. Let linesABandCDmeet atQ, and linesDAandCBmeet atR. Then the Miquel point ofABCD( , the second intersection point of the circumcirclesofADQandABR) lies on the lineQRif and only ifABCDis Fact 3. (This is again just easy angle chasing.)4 An Important Result about Spiral SimilaritiesAspiral similarity2about a pointO(known as the center of the spiral similarity) is a compositionof a rotation and a dilation, both centered 4: An example of a spiral instance, in the complex plane, ifO= 0, then spiral similarities are described by multi-plication by a nonzero complex number.

5 That is, spiral similarities have the formz7 z, where C\ {0}. Here| |is the dilation factor, and arg is the angle of rotation. It is easy todeduce from here that if the center of the spiral similarity is some other point, sayz0, then thetransformation is given byz7 z0+ (z z0) (why?).Fact ,B,C,Dbe four distinct point in the plane, such thatABCDis not a there exists a unique spiral similarity that sendsAtoB, ,b,c,dbe the corresponding complex numbers for the pointsA,B,C,D. We know thata spiral similarity has the formT(z) =z0+ (z z0), wherez0is the center of the spiral similarity,and is data on the rotation and dilation. So we would like to find andz0such thatT(a) =candT(b) =d. This amount to solving the systemz0+ (a z0) =c, z0+ (b z0) = it, we see that the unique solution is =c da b, z0=ad bca b c+ not a parallelogram, we see thata b c+d6= 0, so that this is the unique solutionto the system.

6 Hence there exists a unique spiral similarity that you want to impress your friends with your mathematical vocabulary, a spiral similarity is sometimes called asimilitude, and a dilation is sometimes called ahomothety. (Actually, they are not quite exactly the same thing, butshhh!)4 Winter Camp 2009 Cyclic QuadrilateralsYufei ZhaoExercise can you quickly determine the value of in the above proof without even needingto set up the system of equations?Exercise a geometric argument why the spiral similarity, if it exists, must be unique.(Hint: suppose thatT1andT2are two such spiral similarities, then what can you say aboutT1 T 12?)Now we come to the key result of this section. It gives a very simple and useful descriptionof the center of a spiral similarity.

7 It can be very useful in locating very subtle spiral similaritieshidden in a geometry problem. Remember this fact!(Very Useful) Fact ,B,C,Dbe four distinct point in the plane, such thatACis notparallel toBD. Let linesACandBDmeet atX. Let the circumcircles ofABXandCDXmeetagain atO. ThenOis the center of the unique spiral similarity that 5: Diagram for Fact give the proof only for the configuration shown above. SinceABXOandCDOX arecyclic, we have OBD= OACand OCA= ODB. It follows that trianglesAOCandBODare similar. Therefore, the spiral similarity centered atOthat carriesAtoCmust also the above proof using directed angles mod so that it works for all , it is is worth mentioning that spiral similarities often comes in pairs. If we can sendABtoCD, then we can just as easily the center of the spiral similarity that sendsAtoCandBtoD, thenOis also thecenter of the spiral similarity that spiral similarity preserves angles atO, we have AOB= COD.

8 Also, the dilationratio of the first spiral similarity isOC/OA=OD/OB. So the rotation about with angle AOB= CODand dilation with ratioOB/OA=OD/OCsendsAtoB, andCtoD, as Fact 6 from Facts 2 and , let us apply these results to our configuration in Section the Miquel point of quadrilateralABCD. ThenMis the center of spiralsimilarity that sendsABtoDC, as well as the center of the spiral similarity that Camp 2009 Cyclic QuadrilateralsYufei ZhaoExercise Fact us specialize to a Cyclic quadrilateral, and continue the configuration in Fact 3 Fact a Cyclic quadrilateral with circumcenterO. Let linesABandCDmeet atQ, and linesDAandCBmeet atR. LetMbe the Miquel point ofABCD(which lies on lineQR,due to Fact 3). ThenOMis perpendicular 6: Diagram for the proof of Fact the spiral similarity centered atMwhich sendsAtoDandBtoC(Fact 7).

9 LetM1andM2be the midpoint ofABandDC, respectively. ThenTmust sendM1toM2. SoMis the center of unique spiral similarity that sendsAtoM1andDtoM2(Fact 6), and thus itfollows thatM,M1,M2,Qare concyclic (Fact 5).SinceM1andM2are the midpoints of the chordsABandCD, we have OM2Q= OM1Q,and soO,M1,M2,Qare concyclic, andOQis the diameter of the common circle. It follows thatO,M,M1,M2,Qall lie on the circle with diameterOQ. In particular, OMQ= 90 , as A Criterion for OrthogonalityIn this section, we give another proof of Fact 8 and introduce a very useful computational criterionfor orthogonality.(Very Useful) Fact ,B,C,Dbe points in the plane. Assume thatA6=BandC6= linesABandCDare perpendicular if and only ifAC2+BD2=AD2+ result follows immediately from the following identity.

10 (~A ~C) (~A ~C) + (~B ~D) (~B ~D) (~A ~D) (~A ~D) (~B ~C) (~B ~C) = 2(~B ~A) (~C ~D).Note that the LHS is zero iffAC2+BD2=AD2+BC2and the RHS is zero iffAB proof of Fact the circumradius ofABCD. Using Power of a Point on thecircumcircles ofABCDandABRM, we getQO2 r2=QA QB=QM QR=QM MR+QM26 Winter Camp 2009 Cyclic QuadrilateralsYufei Zhao(the strategy here is to transfer all the data onto the lineQR). Similarly, we haveRO2 r2=RA RD=RM RQ=QM MR+ the two relations, we getQO2 RO2=QM2 RM2,and it thus follows from Fact 9 thatOMis perpendicular Radical AxisGiven two circles in the plane, theirradical axisis the locus of points of equal power to the twocircles. It turns out that this is always a line. If the two circles intersect, then the radical axis isthe line passing through the two intersection points ( , the common chord).