Projective Geometry. Stoienescu Paul. International Computer High School Of Bucharest,


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1 Projective Geometry Stoienescu Paul International Computer High School Of Bucharest, Abstract. In this note, I will present some olympiad problems which can be solved using projective geometry arguments. Keywords: poles, polars, harmonic divisions 1 Definitions Let d be a line and A, C,B and D four points which lie in this order on it.the quadrupet (ACBD) is called harmonic division(or is simply described as being harmonic) if and only if CA CB = DA DB Keep in mind that the lenghts are directed. Let P be a point not collinear with A,B,C,D; We define the pencil P (A, B, C, D) to be made up of 4 lines P A, P B, P C, P D. P (A, B, C, D) is called harmonic when (A, B; C, D) is harmonic. A quadrilateral ABCD is called harmonic if it is cyclic and BA BC = DA DC. Let w be a circle and let P be a point in the plane of w. Furthermore, let XY be an arbitrary chord passing through P with X and Y on w. The polar of a point P with respect to the circle w is defined as the locus of the points Q x y in plane so that the quadrupet (P XQ x yy ) is harmonic ( the chord XY passing through P is the variable here). 2 Useful Lemmas Lemma 1 A pencil P (A, B, C, D) is given. The lines P A, P B, P C, P D intersect a line l at X, Y, Z, T respectively. The quadruplet (X, Y, Z, T ) is harmonic if and only if (A, B, C, D) is harmonic. Lemma 2 In ABC, points D, E, F are on sides BC, CA, AB. Let F E intersect BC at G. Then (BCDG) is harmonic iff AD, BE, CF are concurrent. Lemma 3 A point P is outside or on a circle w. Let P C, P D be tangents to w, and l be a line through P intersecting w at A, B (so that P, A, B are collinear in this order). Let AB intersect CD at Q. Then ACBD is a harmonic quadrilateral and (P QAB) is harmonic. (1)
2 Lemma 4 Points A, C, B, D lie on a line in this order. P is a point not on this line.then any of the following conditions imply the third: 1.(ABCD) is harmonic. 2.P B is the angle bisector of CP D. 3.AP P B Lemma 5 M is the midpoint of a line segment AB. Let P be a point at infinity on line AB. Then (M, P ; A, B) is harmonic. Lemma 6 A point X lies on the polar of a point Y with respect to a circle w. Then Y lies on the polar of X with respect to w. Lemma 7 The points A, B, C, D lie in this order on the circle w with center O. AC and BD intersect at P, AB and DC intersect at Q, AD and BC intersect at R. Then O is the orthocenter of triangle P QR. Furthermore, QR is the polar of P, P Q is the polar of R, and P R is the polar of Q with respect to w. Lemma 8 Let X, Y, Z be three points and w a circle. Let x, y, z be the polars of X, Y, Z with respect to w. Then the points X, Y, Z are collinear if and only if the polars x, y, z are conccurent. 3 Useful theorems Pascal s Theorem Given a hexagon ABCDEF inscribed in a circle, let P = AB BD, Q = BC EF, R = CD AF. Then P, Q, R are collinear. Points A, B, C, D, E, F do not have to lie on the circle in this order. Sometimes, is it useful to use degenerate versions of this nice theorem. For example if A and B coincide than AB becomes the tangent to the circle at A. Desargues Theorem Given two triangles A 1 B 1 C 1 and A 2 B 2 C 2 we say that they are perspective with respect to a point when A 1 B 1 A 2 B 2, A 1 C 1 A 2 C 2, C 1 B 1 C 2 B 2 are collinear. Then two triangles are perspective with respect ro a point iff they are perspective with respect to a line. Pappus s Hexagon Theorem In mathematics, Pappus hexagon theorem (attributed to Pappus of Alexandria) states that given one set of collinear points A, B, C, and another set of collinear points a, b, c, then the intersection points X, Y, Z of line pairs Ab and ab, Ac and ac, Bc and bc are collinear, lying on the Pappus line.
3 4 Let s solve some problems 4.1 Romania TST 2009 The quadrilateral ABCD inscribed in a circle wich has diameter AC. Let A, D are symmetric to A, D with respect to the line BD and AC respectively. If A C BD = F and AC BD = E then prove that EF BD. Solution It is easy to check out that the lines BA, BD, BC and BE form a harmonical pencil (by lemma 4, BC is the bisector of angle DBE ).This means that the pencil form by F A, F R, F C, F E is also harmonical. Again by easy check, F B is the bisector line of angle AF C, so from lemma 4 we have BF E = The ClockTower School Seniors Competition 2010 Let ABCD be a cyclic quadrilateral.the lines AD, BC meet at P ; AB, CD at Q; and AC, BD at R. The perpendicular bisectors of AB, respectively BC, meet
4 P R at X, respectively QR at Y. Prove that XY passes through B. Solution All poles and polars are considered with respect to the given circle of ABCD. To start with, notice that line q = P R is the polar of Q and line p = QR is the polar of P. As Y lies on the polar of P, it follows that P lies on the polar y of Y. The pole of the line OY is the point at infinity on the direction BC, as OY is a diameter line. Thus the polar y of Y is the line P = BC, implying that Y B is tangent to the given circle at B. Similar considerations show that XB is tangent to the given circle at B, hence proving the thesis. 5 Problems to try Some of these problems can be done without using projective geometry, however try to use it in your solution.
5 5.1 Romania TST 2012 Let γ be a circle and l a line in its plane. Let K be a point on l, located outside of γ. Let KA and KB be the tangents from K to γ, where A and B are distinct points on γ. Let P and Q be two points on γ. Lines P A and P B intersect line l in two points R and respectively S. Lines QR and QS intersect the second time circle γ in points C and D. Prove that the tangents from C and D to γ are concurrent on line l. 5.2 Romania TST 2009 Let ABC be a nonisosceles triangle, in which X, Y, and Z are the tangency points of the incircle of center I with sides BC, CA and AB respectively. Denoting by O the circumcircle of ABC, line OI meets BC at a point D. The perpendicular dropped from X to Y Z intersects AD at E. Prove that Y Z is the perpendicular bisector of [EX]. 5.3 USA TST 2005 Let ABC be an acute scalene triangle with O as its circumcenter. Point P lies inside triangle ABC with P AB = P BC and P AC = P CB. Point Q lies on line BC with QA = QP. Prove that AQP = 2 OQB. 5.4 Iran TST 2009 ABC is a triangle and AA, BB and CC are three altitudes of this triangle. Let P be the feet of perpendicular from C to A B, and Q is a point on A B such that QA = QB. Prove that : P BQ = P AQ = P C C 5.5 Tuymaada 2011 Circles ω 1 and ω 2 intersect at points A and B, and M is the midpoint of AB. Points S 1 and S 2 lie on the line AB (but not between A and B). The tangents drawn from S 1 to ω 1 touch it at X 1 and Y 1, and the tangents drawn from S 2 to ω 2 touch it at X 2 and Y 2. Prove that if the line X 1 X 2 passes through M, then line Y 1 Y 2 also passes through M. 5.6 IMO SHORTLIST 2006 Circles w 1 and w 2 with centres O 1 and O 2 are externally tangent at point D and internally tangent to a circle w at points E and F respectively. Line t is the common tangent of w 1 and w 2 at D. Let AB be the diameter of w perpendicular to t, so that A, E, O 1 are on the same side of t. Prove that lines AO 1, BO 2, EF and t are concurrent.
6 5.7 Iran TST 2012 Consider ω is circumcircle of an acute triangle ABC. D is midpoint of arc BAC and I is incenter of triangle ABC. Let DI intersect BC in E and ω for second time in F. Let P be a point on line AF such that P E is parallel to AI. Prove that P E is bisector of angle BP C. 5.8 Iran TST 2008 Suppose that I is incenter of triangle ABC and l is a line tangent to the incircle. Let l be another line such that intersects AB, AC, BC respectively at C, B, A. We draw a tangent from A to the incircle other than BC, and this line intersects with l at A 1. B 1, C 1 are similarly defined. Prove that AA 1, BB 1, CC 1 are concurrent.
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