# Projective Geometry - Part 2

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1 Projective Geometry - Part 2 Alexander Remorov Review Four collinear points A, B, C, D form a harmonic bundle (A, C; B, D) when CA : DA CB DB = 1. A pencil P (A, B, C, D) is the set of four lines P A, P B, P C, P D. It is harmonic iff (A, B; C, D) is harmonic. Intersecting a harmonic pencil with any line produces a harmonic bundle. In ABC, points D, E, F are on sides BC, CA, AB. Let F E intersect BC at G. Then (B, C; D, G) is harmonic iff AD, BE, CF are concurrent. A point P is outside or on a circle ω. Let P C, P D be tangents to ω, and l be a line through P intersecting ω at A, B (so that P, A, B are collinear in this order). Let AB intersect CD at AC Q. Then ACBD is a harmonic quadrilateral (i.e. ) and (P, Q; A, B) is harmonic. CB = AD DB Points A, C, B, D lie on a line in this order, and M is the midpoint of CD. Then (A, B; C, D) is harmonic iff AC AD = AB AM. Furthermore, if (A, B; C, D) is harmonic then MD 2 = MA MB. Points A, C, B, D lie on a line in this order. P is 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. P B is the angle bisector of CP D. 3. AP P B. Given a circle ω with center O and radius r and any point A O. Let A be the point on ray OA such that OA OA = r 2. The line l through A perpendicular to OA is called the polar of A with respect to ω. A is called the pole of l with respect to ω. Consider a circle ω and a point P outside it. Let P C and P D be the tangents from P to ω. Then ST is the polar of P with respect to ω. La Hire s Theorem: A point X lies on the polar of a point Y with respect to a circle ω. Then Y lies on the polar of X with respect to ω. Brokard s Theorem: The points A, B, C, D lie in this order on a circle ω 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 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 ω. 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. 1

2 Heavy Machinery For a point P and a circle ω with center O, radius r, define the power of a point P with respect to ω by d(p, ω) = P O 2 r 2. For two circles ω 1, ω 2 there exists a unique line l, called the radical axis, such that the powers of any point on this line with respect to ω 1, ω 2 are equal. In particular, if ω 1 ω 2 = {P, Q} then line P Q is the radical axis of ω 1, ω 2. Radical Axis Theorem: Given three circles ω 1, ω 2, ω 3, let l, m, n be the radical axes of ω 1, ω 2 ; ω 1, ω 3 ; ω 2, ω 3 respectively. Then l, m, n are concurrent at a point called the radical centre of the three circles. Pascal s Theorem: Given a hexagon ABCDEF inscribed in a circle, let P = AB ED, Q = BC EF, R = CD AF. Then P, Q, R are collinear. (An easy way to remember - the three points of intersection of pairs of opposite sides are collinear). Note: Points A, B, C, D, E, F do not have to lie on the circle in this order. Note: It is sometimes useful to use degenerate versions of Pascal s Theorem. For example if C D then line CD becomes the tangent to the circle at C. Brianchon s Theorem: Given a hexagon ABCDEF circumscribed about a circle, the three diagonals joining pairs of opposite points are concurrent, i.e. AD, BE, CF are concurrent. Note: It is sometimes useful to use degenerate versions of Brianchon s Theorem. For example if ABCD is a quadrilateral circumscribed about a circle tangent to BC, AD at P, Q then P Q, AC, BD are concurrent. 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 A 2, B 1 B 2, C 1 C 2 are concurrent. We say that they are perspective with respect to a line 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 to a point iff they are perspective with respect to a line. Sawayama-Thebault s Theorem: A point D is on side BC of ABC. A circle ω 1 with centre O 1 is tangent to AD, BD and Γ, the circumcircle of ABC. A circle ω 2 with centre O 2 is tangent to AD, DC and Γ. Let I be the incentre of ABC. Then O 1, I, O 2 are collinear. It is unlikely that a problem using this theorem will come up on IMO, however it is a nice result and is a good exercise to prove. See problem 5. Homothety Looking at geometric configurations in terms of various geometric transformations often offers great insight in the problem. You should be able to recognize configurations where transformations can be applied, such as homothety, reflections, spiral similarities, and projective transformations. Today we will be focusing on homothety. The powerful thing about homothety is that it preserves angles and tangency. Consider two circles ω 1, ω 2 with centres O 1, O 2. There are two unique points P, Q, such that a homothety with centre P and positive coefficient carries ω 1 to ω 2, and a a homothety with centre Q and negative coefficient carries ω 1 to ω 2. P is called the exsimilicentre, and Q is called the insimilicentre of ω 1, ω 2. Some useful facts: 2

3 1. P is the intersection of external tangents to ω 1, ω 2. Q is the intersection of internal tangents to ω 1, ω Let ω 1, ω 2 intersect at S, R; P A 1, P A 2 are tangents to ω 1, ω 2 so that A 1, A 2 are on the same side of O 1 O 2 as S. Then P R is tangent to the circumcircle of A 1 RA (P, R; O 1, O 2 ) is harmonic. 4. Monge s Theorem: Given three circles ω 1, ω 2, ω 3. Then the exsimilicentres of ω 1 and ω 2, of ω 1 and ω 3, and of ω 2 and ω 3 are collinear. Proof: Let O 1, O 2, O 3 be the centres of the circles. Let K 1 be the intersection of the common tangents of ω 1, ω 2 and ω 1, ω 3. Define K 2, K 3 similarly. Then K i A i is the angle bisector of K i in K 1 K 2 K 3. Hence K 1 A 1, K 2 A 2, K 3 A 3 are concurrent. The result follows by Desargues theorem. A proof without using Desargues theorem: let X 3 be the exsimilicentre of ω 1, ω 2 ; define X 1, X 2 similarly. Apply Menelaus Theorem to X 1 X 2 O Monge-d Alembert Theorem: Given three circles ω 1, ω 2, ω 3. Then the exsimilicentre of ω 1, ω 2, the insimilicentre of ω 1, ω 3 and the insimilicentre of ω 2, ω 3 are collinear. Proof: Let O 1, O 2, O 3 be the centres of the circles; X 3 be the exsimilicentre of ω 1, ω 2 ; define X 1, X 2 similarly. Apply Menelaus Theorem to O 1 O 2 O 3. Problems Some of these problems are lemmas from Yufei Zhao s handout on Lemmas in Eucledian Geometry. The lemmas cannot be quoted on a math contest, so make sure to know their proofs! 1. The incircle ω of ABC has centre I and touches BC at D. DE is the diameter of ω. If AE intersects BC at F, prove that BD = F C. 2. The incircle of ABC touches BC at E. AD is the altitude in ABC; M is the midpoint of AD. Let I a be the centre of the excircle opposite to A of ABC. Prove that M, E, I a are collinear. 3. A circle ω is internally tangent to a circle Γ at P. A and B are points on Γ such that AB is tangent to ω at K. Show that P K bisects the arc AB not containing point P. 4. Let Γ be the circumcircle of ABC and D an arbitrary point on side BC. The circle ω is tangent to AD, DC, Γ at F, E, K respectively. Prove that the increntre I of ABC lies on EF. 5. Prove the Sawayama-Thebault s Theorem. 6. Γ is the circumcircle of ABC. The incircle ω is tangent to BC, CA, AB at D, E, F respectively. A circle ω A is tangent to BC at D and to Γ at A, so that A and A are on different sides of BC. Define B, C similarly. Prove that DA, EB, F C are concurrent. 7. (Romania TST 2004) The incicrle of a non-isosceles ABC is tangent to sies BC, CA, AB at A, B, C. Lines AA, BB intersect at P, AC and A C at M, and lines B C and BC at N. Prove that IP MN. 3

4 These are very non-trivial problems; the last few are very hard. 8. (Iran TST 2007) The incircle ω of ABC is tangent to AC, AB at E, F respectively. Points P, Q are on AB, AC such that P Q is parallel to BC and is tangent to ω. Prove that if M is the midpoint of P Q, and T the intersection point of EF and BC, then T M is tangent to ω. 9. (Romania TST 2007) The incircle ω of ABC is tangent to AB, AC at F, E respectively. M is the midpoint of BC and N is the intersection of AM and EF. A circle Γ with diameter BC intersects BI, CI at X, Y respectively. Prove that NX NY = AC AB. 10. (Romania TST 2007) ω a, ω b, ω c are circles inside ABC, that are tangent (externally) to each other, and ω a is tangent to AB and AC, ω b is tangent to BA and BC, and ω c is tangent to CA and CB. Let D be the common point of ω b and ω c, E the common point of ω c and ω a, and F the common point of ω a and ω b. Show that AD, BE, CF are concurrent. 11. (Romania TST 2006) Let ABC be an acute triangle with AB AC. Let D be the foot of the altitude from A and Γ the circumcircle of the triangle. Let ω 1 be the circle tangent to AD, BD and Γ. Let ω 2 be the circle tangent to AD, CD and Γ. Let l be the interior common tangent to both ω 1 and ω 2, different from AD. Prove that l passes through the midpoint of BC iff AB + AC = 2BC. 12. (China TST 2006 Generalized) In a cyclic quadrilateral ABCD circumscribed about a circle with centre O, the diagonals AC, BD intersect at E. P is an arbitrary point inside ABCD and X, Y, Z, W are the circumcentres of triangles ABP, BCP, CDP, DAP respectively. Show that XZ, Y W, OE are concurrent. 13. (Iran TST 2009) The incircle of ABC is tangent to BC, CA, AB at D, E, F respectively. Let M be the foot of the perpendicular from D to EF and P be the midpoint of DM. If H is the orthocenter of BIC, prove that P H bisects EF. 14. (SL 2007 G8) Point P lies on side AB of a convex quadrilateral ABCD. Let ω be the incircle of CP D, and let I be its incenter. Suppose that ω is tangent to the incircles of triangles AP D and BP C at points K and L, respectively. The lines AC and BD meet at E, and let lines AK and BL meet at F. Prove that points E, I, and F are collinear. 15. (SL 2008 G7) Let ABCD be a convex quadrilateral with BA BC. Denote the incircles of ABC and ADC by k 1 and k 2 respectively. Suppose that there exists a circle k tangent to lines AD, CD, to ray BA beyond A and to the ray BC beyond C. Prove that the common external tangents to k 1 and k 2 intersect on k. 16. (Iran TST 2010) Circles ω 1, ω 2 intersect at P, K. Points X, Y are on ω 1, ω 2 respectively so that XY is tangent externally to both circles and XY is closer to P than K. XP intersects ω 2 for the second time at C and Y P intersects ω 1 for the second time at B. BX and CY intersect at A. Prove that if Q is the second intersection point of circumcircles of ABC and AXY then QXA = QKP. 4

5 Hints 1-3. Straightforward. 4. Extend KE to meet Γ at M. What can you say about A, I, M? 5. Use problem 4. What can you say about O 1 DO 2? 6. Lots of circles and points of tangency... Which theorem to use? 7. Poles and Polars are BACK. 8. Let ω be tangent to BC at D. Let S be the point of intersection of AD with ω. What can you say about the relation between T and AD with respect to ω? 9. Prove that X, Y lie on EF. 10. Draw the centres of the circles. Which theorem(s) should you be using here? 11. It is obvious which theorem to use here. What can you say about O 1, D, M, O 2? 12. Let M be the point of intersection of circumcircles of BP C, AP D and N the point of intersection of circumcircles of BP A, CP D. Consider the circumcentre of P NM. 13. Harmonic division. 14. Consider the circle tangent to AB, BC, DA. Find two more circles. Again lots of circles Let ω 1, ω 2 be tangent to AC at J, L. Prove that AJ = CL. Draw some excircles. Draw some lines parallel to AC. 16. This is a hard and beautiful problem. Let O be the intersection of AQ and XY. We want to use the radical axis theorem... Where is the third circle? References 1 Yufei Zhao, Lemmas in Euclidean Geometry, 2 Various MathLinks Forum Posts; in particular posts by Cosmin Pohoata and luisgeometria, 5

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