IMO 2016 Training Camp 3

Transcription

1 IMO 2016 Training amp 3 ig guns in geometry 5 March 2016 At this stage you should be familiar with ideas and tricks like angle chasing, similar triangles and cyclic quadrilaterals (with possibly some ratio hacks). ut at the IMO level there is no guarantee that these techniques are sufficient to solve contest geometry problems. It is therefore timely to learn more identities and tricks to aid our missions. 1 Harmonics. Ever thought of the complete quadrilateral below? P E F Q A D y eva s theorem we have E EP F P F A AD E = 1 and by Menelaus theorem, D EP F P F A A = 1 (note: the negative ratio simply denotes that is not on segment A.) We therefore have: AD D : A = 1.( ) We call any family of four collinear points (A, ;, D) satisfying (*) a harmonic bundle. A pencil P (A,,, D) is the collection of lines P A, P, P, P D. We name it a harmonic pencil if (A, ;, D) is harmonic (so the line P (A,,, D) above is indeed a harmonic pencil). As AD D = P A sin AP D A and P sin P D = P A sin AP we know that P sin P 1

2 AD D : A sin AP D sin AP sin AP D sin AP = : we know that = iff P (A,,, D) sin P D sin P sin P D sin P is harmonic pencil (assumming that P and P D are different lines, of course). These imply that for a harmonic bundle (A, ;, D) and a point P not on the line connecting A,,, D, if a line l intersects P A, P, P, P D at A,,, D, then (A, ;, D ) is harmonic. Let s turn back to the diagram above. We can see that if P D, F, AE are concurrent at Q and A, F, E are collinear then (A, ;, D) is harmonic. We can generalize above to the following: Let triangle P A with points D, E, F on sides A, P, P A. Let be a point on line A. Then any two of the three below imply the third: 1. (A, ;, D) is harmonic. 2. P D, F, AE are concurrent. 3., F, E are collinear. Notice that if EF is parallel to A then is actually point of infinity, so A 1. D is therefore the midpoint of A if (A, ;, D) is harmonic. Two other lemmas. If, A, D, are collinear in that order and M is th midpoint of A, then (A, ;, D) is harmonic A = D M MA 2 = M MD. Points (, A,, 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, ;, D) is harmonic. 2. P D is the angle bisector of P D. 3. AP P. Harmonic quadrilateral. A cyclic quadrilateral AD is named harmonic quadrilateral if A = AD. Some facts about this quadrilateral: D 1. Let ω be the circle circumsribing AD. Then point of intersection of tangents to ω at A and lie on D. Likewise, the point of intersection of tangents to ω at and D lie on A. 2. Let P be any point on ω. Then 1 = A : AD D P (A, ;, D) is harmonic. = sin AP sin P : sin AP D sin DP. Therefore, 3. Let Q be the common point of line A, tangent to ω at and tangent to ω at D.Let R = A D. Then (Q, R; A, ) is harmonic. 2 Poles and polars. 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. Notice that if A lies outside ω and and D are on ω such that A and AD are tangent to ω, then D l, the polar of A. 2

3 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. From this identity, we know that if points A and have polars l A and l, then the pole of line A is the point of intersection of l A and l. rokard s Theorem: The points A,,, D lie in this order on a circle ω with center O. A and D intersect at P, A and D intersect at Q, AD and 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. 3 Other heavy machineries. Pascal s Theorem: Given a hexagon ADEF inscribed in a circle, let P = A ED, Q = EF, R = D 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,,, D, E, F do not have to lie on the circle in this order. textbf Note: It is sometimes useful to use degenerate versions of Pascal s Theorem. For example if D then line D becomes the tangent to the circle at. rianchon s Theorem: Given a hexagon ADEF circumscribed about a circle, the three diagonals joining pairs of opposite points are concurrent, i.e. AD, E, F are concurrent. Note: It is sometimes useful to use degenerate versions of rianchon s Theorem. For example if AD is a quadrilateral circumscribed about a circle tangent to, AD at P, Q then P Q, A, D are concurrent. Desargues Theorem: Given two triangles A and A we say that they are perspective with respect to a point when A 1 A 2, 1 2, 1 2 are concurrent. We say that they are perspective with respect to a line when A 1 1 A 2 2, A 1 1 A 2 2, are collinear. Then two triangles are perspective with respect to a point iff they are perspective with respect to a line. Proof? Use Menelaus theorem for a few times. Sawayama-Thebault s Theorem: A point D is on side of A. A circle ω 1 with centre O 1 is tangent to AD, D and Γ, the circumcircle of A. A circle ω 2 with centre O 2 is tangent to AD, D and Γ. Let I be the incentre of A. Then O 1, I, O 2 are collinear. 4 Homothety: triangles and circles. Refer to Desargues theorem above: what happened if A 1 1 A 2 2, A 1 1 A 2 2, ? The fact is, as long as the triangles are not congruent, the lines A 1 A 2, 1 2, 1 2 are still concurrent. (If congruent and are of similar orientation then the three lines are parallel!) The point P at which they concur is called the centre of homothety. Definition. Homothety under a center P and ratio k is the mapping of any point A to A in a plane such that A P = k AP, and A, P, P collinear. (Again, if k < 0 then A and A are at the different side of P ). Fun fact 1: If A is the image of A under some homothety, then all relevant points of A (e.g. centroid, orthocentre, and even circumcircle) is the image of of the corresponding point of A. Fun fact 2: Denote A and A as above. Then if the centre of homothety lies on one relevant line (e.g. Euler s line, A altitude) of A, then this corresponding relevant line of A coincides with the former; otherwise, two such lines are parallel to each other. onsider 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 coeffcient carries ω 1 to ω 2, and a a homothety with centre Q and negative coeffcient carries ω 1 to ω 2. P is called the exsimilicentre, and Q is called 3

4 the insimilicentre ofω 1, ω 2.We assumed that the two circles are not concentric (otherwise the only centre of simlarity is the centre!) and not congruent (otherwise the exsimilicenter is the point of infinity). A H I O A H I O Figure showing how triangles, relevant points and circles can be homothetic. P Some useful facts: 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 to ω 2 intersect at S;, R, and P A 1, P A 2 are tangents toω 1 to 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 A 1 RA (P, Q; O 1, O 2 ) is harmonic. (Why?) 4. Monge s Theorem: Given three circles ω 1, ω 2, ω 3. Then the exsimilicentres of ω 1, ω 2, of ω 1, ω 3, and of ω 2, 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 Deargues theorem: use Menelaus theorem to the triangle formed by the centres of the three circles. 5. Monge-d Alembert s theorem: Given three circles ω 1, ω 2, ω 3. Then the insimilicentres of ω 1, ω 2, of ω 1, ω 3, and the exsimilicentre of ω 2, ω 3 are collinear. 6. If the exsimilicentre of two circles ω 1 and ω 2 lies on the circumference of ω 1 then ω 1 is internally tangent to ω 2, and the exsimicenter is the point of tangency of the circles. 7. If the insimilicentre of two circles ω 1 and ω 2 lies on the circumference of ω 1 then ω 1 is externally tangent to ω 2, and the insimicenter is the point of tangency of the circles. 4

5 5 Examples. 1. IMO 2012, #5. Let A be a triangle with A = 90, and let D be the foot of the altitude from. Let X be a point in the interior of the segment D. Let K be the point on the segment AX such that K =. Similarly, let L be the point on the segment X such that AL = A. Let M be the point of intersection of AL and K. Show that MK = ML. Solution. Denote ω A as the circle with centre A and radius A, and ω as the circle with centre and radius. Now considering ω A and letting the pole of X be R, we know that the pole of D is, so by La Hire s theorem R lies on D. y La Hire s theorem the polar of X contains both (as X D) and R (polar of R is X) and by definition this polar is perpendiular to AX. We therefore have RX A and R AX, so R is the orthocentre of triangle AX, and RL is tangent to X (by definition again). y similar reasoning, R is the pole of AX with respect to ω, and therefore RK is tangent to ω. Now that R is on D, the radical axis of ω A and ω. Thus the lengths of tangent (i.e. the square root of power of point from R to the two circles) are equal, hence RK = RL. Now MK 2 = RM 2 RK 2 = RM 2 RL 2 = ML 2, or MK = ML. R A K X D M L 2. IMO 2004, G8. Given a cyclic quadrilateral AD, let M be the midpoint of the side D, and let N be a point on the circumcircle of triangle AM. Assume that the point N is different from the point M and satisfies AN N = AM M. Prove that the points E, F, N are collinear, where E = A D and F = DA. Solution. Let P be the second point of intersection between D and the circle (AM), and let G = A D. A very simple computation, based on the fact that GD G = GA G = GM GP and M is the midpoint of D will show that P is, in fact, the harmonic conjugate of, D art G, so it belongs to EF. ANM is a harmonic quadrilateral, so (P A, P ; P N, P M), or P (A, ; N, G) is harmonic pencil. Now let EF A = H, we have (G, H; A, ) harmonic, so P (G, H; A, ), or P (G, E; A, ) is a harmonic pencil too. This means lines P H and P N coincide, and we are done. 5

6 3. JOM 2013, G7. Given a triangle A, let l be the median corresponding to the vertex A. Let E, F be the feet of the perpendiculars from, to A, A. Reflect the points E, F across l to the points P, Q. Let AP, AQ intersect at X, Y. Let Γ 1, Γ 2 be the circumcircles of EQY, F P X. Prove that A lies on the line connecting the centers of Γ 1, Γ 2. Solution. Here a b refers to the intersection between objects a and b. Let Γ 1 = U, Y, Γ 1 P Q = V, Q. Let Γ 2 = H, X, Γ 2 P Q = I, P. Let O 1, O 2 be the centers of the circles Γ 1, Γ 2. Let D be the midpoint of. Let P Q =G. A G U V R P I S E Q F HX D Y Firstly, notice that EP AD and AE E, thus EAD = 90 AEP = P E = AF P = AQE = EUY = EUD, from which it follows that A, E, D, U are concyclic. It follows immediately that AUY = AUD = DE = DE = AF E = AQP = V UY which proves that A, U, V are collinear. Now note that if UQ V Y = R, then since the complete quadrilateral formed by U, V, Q, Y, G, A has V, U, Q, Y concyclic on Γ 1, then by rokard s Theorem, GR is the polar of A with respect to Γ 1. Analogously, if IX P H = S, then GS is the polar of A with respect to Γ 2. ut since (GA, GP ; GS, GX) and (GA, GQ; GR, GY ) are both harmonic pencils, G, R, S are collinear. Since by definition AO 1 GR, AO 2 GS, it follows immediately that AO 1 O 2 must be collinear, and we re done. 4. APMO 2013, #5. Let AD be a quadrilateral inscribed in a circle ω, and let P be a point on the extension of A such that P and P D are tangent to ω. The tangent at intersects P D at Q and the line AD at R. Let E be the second point of intersection between AQ and ω. Prove that, E, R are collinear. Solution. The problem is equivalent to proving E, AD, Q are collinear. If we denote X = DE A, Y = AE D then by rokard s theorem Y is the polar of the point E AD so the problem statement of proving that Q contains this point is equivalent of proving that XY contains the pole of Q, which is (all pole and polar relations are w.r.t. ω.) 6

7 A Y E Q D P X Recall from the trigonometry notes that (don t do this is contests the trigonometric sin AXY sin AX identity isn t well known!) it suffices to prove that = sin DXY sin DX. Now sin AXY AY sin XAY sin ADY from XAY = XDY we have = = sin DXY DY sin XDY sin DAY = A DE. sin AX A sin XA A = =. Therefore it is clear that we what we need is sin DX D sin XD D E A DE = D E A. Now since AD and AED are harmonic quadilaterals, D A= AD and A ED=AD E. Therefore A DE=AD E=A D E. 5. IMO 2007, G8. Point P lies on side A of a convex quadrilateral AD. Let ω be the incircle of triangle P D, and let I be its incenter. Suppose that ω is tangent to the incircles of triangles AP D and P at points K and L, respectively. Let lines A and D meet at E, and let lines AK and L meet at F. Prove that points E, I, and F are collinear. Solution. Let J be the centre of the circle Γ tangent to sides A,, AD (or the intersection of internal angle bisectors of AD and A). Then we prove that E and F are the exsimilicentres and insimilicentre of ω and Γ, so they have to lie on line IJ! Denote incircles of AP D and P as ω 1 and ω 2, respectively. As incircles of triangles AP D and P D are tangent to each other, there exists a circle tangent to sides AP, P, D, DA (the reverse of this identity holds true, and is well known. However, as an exercise try to prove it!) Name the circle Γ 1. Similarly there is a circle Γ 2 inscribed in quadrilateral P D. Now, A is the exsimilicentre of Γ 1 and Γ; is the exsimilicentre of Γ 1 and ω. Then by Monge s theorem the exsimilicentre of ω and Γ lies on A. Similarly this exsimilicentre lies on D (by considering exsimilicentres of ω, Γ and Γ 2 ). The exsimilicentre must then be A D = E). Next, A is the exsimilicentre of ω 1 and Γ, K is the insimilicentre of ω 1 and ω. y Monge d Almebert s theorem the insimilicentre of ω and Γ lies on AK. Similarly this insimilicentre lies on L. Therefore, AK L is the insimillicentre of ω and Γ. 6. IMO 2008, #6. Let AD be a convex quadrilateral with A. Denote the incircles of triangles A and AD by ω 1 and ω 2 respectively. Suppose that there exists a circle ω tangent to ray A beyond A and to the ray beyond, which is also tangent to the lines AD and D. Prove that the common external tangents to ω 1 and ω 2 intersect 7

8 on ω. Solution. Throughout the problem the pole-polar relation is with respect to ω. Denote G = A D. Also denote the point of tangency of lines A, D, AD, as U, V, W, X respectively. We prove that G is th intersection of UV and W X. Indeed, the polar of A is UW and the polar of is V X, so the pole of A is UW V X, namely P. In a similar way we deduce that the pole of D is UX V W = Q. So the polar of G = A D is P Q. y rokard s theorem we know that the UV and W X intersect at the polar of P Q, which is G. (note the profuse usage of La Hire s theorem!) Also that pole of A (i.e. P ) lies on the pole of Q (i.e. P G), we know that Q lies on A (La Hire s theorem again). O 1 A G O 2 U V D Z P W X Q O y Monge D Alembert s theorem we get that exsimilicenter of ω and ω 1 (i.e. ), insimilicenter of ω and ω 2 (i.e. D), and insimilicenter of ω 1 and ω 2 is collinear. ut A is a common inner tangent of ω 1 and ω 2, we infer that G is itself the insimilienter of ω 1 and ω 2. Also that, if we denote O 1 and O 2 as centers of ω 1 and ω 2 and Z as the exsimilicenter we desire, then (Z, G; O 1, O 2 ) is harmonic bundle and (OZ, OG; OO 1, OO 2 ) is a harmonic pencil. Since OO 1 UX, OO 2 V W OG P Q (pole and polar relation) and (QG, QP ; QU, QV ) (or (A, QP ; UX, V W )) is harmonic pencil, we have OZ A. and since OP A (P = pole of A), P, O, Z collinear. 8

9 G U V Z V W W P U X Q X Q O Now we try to locate the polars of O 1 and O 2. As, O, O 1 are collinear, the polars of O 1 and (i.e. UX) are collinear. Moreover, the pole of AO 1, U, lies on polar of A which is UW, and with AO 1 bisects A, A OU, AO 1 OU and A OP we have OU bisects UOP. So the polar of O 1 is the line passing through U and parallel to V X. Let this line intersect V X at X and we have UU U P = XX X P = r where r is d the radius of ω and d is the distance OP (so A is of distance r2 from O.) Similarly if d the polar of O 2 intersect V X and UW at V and W respectively then V W V W and V V V P = W W W P = r d. So Q = U X V W is the pole of O 1 O 2. Also notice that U V W X d is the image of UV W X under the homothety at center P and ratio, if we draw the r + d line parallel to A from P (namely l 1 ) and Q (namely l 2 ) we see that l 1, l 2 anda are in that order, with l 1 at distance d from O (remember that A OP and l 1 OP ) and A at distance r2 d from O. So l 2 at distance d + ( r2 d d)( d )=d + (r d) = r from O, r + d i.e. l 2 is tangent to ω. Finally, as OP and l 2 meet at the point of tangency of l 2 and W this point is also on the polar of Q (recall that l 2 passes through Q ). ut with Z O 1 O 2 and O 1 O 2 has pole Q we have the polar of Q contain Z. So Z is indeed the tangency point, or in other words, Z ω. 7. RMM 2012, #6. Let A be a triangle and let I and O denote its incentre and circumcentre respectively. Let ω A be the circle through and which is tangent to the incircle of the triangle A; the circles ω and ω are defined similarly. The circles ω and ω meet at a point A distinct from A; the points and are defined similarly. Prove that the lines AA, and are concurrent at a point on the line IO. Solution. Lines AA,, all pass through the radical centre of ω A, ω and ω ; the fact that the three lines of concurrent will be assumed below. Let T A, T, T be the points of tangency of the incircle of triangle A (namely ω) and circles ω A, ω, ω, respectively. learly the tangent to ω at T A is also tangent to ω A, and is therefore the radical axis of these two circles. Similarly the tangent to ω to T is the radical axis of 9

10 ω and ω. Denote Z as the intersection of the two tangents which is also the radical centre of ω, ω A and ω. Since lies on both ω and ω A, it is itself on the radical axis of these two circles. So, and Z all lie on the radical axis of ω A and ω. Now let L A, L and L be points of tangency of ω to, A and A respectively. We need to following lemma: T A L A, T L and Z are concurrent. Why? y Pascal s theorem we have: Z = T A T A T T, T A L A T L and T A L T L A are concurrent by considering the ordered pairs (T A, T, L ) and (T, T A, L A ). y considering ordered pairs (L A, T, L ) and (L, T A, L A ) we have = L A L A L L, T A L A T L and T A L T L A. Now get it? and Z both lie on the line pssing through T A L A T L and T A L T L A, hence are collinear with T A L A T L. With lines AA,, concurrent we know that the required point is indeed L A T A L T L T (these three lines all concur at the radical canter of ω A, ω and ω. I A T A I L L T I T O L A I A We proceed with the following: Lemma 2: Line L A T A passes through the excentre of A opposite A (namely I A ; define I and I similarly); similar condition for L T and L T. Proof: Now let the perpendicular from I A to be L A, we know that L A and L A are reflections of each other in (why? We have done this in the trigonometry lecture in the previous camp; what you need to see is that I,, and II A all on the circle with diameter II A, namely Γ!) This implies that the midpoint M of L A L A lie on the perpendicular bisector of. Now, let K be another intersection of L A L A and Γ and L another intersection of L A L A and ω then IKI A = 90. We then obtain K as the 10

11 midpoint of LL A (IK the perpendicular bisector of LL A.) Moreover, by the power of point theorem L A L A = KL A L A I A = 2KL A 1 2 L AI A = LL A LL A M, implying that LM is cyclic and with M = M LL A bisects L. Finally, let L and L intersect ω again at X and Y respectively; from th angle bisector condition we know that L A = L A and XY, the tangent to ω at L A. Now triangles LXY and L are similar, so they are homothetic at centre L. Same goes to their circumcircles and we conclude that ω and circle LM is tangent to another. ω A must therefore be the circumcircle of LM and we conclude that T A = L. and T A, L A, I A collinear. (Theoretically, there are two circles passing through both and, and is tangent to ω. However, one such circle has been degenerated to line. We are (finally) ready to present our proof, that is I A L A, I L and I L concur at IO. Now L A L is the polar of w.r.t. ω, hence perpendicular to I. ut I A and I are both perpendicular to I, so I A I L A L. In a similar manner for the other sides we have triangles I A I I and L A L L homothetic, so the whole problem statement becomes proving that the center of homothety lies on IO. Now for triangle I A I I, I is its orthocenter (lies on Euler line) and O is the nine-point circle (also on Euler s line), and the Euler s lines of L A L L is parallel or will coincide with that of I A I I, which is IO. ut they have a common point I since I is the circumcenter of L A L L, the two Euler lines coincide. This means the center of homothety lies on the Euler s line, which is IO. Q.E.D. (Phew!) 6 Practice problems. 1. Let AD be a quadrilateral circumscribed around a circle ω. Let E, F, G, H be the point of tangency of lines A,, D, DA with ω. Prove that lines (diagonals) A, D, EG, F H are concurrent. 2. Starting from the previous practice problem, prove rokard s theorem. 3. Two circles ω 1 and ω 2 are internally tangent at P (with ω 2 lying inside ω 1 ). Let l be a line tangent to ω 2 at Q and intersect ω 1 at points A and. Let P Q intersect ω 1 again at M. Prove that MA = M. 4. JOM 2013, G2. Let ω 1 and ω 2 be two circles, with centres O 1 and O 2 respectively, intersecting at X and Y. Let a line tangent to both ω 1 and ω 2 at A and, respectively. Let E, F be points of O 1 O 2 such that XE tangent to ω 1 and Y F is tangent to ω 2. Let AE ω 1 = A, and F ω 2 =, D. Show that line O 2 is tangent to the circumcircle of AD and line AO 2 is tangent to circumcircle of D. 5. IMO 2005, G6. Let A be a triangle, and M the midpoint of its side. Let γ be the incircle of triangle A. The median AM of triangle A intersects the incircle γ at two points K and L. Let the lines passing through K and L, parallel to, intersect the incircle γ again in two points X and Y. Let the lines AX and AY intersect again at the points P and Q. Prove that P = Q. 6. RMM 2013, #3. Let AD be a quadrilateral inscribed in a circle ω. The lines A and D meet at P, the lines AD and meet at Q, and the diagonals A and D meet at R. Let M be the midpoint of the segment P Q, and let K be the common point of the segment MR and the circle ω. Prove that the circumcircle of the triangle KP Q and ω are tangent to one another. 11

12 7. IMO 2014, G7 (lemma). Let A be a triangle with circumcircle Ω and incentre I. Let the line passing through I and perpendicular to I intersect the segment and the arc (not containing A) of Ω at points U and V, respectively. Let the line passing through U and parallel to AI intersect AV at X, and let the line passing through V and parallel to AI intersect A at Y. Prove that if the points I, X, and Y are collinear, then V A = V. Note: The original problem asks that, if W and Z be the midpoints of AX and, respectively, then the points I, W, and Z are also collinear. Try it if you dare (after proving the lemma above)! 8. RMM 2010, #3. Let A 1 A 2 A 3 A 4 be a quadrilateral with no pair of parallel sides. For each i = 1, 2, 3, 4, define ω 1 to be the circle touching the quadrilateral externally, and which is tangent to the lines A i 1 A i, A i A i+1 and A i+1 A i+2 (indices are considered modulo 4 so A 0 = A 4, A 5 = A 1 and A 6 = A 2 ). Let T i be the point of tangency of ω i with the side A i A i+1. Prove that the lines A 1 A 2, A 3 A 4 and T 2 T 4 are concurrent if and only if the lines A 2 A 3, A 4 A 1 and T 1 T 3 are concurrent. 9. RMM 2011, #3. A triangle A is inscribed in a circle ω. A variable line l chosen parallel to meets segments A, A at points D, E respectively, and meets ω at points K, L (where D lies between K and E). ircle γ 1 is tangent to the segments KD and D and also tangent to ω, while circle γ 2 is tangent to the segments LE and E and also tangent to ω. Determine the locus, as l varies, of the meeting point of the common inner tangents to γ 1 and γ 2. 7 References. 1. Alexander Remorov: 2010, Projective Geometry-Part 2. anadian IMO Training. 2. Lindsey Shorer: 2015, Summary of Some oncepts in Projective Geometry. anadian IMO Training. 3. osmin Phohata: 2012, AMY , Segment 5, Homothety and Inversion. AwesomeMath LL. 7.1 Problem credit. 1. IMO 2004 (shortlist; solution s credit to an AoPS user), 2005 (shortlist), 2007 (shortlist and solution), 2008, 2012, 2014 (shortlist). 2. APMO Romanian Masters in Mathematics (2010, 2011, 2012, 2013). 4. Junior Olympiad of Mathematics 2013 (solution of G7 is creditted to Justin, the problem proposer). c 2016 IMO Malaysia ommittee 12