ON TWO SPEIL POINTS IN TRINGLE KPIL PUSE 1. Introduction Today we will learn about two intriguing special points in triangle that seem to have many interesting properties. You will see many examples where spotting these hidden points often makes the problem very easy. These points do not seem to have a special name in the mathematical folklore. eing informal we take the liberty to call these two points as Humpty-dumpty points. s you wil see, since these points are vertex dependent, we shall call them X -Humpty Dumpty points whenever they correspond to vertex X. 2.1. Humpty point. 2. Some basic facts about Humpty-Dumpty points Definition 1. In triangle the Humpty point P is defined to be a point inside triangle such that P = P and P = P P Figure 1. -Humpty Point 1) lies on median of Facts about P 2)lies on appolonius circle of that is, = P P 1
2 KPIL PUSE 3), P, H, are concyclic, where H is the orthocentre of 4)HP P 2.2. Dumpty point. Definition 2. In triangle, the Dumpty point Q is defined to be a point inside triangle such that Q = Q and Q = Q Q Figure 2. -Dumpty point 1) lies on symmedian facts about Q 2) It is the centre of spiral symilarity sending Q to Q, ie sending to 3), Q, O, are concyclic where O is circumcentre of. 4) OQ Q. lso notice that the humpty Dumpty points are isogonal conjugates of each other. We recommend you to try to prove these facts by yourself. This will help you in befriending the humpty-dumpty points well. If you are done, then lets move forward! 3. Examples Problem 1 (ELMO 2014). In triangle H, O are respectively the orthocentre and circumcentre respectively. ircle O intersects a circle with diameter O at M. M intersects O again at X. Similarly H intersects circle with diameter H at N and N intersects H second time at Y. Prove that MN XY
ON TWO SPEIL POINTS IN TRINGLE 3 M O H N X Y Figure 3. ELMO 2014 Proof. We immediately realize that N, M are Humpty dumpty points of triangle.indeed, HM M and M lies on H implies its Humpty point and similarly for the other. Using the fact that they are isogonal conjugates we chase some angles and see that M Y and N X. This implies that is, M X = N Y M.Y =. = N.X and we are done. Problem 2 (USMO 2008). is a triangle. M, N are respectively the midpts of,. The perpendicular bisector of, meet the median at D, E respectively. D and E meet at F. Prove that, M, F, N are concyclic. Proof. h, yes F is none other than the dumpty point of. To reach the desired conclusion, see that OF F, ON, OM directly implies, M, O, N, F are concyclic with O as diameter. nother way to finish would have been to see that since a spiral similiarity centered at F maps to, it also maps the midpt of to midpoint of that is, M to N. The angle of spiral similarity is F which is 180, and this must also be the MF N and we finish.
4 KPIL PUSE M N F D E Figure 4. USMO 2008 Problem 3 (US TSTST 2015). is a scalene triangle. K a, L a, M a are respectively the intersections with the interior angle bisector,exterior angle bisector and median from. ircle K a L a intersects M a again at X a. Similarly define X b, X c. Prove that the circumcentre of X a X b X c lies on euler line of. H X c X a G X b Figure 5. US TSTST 2015 Proof. The problem looks really monstrous at first sight, but we will see how innocent it is. learly, K a L a is the applonius circle of and so we have X a as the Humpty point of. let us invoke another property of X a, namely
ON TWO SPEIL POINTS IN TRINGLE 5 HX a M a, similarly for b, c. what do we get? Since M a, M b, M c are concurrent at centroid G, which implies X a, X b, X c lies on a circle with diameter HG. So the centre is midpt of HG, lies on euler line indeed. Problem 4 (US TST 2005). P is a point inside triangle such that P = P, P = P. The perpendicular bisector of P meets at Q. If O is the circumcentre of then prove that QP = 2 OQ P O Q M Figure 6. US TST 2005 Proof. Sure enough, it shouldnt take us much time to realize that P is the Humpty point of, hence P is median. Let P intersect at M. The condition of problem requires 90 1 2 QP = 90 OQ QM = QOM QMO is cyclic, that is QO = 90 or Q tangent to. This suggests that we should start working by taking Q such that Q tangent to and try to prove that Q = Q P. We know that = Q 2. also we Q 2 know = P, from properties of Humpty point. Hence we have P P P = Q 2 Q 2. We show that this property leads to Q P tangent to P. Indeed, if tangent from P to P meets at Q then P P = Q 2 Q 2 Q 2 so Q 2 = Q 2 Q 2 which implies Q = Q. Now by power of point, Q 2 = Q.Q = Q P 2. Hence Q = Q P and we are done.
6 KPIL PUSE 4. excercises Given are some problems for the reader to try. Exercise 1. The symmedian of meets its circumcircle at K. Reflection of K in is K. Prove that K is a median. Exercise 2. P is a variable point on side of triangle. M, N are respectively on, such that P M, P N. Prove that as P varies on, MN passes through a fixed point. Exercise 3. M, N are points on semicircle with diameter and centre O. NM meets at X. circles MO, NO meet at K. Prove that XK KO Exercise 4. Q is the dumpty point of. D is altitude from on. Prove that DQ bisects the line segment joining the midpoints of, Exercise 5. P is point on symmedian of. O 1, O 2 are respectively circumcentres of P, P. If O is circumcentre of, prove that O bisects O 1 O 2 Exercise 6. D is an altitude from to in triangle. circle with centre on D is tangent externally to O at X, where O iis circumcentre of. Prove that X is the symmedian.