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Let's Talk About Symmedians!

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Let's Talk About Symmedians! Let's Talk About Symmedians! Sammy Luo and Cosmin Pohoata Abstract We will introduce symmedians from scratch and prove an entire collection of interconnected results that characterize them. Symmedians represent a very important topic in Olympiad Geometry since they have a lot of interesting properties that can be exploited in problems. But first, what are they? Definition. In a triangle ABC, the reflection of the A-median in the A-internal angle bi- sector is called the A-symmedian of triangle ABC. Similarly, we can define the B-symmedian and the C-symmedian of the triangle. A I BC X M Figure 1: The A-symmedian AX Do we always have symmedians? Well, yes, only that we have some weird cases when for example ABC is isosceles. Then, if, say AB = AC, then the A-median and the A-internal angle bisector coincide; thus, the A-symmedian has to coincide with them. Now, symmedians are concurrent from the trigonometric form of Ceva's theorem, since we can just cancel out the sines. This concurrency point is called the symmedian point or the Lemoine point of triangle ABC, and it is usually denoted by K. //As a matter of fact, we have the more general result. Theorem -1. Let P be a point in the plane of triangle ABC. Then, the reflections of the lines AP; BP; CP in the angle bisectors of triangle ABC are concurrent. This concurrency point is called the isogonal conjugate of the point P with respect to the triangle ABC. We won't dwell much on this more general notion here; we just prove a very simple property that will lead us immediately to our first characterization of symmedians. Mathematical Reflections 4 (2013) 1 Theorem 0 (Steiner's Theorem). If D is a point on the sideline BC of triangle ABC, and if the reflection of the line AD in the internal angle bisector of the angle A intersects the line BC at a point E, then BD BE AB2 · = CD CE AC2 A BC DE Figure 2: Theorem 0 Proof. From the Ratio Lemma, we write DB AB sin DAB EB AB sin EAB = · and = · : DC AC sin DAC EC AC sin EAC Thus, keeping in mind that \DAB = \EAC and \DAC = \EAB, by multiplying, we obtain that BD BE AB2 · = ; CD CE AC2 as claimed. As a corollary, we thus get the following result. Characterization 1. In a triangle ABC with X on the side BC, we have that XB AB2 = XC AC2 if and only if AX is the A-symmedian of triangle ABC. This represents what is perhaps the most important characterization of the A-symmedian of the triangle and all the results that we will prove next will return to this, more or less. Characterization 2. Let ABC be a triangle and let X be a point on the side BC. Obviously, for any point P on the line AX, we have that δ(P; AB) δ(X; AB) = δ(P; AC) δ(X; AC) In other words, the ratio of the distances from P to the sides is independent of the point P chosen on AX. Now, the claim is the following. For any point P on AX, we have that δ(P; AB) δ(X; AB) AB = = δ(P; AC) δ(X; AC) AC if and only if AX is the A-symmedian of triangle ABC. Mathematical Reflections 4 (2013) 2 A P BC X Figure 3: Characterization 2 Proof. By Characterization 1, we know that AX is the A-symmedian if and only if XB AB2 = : XC AC2 Hence, by now using the Ratio Lemma, we get that AX is the A-symmedian and only if sin XAB AB = : sin XAC AC But for any point P on AX, we have that δ(P; AB) δ(X; AB) sin XAB = = : δ(P; AC) δ(X; AC) sin XAC Hence, we immediately obtain the conclusion that δ(P; AB) δ(X; AB) AB = = δ(P; AC) δ(X; AC) AC if and only if AX is the A-symmedian of triangle ABC. X V T U A S C B Figure 4: Characterization 3 Mathematical Reflections 4 (2013) 3 Characterization 3. Let ABC be a triangle and let ACUV and ABST be the squares constructed on the sides which are directed towards the exterior of the triangle. Let X be the circumcenter of triangle AT V . Then, the line AX is the A-symmedian of triangle ABC. Proof. The point X, being the circumcenter of triangle AT V , lies on the line bisectors 1 1 of segments AV and AT . Hence, we have that δ(X; AB) = 2 AT and δ(X; AC) = 2 AV ; thus, we get that δ(X; AB) AT AB = = ; δ(X; AC) AV AC and so by Characterization 2, we get that AX is the A-symmedian of triangle ABC. Characterization 4 (BMO 2009). Let MN be a line parallel to the side BC of a triangle ABC, with M on the side AB and N on the side AC. The lines BN and CM meet at point P . The circumcircles of triangles BMP and CNP meet at two distinct points P and Q. Then, the line AQ is the A-symmedian of triangle ABC. A MN P C B Q Figure 5: Characterization 4 Proof. Note that \BQM = \BP M = \CPN = \CQN and \MBQ = \CPQ = \CNQ; thus triangles BQM and NQC are similar. So, we get that δ(Q; AB) δ(Q; MB) BM AB = = = ; δ(Q; AC) δ(Q; NC) CN AC hence AQ is the A-symmedian of ABC, by Characterization 2. Characterization 5 (Lemoine's Pedal Triangle Theorem). The symmedian point K of triangle ABC is the only point in the plane of ABC which is the centroid of its own pedal triangle. Proof. For the direct implication, let D, E, F be the projections of K on the sides BC, CA, AB and take X to be the intersection of DK with EF . We would like to show that X is the midpoint of EF , since after that we could just repeat the argument for EY and FZ and conclude that K is the centroid of DEF . Mathematical Reflections 4 (2013) 4 A E X F K Z Y C B D Figure 6: Characterization 5 By the Ratio Lemma, we know that XE KE sin XKE = · : XF KF sin XKF However, K obviously lies on the A-symmedian, thus by Characterization 2, KE δ(K; AC) AC = = : KF δ(K; AB) AB Furthermore, \XKE = \C and \XKF = \B since the quadrilaterals KDCE and KF BD are cyclic; thus, we conclude that XE AC sin C AC AB = · = · = 1: XF AB sin B AB AC This proves that X is the midpoint of EF and settles the direct implication. As for the converse, things are essentially similar. Now, we know that K is a point having projections D; E; F so that XE KE sin XKE 1 = = · : XF KF sin XKF The equalities \XKE = \C and \XKF = \B coming from the cyclic quadrilaterals KDCE and KF BD are independent of K being the symmedian point; thus, we immediately get that KE AB = : KF AC Hence, by Characterization 2, we conclude that K needs to lie on the A-symmedian, and similarly we can do that for the vertices B and C; thus we get that K is the symmedian point of the triangle. This completes the proof. Characterization 6. Let the tangents at vertices B and C of triangle ABC to the circumcircle meet at a point X. Then, the line AX is the A-symmedian of triangle ABC. Proof. Let T be the intersection of AX with the side BC. Since this point lies in the interior of the segment BC, we notice again that, by Characterization 1, it is enough to TB AB2 show that TC = AC2 . In this case, the line AX would represent the A-symmedian and we Mathematical Reflections 4 (2013) 5 A O C B T X Figure 7: Characterization 6 TB AB2 would be done. So, let's prove that TC = AC2 . This is where the Ratio Lemma comes in. We have that TB XB sin TXB = · ; TC XC sin TXC but XB = XC as they are both tangents from the same point to the circumcircle of ABC; TB sin TXB hence TC = sin TXC . Now, we apply the Law of Sines twice, in triangles XAB and XAC. We get that AB AX AX = = sin TXB sin XBA sin(B + XBC) and AC AX AX = = : sin TXC sin XCA sin(C + XCB) But \XBC = \XCB = \A, since the lines XB and XC are both tangent to the circumcircle of ABC. Hence, it follows that AB AX AC AX = and = : sin TXB sin C sin TXC sin B Therefore, by dividing the two relations, we conclude that TB sin TXB AB sin C AB2 = = · = ; TC sin TXC AC sin B AC2 where the last equality holds because of the Law of Sines applied in triangle ABC. This completes the proof. Mathematical Reflections 4 (2013) 6 Corollary 6'. If D, E, F denote the tangency points of the incircle with the sides BC, CA, AB of triangle ABC, then the lines DA, EB, FC are the symmedians of triangle DEF . Characterization 7. Suppose X is a point of the circumcircle of ABC, different from XB AB the vertex A, such that XC = AC . Then, the line AX is the A-symmedian of triangle ABC. A I O C B T X Figure 8: Characterization 7 Note that this means nothing but that the A-symmedian is the radical axis of the cir- cumcircle of ABC and the A-Appolonius circle.
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