Abstract The symmedian point of a triangle is known to give rise to two circles, obtained by drawing respectively parallels and antiparallels to the

Transcription

1 bstract The symmedian point of a triangle is known to give rise to two circles, obtained by drawing respectively parallels and antiparallels to the sides of the triangle through the symmedian point. In this note we will explore a third circle with a similar construction - discovered by Jean-Pierre Ehrmann [3]. It is obtained by drawing circles through the symmedian point and two vertices of the triangle, and intersecting these circles with the triangle s sides. We prove the existence of this circle and identify its center and radius. 1

2 Ehrmann s third Lemoine circle Darij Grinberg written 005; edited The first two Lemoine circles Let us remind the reader about some classical triangle geometry first. Let L be the symmedian point of a triangle B. Then, the following two results 1 are well-known ([], hapter 9): B L U Fig. 1 Theorem 1 Let the parallels to the lines B, B,,, B, B through L meet the lines, B, B, B, B, at six points. These six points lie on one circle, 1 For the definition of a Tucker circle, see below.

3 the so-called first Lemoine circle of triangle B; this circle is a Tucker circle, and its center is the midpoint of the segment UL, where U is the circumcenter of triangle B. (See Fig. 1.) [The somewhat uncommon formulation "Let the parallels to the lines B, B,,, B, B through L meet the lines, B, B, B, B, at six points" means the following: Take the point where the parallel to B through L meets, the point where the parallel to B through L meets B, the point where the parallel to through L meets B, the point where the parallel to through L meets B, the point where the parallel to B through L meets B, and the point where the parallel to B through L meets.] B L Fig. Furthermore (see [] for this as well): Theorem Let the antiparallels to the lines B, B,,, B, B through L meet the lines, B, B, B, B, at six points. These six points lie on one circle, the so-called second Lemoine circle (also known as the cosine circle) of triangle B; this circle is a Tucker circle, and its center is L. (See Fig..) We have been using the notion of a Tucker circle here. This can be defined as follows: 3

4 Theorem 3 Let B be a triangle. Let Q a and R a be two points on the line B. Let R b and P b be two points on the line. Let P c and Q c be two points on the line B. ssume that the following six conditions hold: The lines Q b R c, R c P a, P b Q a are parallel to the lines B,, B, respectively; the lines P b P c, Q c Q a, R a R b are antiparallel to the sidelines B,, B of triangle B, respectively. (ctually, requiring five of these conditions is enough, since any five of them imply the sixth one, as one can show.) Then, the points Q a, R a, R b, P b, P c and Q c lie on one circle. Such circles are called Tucker circles of triangle B. The center of each such circle lies on the line UL, where U is the circumcenter and L the symmedian point of triangle B. Notable Tucker circles are the circumcircle of triangle B, its first and second Lemoine circles (and the third one we will define below), and its Taylor circle.. The third Lemoine circle Far less known than these two results is the existence of a third member can be added to this family of Tucker circles related to the symmedian point L. s far as I know, it has been first discovered by Jean-Pierre Ehrmann in 00 [3]: Theorem 4 Let the circumcircle of triangle BL meet the lines and B at the points b and c (apart from and B). Let the circumcircle of triangle L meet the lines B and B at the point B c and B a (apart from and ). Let the circumcircle of triangle LB meet the lines B and at the points a and b (apart from B and ). Then, the six points b, c, B c, B a, a, b lie on one circle. This circle is a Tucker circle, and its midpoint M lies on the line UL and satisfies LM = 1 LU (where the segments are directed). The radius of this circle is 1 9r 1 + r, where r is the circumradius and r 1 is the radius of the second Lemoine circle of triangle B. We propose to denote the circle through the points b, c, B c, B a, a, b as the third Lemoine circle of triangle B. (See Fig. 3.) 4

5 a B c M L U B a b B c b Fig. 3 The rest of this note will be about proving this theorem. First we will give a complete proof of Theorem 4 in 3-5; this proof will use four auxiliary facts (Theorems 5, 6, 7 and 8). Then, in 6 and 7, we will give a new argument to show the part of Theorem 4 which claims that the six points b, c, B c, B a, a, b lie on one circle; this argument will not give us any information about the center of this circle (so that it doesn t extend to a complete second proof of Theorem 4, apparently), but it has the advantage of showing a converse to Theorem 4 (which we formulate as Theorem 10 in the final paragraph 8). 3. lemma In triangle geometry, most nontrivial proofs begin by deducing further (and easier) properties of the configuration. These properties are then used as lemmas (and even if they don t turn out directly useful, they are often interesting for themselves). In the case of Theorem 4, the following result plays the role of such a lemma: Theorem 5 The point L is the centroid of each of the three triangles b c, B a BB c, a b. (See Fig. 4.) 5

6 a B c L B a b B c b Fig. 4 ctually this result isn t as much about symmedian points and centroids, as it generalizes to arbitrary isogonal conjugates: Theorem 6 Let P and Q be two points isogonally conjugate to each other with respect to triangle B. Let the circumcircle of triangle P meet the lines B and B at the points B c and B a (apart from and ). Then, the triangles B a BB c and B are oppositely similar, and the points P and Q are corresponding points in the triangles B a BB c and B. (See Fig. 5.) Remark. Two points P 1 and P are said to be corresponding points in two similar triangles 1 and if the similitude transformation that maps triangle 1 to triangle maps the point P 1 to the point P. 6

7 B w γ P B a w α B c Q Fig. 5 Proof of Theorem 6. We will use directed angles modulo 180. very readable introduction into this kind of angles can be found in [1]. list of their important properties has also been given in [4]. The point Q is the isogonal conjugate of the point P with respect to triangle B; thus, P B = Q. Since,, B c, B a are concyclic points, we have B a B c = B c, so that BB a B c = B. Furthermore, B c BB a = B. Thus, the triangles B a BB c and B are oppositely similar (having two pairs of oppositely equal angles). By the chordal angle theorem, P B a B c = P B c = P B = Q = Q. Similarly, P B c P a = Q. These two equations show that the triangles B a P B c and Q are oppositely similar. ombining this with the opposite similarity of triangles B a BB c and B, we obtain that the quadrilaterals B a BB c P and BQ are oppositely similar. Hence, P and Q are corresponding points in the triangles B a BB c and B. (See Fig. 6.) Theorem 6 is thus proven. 7

8 B P B a B c P' Fig. 6 Proof of Theorem 5. Now return to the configuration of Theorem 4. To prove Theorem 5, we apply Theorem 6 to the case when P is the symmedian point of triangle B; the isogonal conjugate Q of P is, in this case, the centroid of triangle B. Now, Theorem 6 says that the points P and Q are corresponding points in the triangles B a BB c and B. Since Q is the centroid of triangle B, this means that P is the centroid of triangle B a BB c. But P = L; thus, we have shown that L is the centroid of triangle B a BB c. Similarly, L is the centroid of triangles b c and a b, and Theorem 5 follows. 4. ntiparallels Theorem 5 was the first piece of our jigsaw. Next we are going to chase some angles. Since the points B,, b, c are concyclic, we have b c = B c, so that b c = B. Thus, the line b c is antiparallel to B in triangle B. Similarly, the lines B c B a and a b are antiparallel to and B. We have thus shown: Theorem 7 In the configuration of Theorem 4, the lines b c, B c B a, a b are antiparallel to B,, B in triangle B. Now let X b, X c, Y c, Y a, Z a, Z b be the points where the antiparallels to the lines B, B,,, B, B through L meet the lines, B, B, B, B,. ccording to Theorem, these points X b, X c, Y c, Y a, Z a, Z b lie on one circle around 8

9 L; however, to keep this note self-contained, we do not want to depend on Theorem here, but rather prove the necessary facts on our own (Fig. 7): Partial proof of Theorem. B X c Z a Y a Y c L Z b X b Fig. 7 Since symmedians bisect antiparallels, and since the symmedian point L of triangle B lies on all three symmedians, the point L must bisect the three antiparallels X b X c, Y c Y a, Z a Z b. This means that LX b = LX c, LY c = LY a and LZ a = LZ b. Furthermore, X c X b = B (since X b X c is antiparallel to B), thus Y c X c L = B; similarly, X c Y c L = B, thus LY c X c = X c Y c L = B = B = Y c X c L. Hence, triangle X c LY c is isosceles, so that LX c = LY c. Similarly, LZ b = LX b and LY a = LZ a. Hence, LX b = LX c = LY c = LY a = LZ a = LZ b. This shows that the points X b, X c, Y c, Y a, Z a, Z b lie on one circle around L. This circle is the so-called second Lemoine circle of triangle B. Its radius is r 1 = LX b = LX c = LY c = LY a = LZ a = LZ b. We thus have incidentally proven most of Theorem (to complete the proof of Theorem, we would only have to show that this circle is a Tucker circle, which is easy); but we have also made headway to the proof of Theorem 4. 9

10 B M Y a Y c L U B a B m B c Fig. 8 Proof of Theorem 9, part 1. Now consider Fig. 8. Let B m be the midpoint of the segment B c B a. ccording to Theorem 5, the point L is the centroid of triangle B a BB c ; thus, it lies on the median BB m and divides it in the ratio : 1. Hence, BL : LB m = (with directed segments). Let M be the point on the line UL such that LM = 1 LU (with directed segments); then, LU = LM, so that UL = LU = LM, and thus UL : LM = = BL : LB m. By the converse of Thales theorem, this yields B m M BU. The lines Y c Y a and B c B a are both antiparallel to, and thus parallel to each other. Hence, Thales theorem yields B m B c : LY c = BB m : BL. But since BL : LB m =, we have BL = LB m, so that LB m = 1 BL and therefore BB m = BL + LB m = BL + 1 BL = 3 BL and BB m : BL = 3. onsequently, B mb c : LY c = 3 and B m B c = 3 LY c = 3r 1. It is a known fact that every line antiparallel to the side of triangle B is perpendicular to the line BU (where, as we remind, U is the circumcenter of triangle B). Thus, the line B c B a (being antiparallel to ) is perpendicular to the line BU. Since B m M BU, this yields B c B a B m M. Furthermore, B m M BU yields BU : B m M = BL : LB m (by Thales), so that BU : B m M = and BU = B m M, and thus B m M = 1 BU = 1 r, where r is the circumradius of triangle B. Now, Pythagoras theorem in the right-angled triangle MB m B c yields MB c = B m B c + B m M = (3 r 1 ) + ( ) 1 9 r = 4 r r = 1 9r1 + r. Similarly, we obtain the same value 1 9r 1 + r for each of the lengths MB a, M a, 10

11 M b, M b and M c. Hence, the points b, c, B c, B a, a, b all lie on the circle with center M and radius 1 9r 1 + r. The point M, in turn, lies on the line UL and satisfies LM = 1 LU. This already proves most of Theorem 4. The only part that has yet to be shown is that this circle is a Tucker circle. We will do this next. 5. Parallels a c B L B a b B c b Fig. 9 Since the points b, B a, a, b are concyclic, we have a B a b = a b b, thus B a b = a b. But since a b is antiparallel to B, we have b a = B, so that a b = b a = B, thus B a b = B; consequently, b B a B. Similarly, B c b B and a c. ltogether, we have thus seen: Theorem 8 The lines B c b, a c, b B a are parallel to B,, B. (See Fig. 9.) Proof of Theorem 4, part. If we now combine Theorem 7 and Theorem 8, we conclude that the sides of the hexagon b c a b B c B a are alternately antiparallel and parallel to the sides of triangle B. Thus, b c a b B c B a is a Tucker hexagon, and 11

12 the circle passing through its vertices b, c, B c, B a, a, b is a Tucker circle. This concludes the proof of Theorem 4. One remark: It is known that the radius r 1 of the second Lemoine circle B is r tan ω, where ω is the Brocard angle of triangle B. Thus, the radius of the third Lemoine circle is 1 9r 1 + r = 1 9 (r tan ω) + r = 1 9r tan ω + r = r 9 tan ω different approach: a lemma about four points t this point, we are done with our job: Theorem 4 is proven. However, our proof depended on the construction of a number of auxiliary points (not only B m, but also the six points X b, X c, Y c, Y a, Z a, Z b ). One might wonder whether there isn t also a (possibly more complicated, but) more straightforward approach to proving the concyclicity of the points b, c, B c, B a, a, b without auxiliary constructions. We will show such an approach now. It will not yield the complete Theorem 4, but on the upside, it helps proving a kind of converse. For a quick proof of this fact, let X, Y, Z denote the feet of the perpendiculars from the point L to the sides B,, B of triangle B. Then, the radius r 1 of the second Lemoine circle is r 1 = LY a. Working without directed angles now, we see that LY a = LX (from the right-angled sin triangle LXY a ), so that r 1 = LY a = rewrites as r 1 = ( LX a ) = r LX sin. Since sin = a r LX r. This rewrites as a a r r 1 = a LX. Similarly, c r r 1 = c LZ. dding these three equations together, we obtain a r r 1 + b r r 1 + c r r 1 = a LX + b LY + c LZ. by the extended law of sines, this b r r 1 = b LY and On the other hand, let S denote the area of triangle B. But the area of triangle BL is 1 a LX (since a is a sidelength of triangle BL, and LX is the corresponding altitude), and similarly the areas of triangles L and LB are 1 b LY and 1 c LZ. Thus, 1 a LX + 1 b LY + 1 c LZ = (area of triangle BL) + (area of triangle L) + (area of triangle LB) = (area of triangle B) = S, so that a LX + b LY + c LZ = S. The equation a r r 1 + b r r 1 + c r r 1 = a LX + b LY + c LZ thus becomes a r r 1 + b r r 1 + c r r 1 = S, so that qed. S 4rS r 1 = a = r r 1 + b r r 1 + c r r a + b + c = r a + b + c = r cot ω = r tan ω, } 4S {{} 1 =cot ω 1

13 B D Fig. 10 Let us use directed areas and powers of points with respect to circles. Our main vehicle is the following fact: Theorem 9 Let, B,, D be four points. Let p, p B, p, p D denote the powers of the points, B,, D with respect to the circumcircles of triangles BD, D, DB, B. Furthermore, we denote by [P 1 P P 3 ] the directed area of any triangle P 1 P P 3. Then, p [BD] = p B [D] = p [DB] = p D [B]. (1) (See Fig. 10.) Proof of Theorem 9.. In the following, all angles are directed angles modulo 180, and all segments and areas are directed. Let the circumcircle of triangle BD meet the line at a point (apart from ). Let the circumcircle of triangle D meet the line BD at a point B (apart from D). Let the lines and BD intersect at P. Since the points, D,, B are concyclic, we have DB = D, thus P B = D. But since the points, D, B, are concyclic, we have DB = D, so that P B = D. Therefore, P B = P B, which leads to B B. Hence, by the Thales theorem, BB = P P B. 13

14 B B' P D ' Fig. 11 The power p of the point with respect to the circumcircle of triangle BD is ; similarly, p B = BB BD. Thus, p [BD] p B [D] = [BD] BB BD [D] = P P B BD [BD] [D]. = [BD] BB BD [D] = BB BD [BD] [D] But it is a known fact that whenever, B, are three collinear points and P is a point not collinear with, B,, then we have B = [PB]. This fact yields [P] P = [DP ] [D] and DP DB = [DP ] [DB], so that P : DP DB = [DP ] [D] : [DP ] [DB] = [DB] [DP ] [DP ] [D] = [BD] [DP ] [DP ] [D] = [BD] [D], 14

15 and thus p [BD] p B [D] = P P B BD [BD] [D] = P = P P B BD ( P : P D BD ( P : DP ) DB P B BD ) = P P B BD P BD P D = P P P B P D. But the intersecting chord theorem yields P P = P B P D, so that p [BD] p B [D] = 1; in other words, p [BD] = p B [D]. Similarly, p B [D] = p [DB], so that p B [D] = p [DB], and p [DB] = p D [B]. ombining these equalities, we get (1). This proves Theorem pplication to the triangle Now the promised alternative proof of the fact that the points b, c, B c, B a, a, b are concyclic: Proof. The identity (1) can be rewritten as p [BD] = p B [D] = p [DB] = p D [B] (since [BD] = [BD] and [DB] = [DB]). pplying this equation to the case when D is the symmedian point L of triangle B, we get p [BL] = p B [L] = p [LB] (we have dropped the third equality sign since we don t need it), where p, p B, p are the powers of the points, B, with respect to the circumcircles of triangles BL, L, LB. But we have p = b and p B = B BB a (Fig. 4); thus, p [BL] = p B [L] becomes b [BL] = B BB a [L], so that b = B BB a [L] [BL]. () 15

16 B X Z L Y Fig. 1 Now let X, Y and Z be the feet of the perpendiculars from the symmedian point L onto the lines B,, B. It is a known fact that the distances from the symmedian point of a triangle to its sides are proportional to these sides; thus, LX : LY : LZ = B : : B. 3. But since the area of a triangle equals 1 (one of its sidelengths) (corresponding altitude), we have [BL] = 1 B LX, [L] = 1 LY and [LB] = 1 B LZ. Thus, () becomes b = B BB a 1 1 LY B = B LX B LY LX = B B B = B B = B B = B. By the converse of Thales theorem, this yields b B a B. Similarly, B c b B and a c ; this proves Theorem 8. The proof of Theorem 7 can be done in the same way as 4 (it was too trivial to have any reasonable alternative). ombined, this yields that the sides of the hexagon b c a b B c B a b are alternately antiparallel and parallel to the sides of triangle B. onsequently, this hexagon is a Tucker hexagon, and since every Tucker hexagon is known to be cyclic, we thus conclude that the points b, c, B c, B a, a, b lie on one circle. This way we have reproven a part of Theorem 4. Here is an alternative way to show that the points b, c, B c, B a, a, b lie on one circle, without using the theory of Tucker hexagons: Proof. (See Fig. 9.) Since a b is antiparallel to B, we have b a = 3 ctually, there is no need to refer to this known fact here, because we have almost completely LX r proven it above. Indeed, while proving that r 1 = r tan ω, we showed that r 1 =, so that LX = a r 1 r a = r 1 r B. Similarly, LY = r 1 r and LZ = r 1 B, so that LX : LY : LZ = B : : B. r 16

17 B, and thus b a c = ( a b ; a c ) = ( a b ; ) = a b = b a = B. (since a c ) Since b c is antiparallel to B, we have b c = B, so that b b c = b c = B = B. Therefore, b b c = b a c, so that the points a, b, b, c lie on one circle. The point B a also lies on this circle, since a B a b = (B; b B a ) = (B; B) (since b B a B) = B = b a (since b a = B was shown above) = a b b. Similarly, the point B c lies on this circle as well. This shows that all six points b, c, B c, B a, a, b lie on one circle. This argument did never use anything but the facts that the lines b c, B c B a, a b are antiparallel to B,, B and that the lines B c b, a c, b B a are parallel to B,, B. It can therefore be used as a general argument why Tucker hexagons are cyclic. 8. converse The alternative proof in 7 allows us to show a converse of Theorem 4: Theorem 10 Let P be a point in the plane of a triangle B but not on its circumcircle. Let the circumcircle of triangle BP meet the lines and B at the points b and c (apart from and B). Let the circumcircle of triangle P meet the lines B and B at the point B c and B a (apart from and ). Let the circumcircle of triangle P B meet the lines B and at the points a and b (apart from B and ). If the six points b, c, B c, B a, a, b lie on one circle, then P is the symmedian point of triangle B. We will not give a complete proof of this theorem here, but we only sketch its path: First, it is easy to see that the lines b c, B c B a, a b are antiparallel to B,, B. Now, we can reverse the argument from 7 to show that the lines B c b, a c, b B a are parallel to B,, B, and use this to conclude that the distances from P to the sidelines B,, B of triangle B are proportional to the lengths of B,, B. But this implies that P is the symmedian point of triangle B. References [1] R.. Johnson, Directed ngles in Elementary Geometry, merican Mathematical Monthly vol. 4 (1917), #3, pp [] Ross Honsberger, Episodes in Nineteenth and Twentieth entury Euclidean Geometry, US [3] Jean-Pierre Ehrmann, Hyacinthos message # [4] Darij Grinberg, The Neuberg-Mineur circle, Mathematical Reflections 3/