EHRMANN S THIRD LEMOINE CIRCLE

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Abner Wiggins
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1 EHRMNN S THIRD EMOINE IRE DRIJ GRINERG 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 [1]. 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. The first two emoine circles et us remind the reader about some classical triangle geometry first. et be the symmedian point of a triangle. Then, the following two results are well-known ([3], hapter 9): O Fig. 1. Theorem 1. et the parallels to the lines,,,,, through meet the lines,,,,, at six points. These six points lie on one circle, the so-called first emoine circle of triangle ; this circle is a Tucker circle, and its center is the midpoint of the segment O, where O is the circumcenter of triangle. (See Fig. 1) The somewhat uncommon formulation et the parallels to the lines,,,,, through meet the lines,,,,, at six points means the following: Take the point where the parallel to through 40

2 EHRMNN S THIRD EMOINE IRE 41 meets, the point where the parallel to through meets, the point where the parallel to through meets, the point where the parallel to through meets, the point where the parallel to through meets, and the point where the parallel to through meets. Fig.. Furthermore (see [3] for this as well): Theorem. et the antiparallels to the lines,,,,, through meet the lines,,,,, at six points. These six points lie on one circle, the so-called second emoine circle (also known as the cosine circle) of triangle ; this circle is a Tucker circle, and its center is. (See Fig..) We have been using the notion of a Tucker circle here. This can be defined as follows: Theorem 3. et be a triangle. et Q a and R a be two points on the line. et R b and P b be two points on the line. et P c and Q c be two points on the line. 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,,, respectively; the lines P b P c, Q c Q a, R a R b are antiparallel to the sidelines,, of triangle, 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. The center of each such circle lies on the line O, where O is the circumcenter and the symmedian point of triangle. Notable Tucker circles are the circumcircle of triangle, its first and second emoine circles (and the third one we will define below), and its Taylor circle.. The third emoine 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. s far as I know, it has been first discovered by Jean-Pierre Ehrmann in 00 [1]: Theorem 4. et the circumcircle of triangle meet the lines and at the points b and c (apart from and ). et the circumcircle of triangle

3 4 DRIJ GRINERG meet the lines and at the point c and a (apart from and ). et the circumcircle of triangle meet the lines and at the points a and b (apart from and ). Then, the six points b, c, c, a, a, b lie on one circle. This circle is a Tucker circle, and its midpoint M lies on the line O and satisfies M = 1 O (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 emoine circle of triangle. We propose to denote the circle through the points b, c, c, a, a, b as the third emoine circle of triangle. (See Fig. 3) a c c O M b a 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 sections 3-5; this proof will use four auxiliary facts (Theorems 5, 6, 7 and 8). Then, in sections 6 and 7, we will give a new argument to show the part of Theorem 4 which claims that the six points b, c, c, 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 section 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 is the centroid of each of the three triangles b c, a c, a b. (See Fig. 4)

4 EHRMNN S THIRD EMOINE IRE 43 a c c b a b Fig. 4. ctually this result isn t as much about symmedian points and centroids, as it generalizes to arbitrary isogonal conjugates: Theorem 6. et P and Q be two points isogonally conjugate to each other with respect to triangle. et the circumcircle of triangle P meet the lines and at the points c and a (apart from and ). Then, the triangles a c and are oppositely similar, and the points P and Q are corresponding points in the triangles a c and. (See Fig. 5) Remark 1. 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. c P Q a 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 [4]. list of their important properties has also been given in [].

5 44 DRIJ GRINERG The point Q is the isogonal conjugate of the point P with respect to triangle ; thus, P = Q. Since,, c, a are concyclic points, we have a c = c, so that a c =. Furthermore, c a =. Thus, the triangles a c and are oppositely similar (having two pairs of oppositely equal angles). y the chordal angle theorem, P a c = P c = P = Q = Q. Similarly, P c P a = Q. These two equations show that the triangles a P c and Q are oppositely similar. ombining this with the opposite similarity of triangles a c and, we obtain that the quadrilaterals a c P and Q are oppositely similar. Hence, P and Q are corresponding points in the triangles a c and. (See Fig. 6.) Theorem 6 is thus proven. 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 ; the isogonal conjugate Q of P is, in this case, the centroid of triangle. Now, Theorem 6 says that the points P and Q are corresponding points in the triangles a c and. Since Q is the centroid of triangle, this means that P is the centroid of triangle a c. ut P = ; thus, we have shown that is the centroid of triangle a c. Similarly, 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, c are concyclic, we have b c = c, so that b c =. Thus, the line b c is antiparallel to in triangle. Similarly, the lines c a and a b are antiparallel to and. We have thus shown: Theorem 7. In the configuration of Theorem 4, the lines b c, c a, a b are antiparallel to,, in triangle. Now let X b, X c, Y c, Y a, Z a, Z b be the points where the antiparallels to the lines,,,,, through meet the lines,,,,,. ccording to Theorem, these points X b, X c, Y c, Y a, Z a, Z b lie on one circle around ; 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. 6):

6 EHRMNN S THIRD EMOINE IRE 45 X c Z a Y c Y a Z b Fig. 6. Partial proof of Theorem. Since symmedians bisect antiparallels, and since the symmedian point of triangle lies on all three symmedians, the point must bisect the three antiparallels X b X c, Y c Y a, Z a Z b. This means that X b = X c, Y c = Y a and Z a = Z b. Furthermore, X c X b = (since X b X c is antiparallel to ), thus Y c X c = ; similarly, X c Y c =, thus Y c X c = X c Y c = = = Y c X c. Hence, triangle X c Y c is isosceles, so that X c = Y c. Similarly, Z b = X b and Y a = Z a. Hence, X b X b = X c = Y c = Y a = Z a = Z b. This shows that the points X b, X c, Y c, Y a, Z a, Z b lie on one circle around. This circle is the so-called second emoine circle of triangle. Its radius is r 1 = X b = X c = Y c = Y a = Z a = Z 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. Y c c O M m Y a a Fig. 7. Proof of Theorem 9, part 1. Now consider Fig. 8. et m be the midpoint of the segment c a. ccording to Theorem 5, the point is the centroid of triangle

7 46 DRIJ GRINERG a c ; thus, it lies on the median m and divides it in the ratio : 1. Hence, : m = (with directed segments). et M be the point on the line O such that M = 1 O (with directed segments); then, O = M, so that O = O = M, and thus O : M = = : m. y the converse of Thales theorem, this yields m M O. The lines Y c Y a and c a are both antiparallel to, and thus parallel to each other. Hence, Thales theorem yields m c : Y c = m :. ut since : m =, we have = m, so that m = 1 and therefore m = + m = + 1 = 3 and m : = 3. onsequently, m c : Y c = 3 and m c = 3 Y c = 3 r 1. It is a known fact that every line antiparallel to the side of triangle is perpendicular to the line O (where, as we remind, O is the circumcenter of triangle ). Thus, the line c a (being antiparallel to ) is perpendicular to the line O. Since m M O, this yields c a m M. Furthermore, m M O yields O : m M = : m (by Thales), so that O : m M = and O = m M, and thus m M = 1 O = 1 r, where r is the circumradius of triangle. Now, Pythagoras theorem in the right-angled triangle M m c yields M c = m c + 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 M a, M a, M b, M b and M c. Hence, the points b, c, c, 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 O and satisfies M = 1 O. 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 Since the points b, a, a, b are concyclic, we have a a b = a b b, thus a b = a b. ut since a b is antiparallel to, we have b a =, so that a b = b a =, thus a b = ; consequently, b a. Similarly, c b and a c. ltogether, we have thus seen: Theorem 8. The lines c b, a c, b a are parallel to,,. (See Fig. 8)

8 EHRMNN S THIRD EMOINE IRE 47 a c c b a b Fig. 8. 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 c a are alternately antiparallel and parallel to the sides of triangle. Thus, b c a b c a is a Tucker hexagon, and the circle passing through its vertices b, c, c, 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 emoine circle is r tan ω, where ω is the rocard angle of triangle. 1 Thus, the radius of the third emoine circle is 1 9r1 + r = 1 9 (r tan ω) + r = 1 9r tan ω + r = r 9 tan ω For a quick proof of this fact, let X, Y, Z denote the feet of the perpendiculars from the point to the sides,, of triangle. Then, the radius r 1 of the second emoine circle is r 1 = Y a. Working without directed angles now, we see that Y a = X (from the sin right-angled triangle XY a ), so that r 1 = Y a = X sin. Since sin = a by the extended law r of sines, this rewrites as r 1 = ( X X r a ) =. This rewrites as a a r r 1 = a X. Similarly, r b r r 1 = b Y and c r r 1 = c Z. dding these three equations together, we obtain a r r 1 + b r r 1 + c r r 1 = a X + b Y + c Z. On the other hand, let S denote the area of triangle. ut the area of triangle is 1 a X (since a is a sidelength of triangle, and X is the corresponding altitude), and similarly the areas of triangles and are 1 b Y and 1 c Z. Thus, 1 a X + 1 b Y + 1 c Z = (area of triangle ) + (area of triangle ) + (area of triangle ) = (area of triangle ) = S,

9 48 DRIJ GRINERG 6. 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 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, c, 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. D Fig. 9. et us use directed areas and powers of points with respect to circles. Our main vehicle is the following fact: Theorem 9. et,,, D be four points. et p, p, p, p D denote the powers of the points,,, D with respect to the circumcircles of triangles D, D, D,. Furthermore, we denote by [P 1 P P 3 ] the directed area of any triangle P 1 P P 3. Then, (1) p [D] = p [D] = p [D] = p D []. (See Fig. 10.) so that a X + b Y + c Z = S. The equation a r r 1 + b r r 1 + c r r 1 = a X + b Y + c Z 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

10 EHRMNN S THIRD EMOINE IRE 49 Proof of Theorem 9. In the following, all angles are directed angles modulo 180, and all segments and areas are directed. et the circumcircle of triangle D meet the line at a point (apart from ). et the circumcircle of triangle D meet the line D at a point (apart from D). et the lines and D intersect at P. Since the points, D,, are concyclic, we have D = D, thus P = D. ut since the points, D,, are concyclic, we have D = D, so that P = D. Therefore, P = P, which leads to. Hence, by the Thales theorem, = P P. P D Fig. 10. The power p of the point with respect to the circumcircle of triangle D is ; similarly, p = D. Thus, p [D] p [D] = [D] D [D] = P P D [D] [D]. = [D] D [D] = D [D] [D] ut it is a known fact that whenever,, are three collinear points and P is a point not collinear with,,, then we have = [P]. This fact yields [P] P = [DP ] [D] so that P : DP D = [DP ] [D] : [DP ] [D] and = [D] [DP ] [DP ] [D] DP D = [DP ] [D], = [D] [DP ] [DP ] [D] = [D] [D],

11 50 DRIJ GRINERG and thus p [D] p [D] = P P D [D] [D] = P = P P D ( P : P D D ( P : DP ) D P D ) = P P D P D P D = P P P P D. ut the intersecting chord theorem yields P P = P P D, so that p [D] p [D] = 1; in other words, p [D] = p [D]. Similarly, p [D] = p [D], so that p [D] = p [D], and p [D] = p D []. 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, c, a, a, b are concyclic: Proof. The identity (1) can be rewritten as p [D] = p [D] = p [D] = p D [] (since [D] = [D] and [D] = [D]). pplying this equation to the case when D is the symmedian point of triangle, we get p [] = p [] = p [] (we have dropped the third equality sign since we don t need it), where p, p, p are the powers of the points,, with respect to the circumcircles of triangles,,. ut we have p = b and p = a (Fig. 4); thus, p [] = p [] becomes so that () b [] = a [], b = a [] []. Z X Fig. 11. Y

12 EHRMNN S THIRD EMOINE IRE 51 Now let X, Y and Z be the feet of the perpendiculars from the symmedian point onto the lines,,. It is a known fact that the distances from the symmedian point of a triangle to its sides are proportional to these sides; thus, X : Y : Z = : :. ut since the area of a triangle equals 1 (one of its sidelengths) (corresponding altitude), we have [] = 1 X, [] = 1 Y and [] = 1 Z. Thus, () becomes b = a 1 1 Y = X Y X = = = =. y the converse of Thales theorem, this yields b a. Similarly, c b and a c ; this proves Theorem 8. The proof of Theorem 7 can be done in the same way as in section 4 (it was too trivial to have any reasonable alternative). ombined, this yields that the sides of the hexagon b c a b c a b are alternately antiparallel and parallel to the sides of triangle. 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, c, 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, c, a, a, b lie on one circle, without using the theory of Tucker hexagons: Proof. (See Fig. 9.) Since a b is antiparallel to, we have b a =, and thus b a c = ( a b ; a c ) = ( a b ; ) = a b = b a =. (since a c ) Since b c is antiparallel to, we have b c =, so that b b c = b c = =. Therefore, b b c = b a c, so that the points a, b, b, c lie on one circle. The point a also lies on this circle, since a a b = (; b a ) = (; ) (since b a ) = = b a (since b a = was shown above) = a b b. Similarly, the point c lies on this circle as well. This shows that all six points b, c, c, a, a, b lie on one circle. ctually, there is no need to refer to this known fact here, because we have almost completely X r proven it above. Indeed, while proving that r 1 = r tan ω, we showed that r 1 =, so that a X = r 1 r a = r 1 r. Similarly, Y = r 1 r and Z = r 1, so that X : Y : Z = : r :.

13 5 DRIJ GRINERG This argument did never use anything but the facts that the lines b c, c a, a b are antiparallel to,, and that the lines c b, a c, b a are parallel to,,. 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, also found by Ehrmann in [1]: Theorem 10. et P be a point in the plane of a triangle but not on its circumcircle. et the circumcircle of triangle P meet the lines and at the points b and c (apart from and ). et the circumcircle of triangle P meet the lines and at the point c and a (apart from and ). et the circumcircle of triangle P meet the lines and at the points a and b (apart from and ). If the six points b, c, c, a, a, b lie on one circle, then P is the symmedian point of triangle. 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, c a, a b are antiparallel to,,. Now, we can reverse the argument from section 7 to show that the lines c b, a c, b a are parallel to,,, and use this to conclude that the distances from P to the sidelines,, of triangle are proportional to the lengths of,,. ut this implies that P is the symmedian point of triangle. References [1] J. P. Ehrmann. Hyacinthos message 6098, Dec 00. [] D. Grinberg. The Neuberg Mineur circle. Mathematical Reflections, [3] R. Honsberger. Episodes in nineteenth and twentieth century Euclidean geometry, volume 37. The Mathematical ssociation of merica, [4] R.. Johnson. Directed angles in elementary geometry. The merican Mathematical Monthly, 4(3): , address: darijgrinberg@gmail.com Massachusetts Institute of Technology

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

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