agge circles revisited Nguyen Van Linh 24/12/2011 bstract In 1907, Karl agge wrote an article on the construction of circles that always pass through the orthocenter of a given triangle. The purpose of his work is to find an extension of the Wallace-Simson theorem when the generating point is not on the circumcircle. Then they was named agge circles. In this paper, we present the new insight into agge circles with many corollaries. Furthermore, we also introduce three problems which are similar to agge circles. 1 agge circles and their corollaries roblem 1 (The construction of agge circles). Given triangle with its circumcircle () and orthocenter. hoose an arbitrary point in the plane. The cevian lines,, meets () again at, 1, 1. Denote 2, 2, 2 the reflections of, 1, 1 with respect to,,, respectively. Then,, 1, 1 are concyclic. The circle (,, 1, 1 ) is called -agge circle. ere we give two proofs for problem 1. oth of them start from the problem which appeared in hina Team Selection Test 2006. roblem 2 (hina TST 2006). Given triangle with its circumcircle () and orthocenter. Let be an arbitrary point in the plane. The cevian lines,, meets () again at, 1, 1. Denote 2, 2, 2 the reflections of, 1, 1 across the midpoints of,,, respectively. Then,, 1, 1 are concyclic. We introduce a lemma: Lemma 1. ( 2 2, 2 2 ) (, )(modπ) 1
2 M 1 Z Y 1 2 ' ' 2 N ' X Denote M, N the reflections of 2, 2 across the midpoint of then M 1 N is the reflection of 2 2 2 across the midpoint of. Since M 2, M = 2 and 2, = 2 we get M is a parallelogram. Similarly, 1 N is a parallelogram too. Let,, be the midpoints of, 1, 1 then, are the midpoints of M, N, respectively. The homothety 1 2 : M, 1, N therefore M 1 N. This means ( 1 M, 1 N) (, )(modπ). n the other side,,, are perpendicular to, 1, 1 hence,,,, are concyclic, which follows that: ( 2 2, 2 2 ) ( 1 M, 1 N) (, ) (, ) (, )(modπ). We are done. ack to problem 2. 3 1 4 2 Z Y 4 1 2 2 X 3 4 3 2
onstruct three lines through, 1, 1 and perpendicular to, 1, 1, respectively. They intersect each other and form triangle 3 3 3, intersect () again at 4, 4, 4, respectively. Note that 4, 4, 4 are diameters of () therefore 4, 4 are the reflections of across the midpoints of,. We get 2 4, 2 4 1 are parallelograms, which follows that 2 4, 2 1 4. ut 1 3 is a cyclic quadrilateral and applying lemma 1 we have ( 2, 2 ) ( 3 1, 3 ) ( 1, ) (, ) ( 2 2, 2 2 )(modπ) or, 2, 2, 2 are concyclic. ur proof is completed. Remark 1: Now come back to problem 1. Let 3, 3, 3 be the reflections of 2, 2, 2 across the midpoints of,,, respectively. Since ( 2 ) is the reflections of () wrt then 3 (). It is easy to see that is the midline of the triangle 2 3 hence 3. We claim that 3, 3, 3 concur at the isogonal conjugate Q of wrt. Then according to problem 2 we obtain, 2, 2, 2 are concyclic. nother proof (Nsato) [9] We introduce a lemma. Lemma 2. Given triangle and its circumcircle (). Let D be an arbitrary point on (). line d through D and d.d () = {G} then G is parallel to the Simson line of point D. G E F D Denote d = {E}, F the projection of D on then EF is the Simson line of D. Since DEF is a cyclic quadrilateral we get G = GD = EF. Therefore G EF. ack to problem 1. 3
Z L Y Q 4 4 2 X 4 K 3 Let Q be the isogonal conjugate of wrt ; X, Y, Z be the reflections of,, wrt Q; 3 3 3 be the circumcevian triangle of Q; 4 4 4 be the median triangle of. The symmetry with center Q takes to X, to Y, to Z then XY Z. ut 4 4 4 is the median triangle of hence there exist a point K which is the center of the homothety with ratio 1 2. This transformation maps XY Z to 4 4 4. n the other hand, let L be the second intersection of 2 and (). pplying lemma 2, L 2. Moreover, L is perpendicular to and 3 is parallel to therefore 3 L, which follows that 3 L or 3 2. From remark 1, 4 is the midpoint of 2 3 hence X 2 K 3 is a parallelogram. This means K 2 3. So K 2 2 or 2 lies on the circle with diamenter K. Similarly we are done. roblem 3. Given triangle with its circumcircle () and orthocenter. Let be an arbitrary point in the plane. The cevian lines,, meets () again at, 1, 1. Denote 2, 2, 2 the reflections of, 1, 1 with respect to,,, respectively. Then 2 2 2 1 1. Firstly we introduce a lemma: Lemma 3. Given triangle and its circumcircle (). Let E, F be two arbitrary points on (). Then the angle between the Simson lines of two points E and F is half the measure of the arc EF. 4
K F E I J L G See on the figure, IJ, G are the Simson lines of two points E and F, respectively. IJ G = {K} Denote L the projection of K on. We have KL F G and F G is a cyclic quadrilateral thus (KG, KL) (GK, GF ) (, F )(modπ)(1) Similarly, (KL, KJ) (E, )(modπ)(2) From (1) and (2) we are done. ack to problem 3. 1 1 2 2 2 ccording to agge circle, we have, 2, 2, 2 are concyclic. Since 2 is the reflections of wrt we get 2 is the Steiner line of, which is parallel to it s Simson line. nalogously, 2 is the Steiner line of 1. Then applying lemma 3, ( 2 2, 2 2 ) ( 2, 2 ) ( 1, 1 1 )(modπ). Do similarly with other direct angles we refer to the similarity of two triangles 1 1 and 2 2 2. roblem 4. Using the same notations as problem 1, let 3, 3, 3 be the second intersections of,, and ( 2 2 2 ), respectively. Then 2 3, 2 3, 2 3 concur at. 5
2 1 3 1 3 2 2 3 ( 2 2, 2 3 ) ( 2, 3 ) ( 2, )+(, ) π/2+( 2, )(modπ). Let L, Z be the projections of 1 on,, respectively then LZ is the Simson line of 1. We get 2 LZ Therefore ( 2, ) (LZ, L) ( 1 Z, 1 )(modπ) ence ( 2 2, 2 3 ) π/2 + ( 1 Z, 1 ) (, 1 ) (, 1 )(modπ) Since, 3, 3, 3 are concyclic then ( 3 3, 3 3 ) ( 3, 3 ) (, )(modπ). Similarly we get 3 3 3. ut from problem 3, 2 2 2 1 1. Thus there exist a similarity transformation which takes 3 2 3 2 3 to 1 and it follows that 2 3, 2 3, 2 3 concur at. Moreover, = 3 2 hence. Similarly,,. This means. ur proof is completed then. roblem 5. Let I be the circumcenter of triangle 2 2 2. K is the reflection of across I. K, K, K cut (I) again at 4, 4, 4. Then 3 4, 3 4, 3 4 are concurrent. 4 3 3 I 4 3 4 K 6
Since K 3 we have ( 3 4, 3 3 ) (K 4, K 3 ) (K, )(modπ). Similarly, ( 3 4, 3 3 ) (K, )(modπ). Do the same with other angles then applying eva-sine theorem for triangle 3 3 3 we are done. roblem 6. Given triangle with its circumcircle (). Let and Q be two isogonal conjugate points, 1, 2 be the circumcenters of -agge and Q-agge circles, respectively. Then Q 1 2 and Q = 1 2. Lemma 4. Given triangle with its orthocenter, circumcenter and let be an arbitrary point in the plane.,, intersect () again at, 1, 1, respectively. Denote 2, 2, 2 the reflections of, 1, 1 across the midpoints of,,, respectively, 2 the center of ( 2 2 2 ). Then 2 is a parallelogram. 1 1 ' 2 ' ' ' Y U Z 2 X 2 2 2 Denote, the reflections of 2, 2 across the midpoint of, respectively. Then 1 is the reflections of 2 2 2 across the midpoint of. Since 2, = 2, 2, 2 = we get is a parallelogram. (1) Similarly, 1 is a parallelogram. (2) Let X, Y, Z be the midpoints of, 1, 1, respectively then X, Y, Z lies on ( ). ut from (1) and (2), X, Z are the midpoints of,. homothety 1 2 : X, 1 Y, Z therefore 1 XY Z symmetry with center is the midpoint of : S :, the orthocenter of triangle, 2 2 2 2 It is easy to show that,, are collinear. ence the homothety 1 2 :, 2 U the circumcenter of triangle XY Z. We conclude that = 2 = 2. Therefore 2 is a parallelogram. We are done. ack to our problem. From the lemma above, 1 = Q, 2 = thus Q = 1 2 7
ur proof is completed. Remark 2. The radius of -agge circle is equal to the distance of Q and. roblem 7. Given triangle, its Nine-point circle (E) and an arbitrary point on (E). Let Q be the isogonal conjugate point of wrt. Then the center I of Q-agge circle lies on (E) too. Moreover, I is the reflection of across point E. The conclusion follows immediately from remark 2. Note that the center of Q-agge circle is always the reflection of wrt E. roblem 8. Given triangle and its circumcenter. and Q are isogonal conjugate wrt triangle such that,, Q are collinear. Then -agge circle and Q-agge circle are tangent. This problem is also a corollary of remark 2. We will leave the proof for the readers. roblem 9. Given triangle and its circumcircle (, R). circle which has center and radius r < R meets,, at, 2, 1, 2, 1, 2, respectively. Let E, F be the Miquel points of triangle wrt (, 1, 1 ) and ( 2, 2, 2 ) respectively. Then E-agge circle and F-agge circle are congruent. We will introduce a lemma. Lemma 5. Given triangle. circle () intersects,, at 6 points, 2, 1, 2, 1, 2, respectively. Let, Q be the Miquel points of wrt (, 1, 1 ) and ( 2, 2, 2 ). Then and Q are isogonal conjugate and = Q. 1 2 X Q 2 Y 1 2 Z Let X, Y, Z be the second intersections of, 1, 1 and (), respectively. We have (ZY, ZX) ( Y, X) ( Y, Y ) + ( Y, X) ( 2, 2 1 ) + ( 2, 1 ) ( 1 2, 1 2 )(modπ) This means XY = 2 2. Similarly, Y Z = 2 2, XZ = 2 2. Therefore XY Z = 2 2 2. n the other side, ( X, Y ) (Q 2, Q 2 )(modπ), ( Y, Z) (Q 2, Q 2 )(modπ) then there exist a rotation with center which takes XY Z to 2 2 2 and to Q. We conclude that = Q. Moreover, (, ) ( 1, 1 1 ) (ZY, Z ) ( 2 2, 2 Q) ( 2, Q)(modπ). Similarly we get and Q are isogonal conjugate wrt. ack to our problem. ccording to lemma 5, and Q are isogonal conjugate wrt and = Q. Then applying remark 2 the conclusion follows. 8
roblem 10. Given triangle with its orthocenter. Let, Q be two isogonal conjugate points wrt. onstruct the diameter K of -agge circle. Let G be the centroid of the triangle. Then Q, G, K are collinear and QG = 1 2 GK. 1 Q 1 2 2 G K X M a 2 Let X be the reflection of wrt Q, M a be the midpoint of then from proof 2 of problem 1, we obtain M a is the midpoint of XK. Then applying Menelaus theorem for the triangle XM a with the line (Q, G, K) we have Q, G, K are collinear. pplying Menelaus theorem again for the triangle QXK with the line (, G, M a ) then QG = 1 2. roblem 11. The centroid G lies on L-agge circle where L is the Symmedian point. 2 1 1 2 3 3 G 3 2 Let 1 1, 2 2 2 be the circumcevian triangles of G and L, respectively; 3, 3, 3 be the reflections of 2, 2, 2 wrt,,, respectively. It is easy to see that 3, 3, 3 lie on G, G, G, respectively. From problem 3, 3 3 3 2 2 2. Then ( 2 2, 2 2 ) ( 3 3, 3 3 )(modπ) n the other side, (G 3, G 3 ) 1 2 ( 1 + ( 2 2, 2 2 ) ( 3 3, 3 3 )(modπ). 1 ) 1 2 ( 2 + 2 ) 9
This means G, 3, 3, 3 are concyclic. We are done. More results. 1. When I the incenter of triangle, -agge circle is called Fuhrmann circle of, after the 19th-century German geometer Wilhelm Fuhrmann. The line segment connecting the orthocenter and the Nagel point N is the diameter of Fuhrmann circle. The solution for this result can be found in [8]. Moreover, denote, I the circumcenter and the incenter of triangle then N = 2I. 2. When lies on the circumcircle of triangle, -agge circle becomes the Steiner line of. 3. When Q lies on the circumcircle of triangle, we have a problem: Given triangle and an arbitrary point on its circumcircle (). Denote,, the reflections of across the midpoints of,,. Then,,, are concyclic. 4. When lies at the infinity, problem 1 can be re-written as below: Given triangle with its circumcircle (), orthocenter. Denote, 1, 1 points on () such that 1 1, 2, 2, 2 the reflections of, 1, 1 across the lines,,, respectively. Then, 2, 2, 2 are concylic. 5. When the orthocenter of, -agge circle becomes point. 6. When the circumcenter of, the center of -agge circle coincides with. 2 Results from the triads of congruent circles Three year ago, Quang Tuan ui, the Vietnamese geometer, wrote an article about two triads of congruent circles from reflections in the magazine Forum Geometricorum. In that note, he constructed two triads of congruent circles through the vertices, one associated with reflections in the altitudes, and the other reflections in the angle bisectors. In 2011, Tran Quang ung, the teacher at igh School for Gifted Student, anoi University of Science, also constructed another triad which associated with reflections in the lines joining the vertices and the Nine-point center. is work was published in a abook called Exploration and reativity. ere we introduce the involvement of agge circles in these problems. roblem 12. Given triangle with its orthocenter, circumcenter. Let a, a be the reflections of and across the line. Similarly, consider the reflections b, b of,, respectively across the line, and c, c of, across the line. Then the circumcircles of three triangles b c, c a, b a and -agge circle are congruent. 10
a 1 a 1 2 c b c b Note. In [2], Quang Tuan ui proved that three circles ( b c ), ( c a ), ( b a ) are congruent with the circle (, ). y remark 2, is the radius of -agge circle. We are done. roblem 13. Let I be the incenter of triangle. onsider the reflections of the vertices in the angles bisectors: a, a of, in I, b, b of, in I, c, c of, in I. Then the circumcircles of three triangles b c, c a, b a and Fuhrmann circle are congruent. b T c 1 1 2 2 I S a c 2 b a Note. In his article, Quang Tuan ui also proved that three circles ( b c ), ( c a ), ( b a ) are congruent with the circle (I, I). y remark 2, I is the radius of Fuhrmann circle and it completes the proof. roblem 14. Given triangle with its Nine-point center E. Let a, a be the reflections of and across the line E. Similarly, consider the reflections b, b of,, respectively across the line E, and c, c of, across the line E. Then the circumcircles of three triangles b c, c a, b a and E-agge circle are congruent. 11
b 1 c c 2 1 J 2 E Q 2 Denote 1 1 the circumcevian triangle of E. 2, 2, 2 the reflections of, 1, 1 across,,, respectively. Let be the orthocenter of triangle. meets the circumcircle () of triangle again at Q, ( c ) be the circumcircle of triangle. Since ( c ) and () are symmetric wrt, 2 is the reflection of 1 wrt we obtain 2 ( c ). It is not hard to see that c lies on E. owever, and c are symmetric wrt E we conclude that c ( c ). We have (, c ) (Q, 1 )(modπ), which follows from the fact that Q, c 1. This means the direct angle (, c ) is half the measure of the arc Q 1 or 2. Therefore c = 2 or 2 c is a isosceles trapezoid. nalogously, 2 b is also a isosceles trapezoid. We claim ( c b ) is the image of ( 2 2 ) across the symmetry whose the axis is the perpendicular bisector of. This means two circles ( c b ) and E-agge circle are congruent. Similarly we are done. Remark 3. ther synthetic proofs of the congruent of three circles ( b c ), ( c a ), ( b a ) can be found in [6] or [7]. We have some properties: 1. The radius of three circles ( b c ), ( c a ), ( b a ) is equal to the distance of and Kosnita point K (the isogonal conjugate of E). 2. The orthocenter of triangle I a I b I c is the reflection of wrt the center of N-agge circle (I a, I b, I c are the circumcenters of three triangles b c, c a, b a, respectively). 3. N is the center of K-agge circle. 3 Near agge circles Finally, we give 3 interesting problems which involve in agge circles. roblem 15. Given triangle, its orthocenter and its circumcircle (). Denote, 1, 1 the reflections of,, wrt, respectively, an arbitrary point in the plane. Let 2 2 2 be the pedal triangle of wrt, 3, 3, 3 be the reflections of, 1, 1 wrt 2, 2, 2, respectively. Then 4 points, 3, 3, 3 are concyclic. 12
1 1 4 4 5 2 3 2 5 5 3 4 2 3 Let 4, 4, 4 be the midpoints of,,, respectively then 4 is the midpoint of 3. Similar with 4, 4. We get 3 = 2 2 4, 3 = 2 2 4, 3 = 2 2 4.(1) onstruct three rectangles 4 2 5, 4 2 5, 4 2 5 hence 5, 5, 5 lie on the circle with diameter.(2) ut 5 = 4 2, 5 = 4 2, 5 = 4 2 (3) From (1), (2) and (3) we claim, 3, 3, 3 are concyclic. Moreover, we can show that 3 3 3. roblem 16. Given triangle with its circumcircle () and its orthocenter. Let a b c be the orthic triangle of triangle. is an arbitrary point on Euler line.,, intersect () again at, 1, 1. Let 2, 2, 2 be the reflections of, 1, 1 wrt a, b, c, respectively. Then, 2, 2, 2 are concyclic ( See [12] or [14]). We will prove this problem in the general case as below. Generalization. Given triangle and its circumcircle (). Let X, Y be two points inside () such that lies on the line segment XY. X, X, X intersect () again at, 1, 1, Y, Y, Y intersect () again at 2, 2, 2. Let 3, 3, 3 be the reflections of 2, 2, 2 wrt the midpoints of X, 1 Y, 1 Z. Then 3, 3, 3, X lie on a circle ( ) and lies on XY. 13
2 T 1 3 Z 1 2 S X Y V Q 3 3 W U 1 2 Let S be the intersection of 2 and 2 1. pplying ascal s theorem for 6 points 1, 2,, 2,, we get S XY. Denote T, U the second intersections of the perpendicular lines through, 1 to 2, 1 2 with (), respectively. pplying ascal s theorem again for 6 points, 2, 1, 2, T, U We conclude that XY. Similarly we obtain the perpendicular line through 1 to 1 2 also passes through, three lines which pass through 2, 2, 2 and perpendicular to 2, 1 2, 1 2, respectively concur at Q XY. onstruct three rectangles 2 V, 1 2 W, Z 2 1 then X 3 2 V and X 3 = 2 = V. Let L be the midpoint of X then L 3 = LV = 1 2 3V. Similarly we get ( 3 3 3 ) is the reflection of (V W Z) wrt L. ut, V, W, Z lie on a circle with diameter Q so X lies on ( 3 3 3 ). Moreover, is the reflection of wrt L, which follows that XY. We are done. roblem 17. Given triangle and its circumcircle (), its orthocenter. Let be an arbitrary point in the plane.,, intersect () again at, 1, 1, respectively. Let 2 2 2 be the pedal triangle of wrt, 3, 3, 3 be the reflections of, 1, 1 wrt 2, 2, 2, respectively. Then 3, 3, 3, are concyclic. We introduce two lemmas. Lemma 6. Three triangles 1 1, 2 2 2, 3 3 3 are similar. 14
1 2 3 2 j 1 3 3 2 4 4 The result 1 1 2 2 2 is trivival. So we only show that 3 3 3 1 1. Let 4, 4 be the reflections of 3, 3 wrt 2, respectively. We get 4 4 3 3 3.(1) Since 2 2 is the midline of triangle 1 3 4 then 1 4 2 2. nalogously, 1 4 2 2. Denote R the intersection of 1 4 and 1 4. We conclude that (R 1, R 1 ) ( 2 2, 2 2 ) ( 1, 1 )(modπ), which follows that R (). ence ( 1, 1 4 ) ( 1, 1 4 )(modπ). n the other side, 1 = 2 2 = 1 4. Therefore 1 4 1 4. 1 2 2 1 4 This means ( 1, 4 ) ( 1, 4 )(modπ), then ( 4, 4 ) ( 1, 1 )(modπ). Similarly we get 1 1 4 4.(2) From (1) and (2) we are done. Lemma 7. Given triangle. Let be the projection of on, 2, 2, 2 be the midpoints of,,, respectively. Let be an arbitrary point in the plane, be the pedal triangle of wrt, F and F be the intersections of ( ) and ( 2 2 2 ). line through and parallel to intersects at. Then the circle with diameter passes through one of two point F, F. '' I' V M 2 L 2 N ' I F' ' 2 ' Let V be the intersection of and 2 2. ccording to the first Fontene s theorem (see [4] or [5]), we obtain, V, F are collinear. 15
n the other side, denote I, L the midpoints of,, respectively, I the center of ( ). ecause is a parallelogram then I, L, I are collinear and II. ut L lies on 2 2 then I is the reflection of I wrt 2 2. Thus (I, I ) is the reflection of (I, I ) wrt 2 2, which follows that two circles intersect at two points M and N which lie on 2 2. Since, (I, I ) we get V F.V = V.V = V M.V N hence F,, M, N are concyclic or (I, I ) passes through F. ack to our problem. T M Z Y 2 ' R 2 L 3 S X N 2 Let L be the intersection of ( 2 2 2 ) and the Nine-point circle of triangle, X, Y, Z be the midpoints of,,, respectively, N be the second intersection of L 2 and (XY Z). Let M be the midpoint of, K be the projection of on, R be the midpoint of. Since X = M then = XM. Denote T the intersection of the lines through 2 and parallel to and K. pplying lemma 7 we obtain ( 2 T ) passes through L. Then ( 2 T, 2 L) (KT, KL) (NM, NL)(modπ), which follows that MN 2 T. ence (MX, MN) (, R)(modπ). ut (NM, NX) (R, ) π/2(modπ) thus XMN = R. Then R = XN, we obtain RNX is a parallelogram. This means RN = X = 1 2. Therefore N is the midpoint of or N is the midline of triangle 3. We have 3 2 L. Similarly, 3 2 L. Then ( 3, 3 ) (L 2, L 2 ) ( 2 2, 2 2 )(modπ). pplying lemma 6 we get ( 3, 3 ) ( 3 3, 3 3 )(modπ) or, 3, 3, 3 are concyclic. ur proof is completed then. 16
References [1] hristopher J. radley and Geoff.Smith, n a onstruction of agge, Forum Geom., 7 (2007) 231-247. [2] Quang Tuan ui, Two triads of congruent circles from reflections, Forum Geom., 8 (2008) 7-12. [3] Nguyen Van Linh, Two similar geometry problems. http://nguyenvanlinh.wordpress.com [4] Wolfram Mathworld, Fontene theorems. http://mathworld.wolfram.com/fontenetheorems.html [5] Nguyen Van Linh, Fontene theorems and some corollaries. http://nguyenvanlinh.wordpress.com [6] Nguyen Van Linh, 3 equal circumcircles. http://nguyenvanlinh.wordpress.com [7] Exploration and reativity 2011, igh School for Gifted Student, anoi University of Science. [8] Ross onsberger, Episodes in Nineteenth and Twentieth entury Euclidean Geometry (New Mathematical Library), The Mathematical ssociation of merica. [9] Shobber, rtofproblemsolving topic 97484 (ircumcircle passes through orthocentre [Fuhrmann extended]). http://www.artofproblemsolving.com/forum/viewtopic.php?t=97484 [10] Livetolove212, rtofproblemsolving topic 319520 (oncyclic). http://www.artofproblemsolving.com/forum/viewtopic.php?t=319520 [11] Livetolove212, rtofproblemsolving topic 321333 (oncyclic again). http://www.artofproblemsolving.com/forum/viewtopic.php?t=321333 [12] vittasko, rtofproblemsolving topic 345850 (arallelism relative to the Euler line of a triangle.) http://www.artofproblemsolving.com/forum/viewtopic.php?t=345850 [13] Nguyen Van Linh, agge circle. http://nguyenvanlinh.wordpress.com [14] Nguyen Van Linh, concyclic problem, yacinthos message 18801. http://tech.groups.yahoo.com/group/yacinthos/message/18801 [15] Jean-Louis yme, Le -ercle De agge, Une preuve purement synthetique. http://perso.orange.fr/jl.ayme Nguyen Van Linh, Foreign Trade University of Vietnam, 91 hua Lang, anoi. E-mail address: lovemathforever@gmail.com 17