Journal for eometry and raphics olume 9 (005) No. 119 1. eometry of Regular Heptagons vonko Čerin Kopernikova 7 10010 agreb roatia urope email: cerin@math.hr bstract. his paper explores some new geometric properties of regular heptagons. e add to the list of results from the ankoff-arfunkel famous paper on regular heptagons 0 years ago enlisting the help from computers. ur idea is to look at the central points (like incenters centroids circumcenters and orthocenters) of certain triangles in the regular heptagon to find new related regular heptagons which have simple constructions with ruler and compass from the original heptagon. In the proofs we use complex numbers and the software Maple. he eleven figures are made with the eometer s Sketchpad. Key ords: regular heptagon heptagonal triangle MS 000: 51N0 51M04 145 14Q05 1. Introduction Leon ankoff and Jack arfunkel in the reference [1] thirty years ago gave the review of several results on regular heptagons and on the associated heptagonal triangle (see ig. 1). In [] [4] and [5] some of these initial theorems in [1] have been improved. In this paper our goal is to continue these investigations. e use again complex numbers to discover new relationships in these geometric configurations. In order to simplify our statements we use the following notation. he midpoint of points and is [; ] while l and l are the parallel and the perpendicular to the line l through the point. Let Θ = be a regular heptagon inscribed into the circle k with the center and the radius R. e now define fourteen regular heptagons associated to Θ. It suffices to describe only their first vertex because the other vertices are obtained by rotations about the point. he first vertices are shown in ig. 1 and are defined as follows: m = [; ] = [; ] = [; ] d = s = m = [ m ; m ] = [ ; ] d m = m m m m d = s m = m m m m s = and let m be the midpoints of the shorter arcs m m. or different points and and a real number r > 0 let γ(; ) and γ(; r) denote circles with the center at which goes through and with the radius r respectively while ε( ) is the complement of the segment in the line. ISSN 14-8157/$.50 c 005 Heldermann erlag
10. Čerin: eometry of Regular Heptagons s m m m d m d s s m m d igure 1: irst vertices of 14 regular heptagons associated to the regular heptagon and one of its heptagonal triangles. New theorems e begin with an improvement of the observations attributed to hébault on the pages 10 and 11 of [1]. he parts (1) (4) are from there while (5) (10) are new. hen investigating regular heptagons it seems natural to look for various regular heptagons associated to it. In our first theorem we discovered four such heptagons. heorem 1 (1) he segment m is the diagonal of the square build on the inradius of Θ. () xtend over to the point so that =. he segment is the diagonal of the square constructed on half the side of the equilateral triangle inscribed to the circle k. () he circle with the center at which is orthogonal to k has m as radius. (4) Let = ( m ) S = [; ] and m = γ(s; ). he points and m are on the circle m and the line m is its tangent (ig. ). (5) Let L = (m ) \ {} H = m ε() I = m ε() J = m ε() K = m ε(). hen m L HIJK is a regular heptagon with the length of sides m. (6) Let H = m ε( ) = m ε( ) L = (m ) \ { } Q = (m ) \ { } K = m ε( ) J = m ε( ). hen H L Q K J is a regular heptagon. (7) he midpoints I H L Q K J of shorter arcs I H H L L Q m K K J J are vertices of a regular heptagon whose sides are parallel with sides of. (8) Let n = γ( ; m). he circles m and n intersect in the points H and L. (9) Let M = n m N = n J P = n ε( L) Q = n ε( K) = n ε( I). hen mmnhp Q is a regular heptagon.
PSfrag replacements. Čerin: eometry of Regular Heptagons 11 I P H J Q n N S m M L m m K k igure : he circles m and n with the regular heptagons inscribed into them (10) he circle with the center at which is orthogonal to n has as radius. Proof: (1) In this proof we shall assume that the complex coordinates (or the affixes) of the vertices of the heptagon are = 1 = f = f 4 = f 6 = f 8 = f 10 = f 1 where f is the 14th root of unity. rom = f 6 + f 4 and m = 1 follows that m where is equal to (f 4 + f 1)p + p 4 p = f 6 f 5 + f 4 f + f f + 1 and p + = f 6 + f 5 + f 4 + f + f + f + 1. ut f 14 1 is (f 1) p p + and p + = 1 + i (1 + cos π 7 ) sin π 7 0. e see that p = 0 so that m = (R cos π 7 ). () he point is on and γ( ; ) if f 4 + f 10 f 8 = 0 and (f 10 + f 8 ) (f 6 + f 4 ) + f 1 + f 11 + f + f = 0. y solving this system we get = f 6 + f 4 f f(f 1) (f + f + )( f + f + 1).
1. Čerin: eometry of Regular Heptagons It is now easy to check that = (f 11 f 8 f 7 + f + ) p = 0. () ne of the intersections of k and the circle with diameter is the point with affix + (f 1 f 1 + f 11 f + f ) f 7 (f + f + f(f 1) (f + f + )(f + f + 1)). he difference m contains p as a factor. (4) he equations of and m are f 6 z + f z = 1 f 10 and z + z = 1. Hence S = f 10 + f 1 f (1 f 4 ). Since both S S and S m S contain p as a factor we infer that and m are on m. lso the lines m and S are perpendicular. (5) Note that f k ( m S) + S for k = 1 4 6 lie on lines while for k = 5 it agrees with. Hence these are the vertices of m KJIH L. Moreover L = m. (6) Now f k ( S) + S for k = 1... 6 lie on lines so that these are the last six vertices of the regular heptagon H L Q K J. (7) his part is more complicated even on a computer so that we only outline main steps. irst find the equation of m and of the line l joining S with the midpoint of the segment I. ne of the points in the intersection m l is I. hen we rotate six times through the angle π 7 to get points H L Q K and J. inally we check that (only one) corresponding sides of I H L Q K J and are parallel. (8) he equations of the circles m and n (ig. ) are and (f 4 1) z z + f 4 (f 6 + f 4 1) z (f 10 + f 8 1) z = 0 4 z z f 4 (f + 1)(f 4 z + z) + f 1 + f 11 + f + f 1 = 0. heir intersections have rather complicated affixes but after some clever manipulation with square roots one can show that they represent points H and L. (9) e rotate the point m about the point six times through the angle π. he third 7 point concides with the point H while the others are on lines m J L K and I so that mmnhp Q is indeed a regular heptagon. (10) Similar to the proof of (). heorem Let the lines and intersect the line in points M and N. Let and denote circumcenters of the triangles M MN and N (ig. ). (1) = and is the reflection of at the line. () is the reflection of at the line and the midpoint of the shorter arc M on the circumcircle of M. () = R and = H = R where H is the intersection of the lines and. lso = = H. (4) If = ( ) = ( ) and = ( ) then is a regular heptagon whose sides are perpendicular to the corresponding sides of.
ag replacements. Čerin: eometry of Regular Heptagons 1 I J H M N K igure : he triangle on circumcenters of the triangles M MN and N and two regular heptagons with sides perpendicular to sides of P (5) If I = ( ) J = (I ) K = (J ) L = (K ) and P = ( ) then IJKLP is a regular heptagon whose sides are perpendicular to the corresponding sides of. Proof: (1) he affixes of the points M N and H are f 10 f 8 + f 6 f 10 + f 6 + f 4 f 4 + f + 1 f 10 + f 6 f 10 + f 6 f 4 + f + 1 f 10 + f 8 + f 6 + f 4 f (f 4 + 1) f 4 + f + 1 f 4 + f + ( ) respectively. Since f = 6 p we infer that =. It is easy to find the f 4 + f + 1 reflection of in the line and check that it coincides with the point. () Since the reflection of a complex number x in the line determined by different complex y(x z) + z(y x) numbers y and z is by direct substitution of affixes we see that is a y z reflection of in the line. () Since = we get = 1 = R. lso since = (f +f 4 +f 6 f 10 f 1 ) ( (f 4 + f + 1) f and factors as 10 + f 8 + f 6 f 4 f ) p + p it follows that = (f 4 + f + 1) = R. In a similar fashion we can also prove that H = = R and that = = H. (4) Let = f 1 + f 8 f 6 1 denote the circumcenter of the triangle. hen f k ( ) + for k = 1... 6 is. Hence is a regular heptagon. Since is perpendicular to its sides are perpendicular to the corresponding sides of. L
14. Čerin: eometry of Regular Heptagons PSfrag replacements 0 0 N 0 0 M 0 N 0 0 0 igure 4: he triangle on centers of the nine-point circles of triangles M MN and N with = R (5) Similar to the proof of (4). here is a similar result for centers of the nine-point circles instead of circumcenters. heorem Let the lines and intersect the line in points M and N. Let 0 0 0 0 0 0 0 and N 0 be midpoints of the segments M M M M M M M and N. Let and be centers of the nine-point circles d 9 m 9 and a 9 of the triangles M MN and N (ig. 4). (1) he point is the midpoint of the segment M and the point is the midpoint of the shorter arc 0 0 of the nine-point circle d 9 of the triangle M. () he point is the reflection of at the line 0 N 0. lso = R and = = R. () he polyhedron 0 0 0 0 0 0 0 is a regular heptagon inscribed into d 9 which is the image of in the homothety h(m 1). (4) Let N be the projection of the point N on the side let K = N let L be the reflection of K in the line 0 N 0 let P = [; N] and S = [M; N]. Let Q = [; ] H = N N 0 0 and = 0 0 0. hen 0 S 0 HN 0 K and N 0 N P Q 0 L are regular heptagons inscribed into m 9 and a 9 related by the homothety h([ ; ] 1) and homothetic with the heptagon.
PSfrag replacements. Čerin: eometry of Regular Heptagons 15 H 0 0 N 0 L K N M S 0 N P Q 0 igure 5: nlarged rectangular part of ig. 4 with two regular heptagons homothetic with the heptagon Proof: (1) he affixes of the points M and N have been given in the proof of the previous theorem. It follows that the midpoints 0 0 N 0 are f 10 f 8 + f 6 + f 4 while are f 10 f 8 + f 6 f 10 + f 8 + f 6 + f 4 (f 4 + f + 1) f 10 f 8 + f 6 f 10 + f 8 + f 6 + f 4 + 1 (f 4 + f + 1) f 10 + f 8 + f 6 + f 4 (f 4 + f + 1) respectively. y direct inspection we see that is the midpoint of M and that is the midpoint of the shorter arc 0 0 of the nine-point circle of M because = 1 and 0 = 0. () Just as easy is to check that is the reflection of at the line 0 N 0. inally since f = 6 + f 4 f 10 f 1 (f 8 + f 6 + f 4 + f + 1) and both 1 and 1 contain the polynomial p as a factor we get that = =. () his part is obvious.
16. Čerin: eometry of Regular Heptagons (4) In a routine fashion we discover that the affixes of the points N K L P Q H S and are f 10 + f 8 + f 6 + f 4 (f 4 + f + 1) f 10 + f 8 + f 6 + f 4 (f 4 + f + 1) f 10 + f 8 + f 6 + f 4 + f + 1 (f 1 + f 4 + f + ) f 10 + f 6 + f 4 + 1 (f 4 + f + 1) f 6 + f f 1 + f 10 + f 8 + f 6 + f 4 + f (f 1 + f 4 + f + ) 5 f 1 + f 10 f 8 f 6 + 5f + 6 ( f 10 + 4f 8 + f 6 f ) f 1 f 10 + f 4 f 1 (f 1 f 8 f 6 f 4 + 1) respectively. Since f k ( 0 ) + for k = 1... 6 is K N 0 H 0 S we infer that 0 S 0 HN 0 K is a regular heptagon inscribed into m 9. Since 0 S is parallel to it is homothetic to (ig. 5). Similarly since f k (N 0 ) + for k = 1... 6 is L 0 Q P N we get that N 0 N P Q 0 L is a regular heptagon inscribed into a 9. Since N is parallel to it is homothetic to. f course the triangles and from the previous two theorems are closely related as the following result clearly shows. Recall that triangles and are orthologic provided the perpendiculars at vertices of onto sides and of are concurrent. It is well-known that the relation of orthology for triangles is reflexive and symmetric. heorem 4 Let and be reflections of and at the line. Let K be the intersection of the lines and. hen any two among the triangles and are orthologic. he triangles and are images under the homotheties h(k ) and h( ) of the triangles and (ig. 6). Proof: Since the points have the affixes f 10 f 8 f 10 + f 6 + f 4 + 1 f 4 + f + 1 f 6 + f 4 f 4 + f + 1 f 10 f 8 + f 6 + f 4 f 10 + f 6 + f 4 + 1 and f 10 + f 8 + f 6 + f 4 + 1 (f 4 + f + 1) (f 4 + f + 1) respectively it is easy to check using heorem 5 in [] that the triangles and are orthologic. It is also easy to verify that are midpoints of the segments K K K and that are midpoints of the segments which proves the claims about homotheties. heorem 5 Let be a regular heptagon inscribed to a circle of radius R. Let the lines and intersect the line in points M and N. If H and K are the de Longchamps points of the triangles M and N then HK = R 11. Proof: Since H = f 10 + f 8 + f 6 and K = f 10 + f 8 + f 6 + f 4 we get that HK 11 is f 4 + f + 1 5 p + p f 8 + f 6 + f 4 + f + 1.
. Čerin: eometry of Regular Heptagons 17 S S K igure 6: he triangles and are orthologic and homothetic with their reflections at the line heorem 6 Let the line in the regular heptagon intersect the line in the point M. Let H I J K be centroids of the triangles M M M M. hen HIJK is a rhombus whose side is R cos π and whose area is /9 of the area of the quadrangle 14. he angles IHK and KJI are equal to 4π so that the regular heptagons build 7 on IH HK and on IJ JK share one side (ig. 7). Proof: Since M = f 10 f 8 + f 6 we easily find that H = f 10 f 8 + f 6 I = f 1 + f 10 f 8 + f 6 and K = f 10 f 8 + f 6 + f J = f 1 + f 10 f 8 + f 6 + f Now we compute HI IJ HI JK and HI KH to discover that they are all zero (this is immediate for the second difference while for the first and the third it follows from the fact that both contain p as a factor). Hence the quadrangle HIJK is the rhombus. he claim about the side and the angles is a consequence of easily verified equalities IJ = 4 9 cos π HP and = cos π where P is the center of the rhombus. inally since the triangles HIJ and have areas i (f 1 + f 10 f 4 f ) 14 HI 7 6 i (f 10 + f 8 f 6 f 4 ) and i (f 1 + f 10 f 8 + f 6 f 4 f ) respectively it follows that 4 4 HIJK = 9. Remark (drian ldknow): he statement about HIJK = has actually nothing to do 9 with heptagons it is just a particular case of the very easily proved result that if M.
18. Čerin: eometry of Regular Heptagons PSfrag replacements S M H I K J igure 7: he quadrangle HIJK on centroids of triangles M M M M is a rhombus is any internal point of any convex quadrangle then the centroids of the triangles M M M and M form a parallelogram with sides 1 and 1 and area. 9 heorem 7 Let be the centroids of the triangles in the regular heptagon Θ =. Let S be the intersection of the lines joining and with the midpoints H and K of the segments and. Let λ = u/v where v = 9 18 cos π 7 and u = cos π cos π + 4 cos π. Let P and Q be intersections of the lines and 7 7 7 with the lines and. (1) If h 1 is the homothety h(s λ) then = h 1 ( ) (ig. 8). () If h and h are the homotheties h(p λ) and h(q λ) then Θ = h (Θ) and Θ = h (Θ) are regular heptagons built on segments and. () If M and N are the centers of Θ 1 Θ and Θ and Φ and Ψ denote the regular heptagons built on the the segments MN and and containing the vertex then the center of Φ is and Ψ is obtained from Φ by the translation for the vector. (4) he point S lies on the side of Ψ. Proof: he centroids are = f 1 + f 10 + f 6 = f 1 + f 6 + f = f 6 + f 4 + f.
PSfrag replacements. Čerin: eometry of Regular Heptagons 19 M P H N Q K S igure 8: he triangle on centroids of triangles is homothetic with the triangle he lines and concur at the point S = H K with the affix f 1 + f 10 + f 8 + f 6 + 1. he conditions for the corresponding sidelines of the triangles f 8 + f 6 + f 4 + f + and to be parallel are satisfied because they are zero (as desired) or are zero because they contain the polynomial p as a factor. he quotient λ = is equal to the square root of f 1 + f 10 + f 8 + f 6 + 1. Let M = f 1 f 8 f 4 be the intersection of the perpendicular 9(f 8 + f 4 + 1) bisector of with the parallel through to the line. he point M + f ( M) lies on the line P. Replacing f with other even powers of f we check that Θ = h (Θ) is the regular heptagon built on the segment. Let = f 1 + f 10 + f 8 + f 6 + f 4 + 1 be the intersection of the perpendicular bisector of with the parallel through M to the line. hen is the center of the regular heptagon built on MN which contains the point. e can now easily check that M N and are parallel segments of the same length. or the last claim we translate points + f 6 (M ) and + f 8 (M ) for and check that the point S lies on the segment joining these translated points. heorem 8 Let be the orthocenters of the triangles in the regular heptagon Θ =. Let be the intersection of the perpendiculars at and to the lines and. Let u µ = with u = cos π w 7 cos π 7 + 4 cos π 7 and w = 1 cos π 7. Let P be the intersection of the perpendicular bisectors p and q of and. Let M and N be intersections of p and q with perpendiculars to at and. Let H and K be the intersections of the lines and with the lines and.
10. Čerin: eometry of Regular Heptagons ag replacements N P M igure 9: he triangle on orthocenters of triangles is homothetic with the triangle (1) If h 4 is the homothety h( µ) then = h 4 ( ) (ig. 9). () If h 5 and h 6 are the homotheties h(h µ) and h(k µ) then Θ 5 = h 5 (Θ) and Θ 6 = h 6 (Θ) are regular heptagons built on segments and with centers M and N. () If Φ denotes the regular heptagon built on the the segment MN containing the vertex and Ψ = h 4 (Θ) then P is a common center of Φ and Ψ. Proof: he orthocenters are = f 1 + f 10 + f 6 = f 1 + f 6 + f = f 6 + f 4 + f. he claims of the theorem are now easily verified in the same way as in the proof of the previous theorem. he triangles and from the previous two theorems are themselves homothetic as the following result shows. heorem 9 he line S goes through the center and the triangle on orthocenters is the image under the homothety h( ) of the triangle on centroids (ig. 10). Proof: Since S = f 1 + f 10 + f 6 + f + 1 and = f 1 + f 10 + f 8 + f 6 + 1 1 f f 6 f 8 f 1 f 8 + f 6 + f 4 + f + the free term of the equation of the line S is zero so that the center (the origin) lies on S. he second claim about the triangles and is clearly true because the complex
. Čerin: eometry of Regular Heptagons 11 PSfrag replacements S igure 10: he triangles on orthocenters and on centroids of triangles are homothetic frag replacements S S igure 11: he triangles on the de Longchamps points and on the centers of the nine-point circles of triangles are related in homothety h( )
1. Čerin: eometry of Regular Heptagons coordinates of the vertices of the second triangle are one third of the complex coordinates of the vertices of the first triangle. here are analogous results for the centers of the nine-point circles and for the de Longchamps points of the triangles. e shall not give precise formulations of these theorems leaving this task as an exercise to the reader. In ig. 11 the isosceles triangles on the de Longchamps points and on the centers of the nine-point circles are shown together. References [1] L. ankoff J. arfunkel: he heptagonal triangle. Mathematics Magazine 46 16 187 (197). []. Čerin: Hyperbolas orthology and antipedal triangles. lasnik Mat. 14 160 (1998). []. [4]. [5]. Čerin: eometrija pravilnog sedmerokuta. Poučak 14 5 14 (00). Čerin: Regular heptagon s intersections circles. lem. Math. (to appear). Čerin: Regular heptagon s midpoints circle. (preprint). [6] R. eaux: Introduction to the geometry of complex numbers. ngar Publishing o. New ork 1956. Received May 4 004; final form ugust 7 005