NNew Sense of a Circle
Mamuka Meskhishvili
Abstract
New condition is found for the set of points in the plane, for which thelocus is a circle.It is proved: the locus of points, such that the sum of the (2 m ) -th powers S (2 m ) n of the distances to the vertexes of fixed regular n -sided polygon isconstant, is a circle if S (2 m ) n > nr m , where m = 1 , , . . . , n − and r is the distance from the center of the regular polygon to the vertex.The radius (cid:96) satisfies: S (2 m ) n = n (cid:34) ( r + (cid:96) ) m + [ m ] (cid:88) k =1 (cid:18) m k (cid:19) ( r + (cid:96) ) m − k ( r(cid:96) ) k (cid:18) kk (cid:19) (cid:35) . Key words and phrases.
Locus, circle, center, regular polygon, vertex,constant sum of squares, constant sum of m -th powers, cosine sum, cosinepower sum. a r X i v : . [ m a t h . G M ] J un ontents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Regular Polygon. Constant Sum of 2nd Powers . . . . . . . . . . . . . . . . . . . . . . . . 4 Regular Triangle. Constant Sum of 4th Powers . . . . . . . . . . . . . . . . . . . . . . . 13 Square. Constant Sum of 4th and 6th Powers . . . . . . . . . . . . . . . . . . . . . . . . 15 Regular Pentagon. Constant Sum of 4th, 6th and 8th Powers . . . . . . . . 18 Circle as Locus of Constant Sum of Powers . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Cosine Multiple Arguments and Cosine Powers Sums . . . . . . . . . . . . . . . . . 30 General Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
Introduction
A locus of points is the set of points, and only those points, that satisfiesgiven conditions. A circle is usually defined as the locus of points (in theplane) at a given distance from a given point. This is well-known definitionof a circle, but not only. It is known definition of a circle by using two fixedpoints – the locus of points, such that the sum of the squares of the distancesto two distinct fixed points is constant, is a circle.More generally, for any collection of n number of points P i , constants λ i and S (2) n , distances d ( X, P i ) – the locus of points X such that n (cid:88) i =1 λ i d ( X, P i ) = S (2) n , is either (a) a circle, a point or empty set if n (cid:88) i =1 λ i (cid:54) = 0 , (b) a line, a plane or the empty set if n (cid:88) i =1 λ i = 0 .Indeed, let ( a i , b i ) be the coordinates of point P i and ( x, y ) the coordi-nates of point X . Then the equation satisfied by X takes the form S (2) n = n (cid:88) i =1 λ i (cid:0) ( x − a i ) + ( y − b i ) (cid:1) == ( x + y ) n (cid:88) i =1 λ i − x · n (cid:88) i =1 λ i a i −− y · n (cid:88) i =1 λ i b i + n (cid:88) i =1 λ i ( a i + b i ) .
3f the coefficient of x + y is nonzero, then this equation determines either acircle, a point or the empty set and if it is zero, then the equation determineseither a line, a plane, or the empty set [1] .In this article we consider under points P i – vertexes of a regular n -sidedpolygon P n and for coefficients λ i = 1 , i = 1 , , . . . , n. We start m = 1 , i.e. we have the constant sum of the second powers ofdistances from point X to the vertexes of regular n -sided polygon. In ournotations – case S (2) n . The case of two points ( n = 2) is well-known [2] ,thus we discuss n > . Theorem 2.1.
The locus of points, such that the sum of thesecond powers S (2) n of the distances to the vertexes of fixed regular n -sided polygon is constant, is a circle if S (2) n > nr , where r is the distance from the center of the regular polygon tothe vertex. The center of the circle is at the center of the regularpolygon and the radius is (cid:115) S (2) n n − r . f S (2) n = nr the locus is a point – the center of polygon andif S (2) n < nr it is the empty set. Firstly discuss the regular polygon with even number of vertexes n . Nu-merate vertexes as shown in Figure 1. Denote the distance from an arbitrarypoint M to the vertex i by x i , respectively, i = 1 , . . . , n and OM = (cid:96) .In even case vertexes and n − , . . . , k and n − k , . . . , n − and n + 1 are symmetrical with respect to the line which passes through center andvertex n , as shown in Figure 1. ÀÀ n O M n - k n -1 1 kn -1 nn +1 r a l Figure 1.5alculate the square of distances in pairs x and x n − , . . . , x k and x n − k , . . . , x n − and x n +1 ,x n and x n . To calculate angles, we use arrows directions (Figure 1) clockwise for , . . . , n − vertexes and counterclockwise for n − , . . . , n + 1 vertexes; x = r + (cid:96) − r(cid:96) cos (cid:16) ◦ n − α (cid:17) and x n − = r + (cid:96) − r(cid:96) cos (cid:16) ◦ n + α (cid:17) , ... x k = r + (cid:96) − r(cid:96) cos (cid:16) k ◦ n − α (cid:17) and x n − k = r + (cid:96) − r(cid:96) cos (cid:16) k ◦ n + α (cid:17) , ... x n − = r + (cid:96) − r(cid:96) cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17) and x n +1 = r + (cid:96) − r(cid:96) cos (cid:16) ( n −
1) 360 ◦ n + α (cid:17) ,x n = r + (cid:96) − r(cid:96) cos(180 ◦ − α ) and x n = r + (cid:96) − r(cid:96) cos α. Summarize each pair: 6 + x n − = 2( r + (cid:96) ) − r(cid:96) · α · cos 360 ◦ n , ... x k + x n − k = 2( r + (cid:96) ) − r(cid:96) · α · cos k ◦ n , ... x n − + x n +1 = 2( r + (cid:96) ) − r(cid:96) · α · cos (cid:16) ( n −
1) 360 ◦ n (cid:17) ,x n + x n = 2( r + (cid:96) ) − r(cid:96) · ◦ · cos(90 ◦ − α ) == 2( r + (cid:96) ) . There are n pairs, so total sum is S (2)2 n = n (cid:88) i =1 x i == 2 n ( r + (cid:96) ) − r(cid:96) cos α n − (cid:88) k =1 cos (cid:16) k ◦ n (cid:17) . n − (cid:88) k =1 cos (cid:16) k ◦ n (cid:17) = n − (cid:88) k =1 cos (cid:16) k ◦ n (cid:17) == cos 180 ◦ n +cos 2 180 ◦ n + · · · +cos( n −
2) 180 ◦ n +cos( n −
1) 180 ◦ n . After grouping 7 − (cid:88) k =1 cos (cid:16) k ◦ n (cid:17) == cos 180 ◦ n +cos( n −
1) 180 ◦ n +cos 2 180 ◦ n +cos( n −
2) 180 ◦ n + · · · += cos 180 ◦ n + cos (cid:16) ◦ − ◦ n (cid:17) + cos 2 180 ◦ n ++ cos (cid:16) ◦ − ◦ n (cid:17) + · · · = 0 , since each pair is . Note. If n is odd we have n − pairs, but if n is even n − pairs (whosesums are ) and one more alone term cos (cid:16) n · ◦ n (cid:17) , which itself is . Sowe prove n − (cid:88) k =1 cos k ◦ n = 0 , and from that it follows S (2) n = n (cid:88) i =1 x i = 2 n ( r + (cid:96) ) . Now discuss the case, when the number of vertexes of a regular polygonis odd – n + 1 .Numerate vertexes as shown in Figure 2. As in even case vertexes and n , . . . , k and n + 1 − k , . . . , n and n + 1 andvertex n + 1 , as shown in Figure 2. ÀÀ n +1 O M n +1- k n knn +1 r l a Figure 2.Calculate the square of distances in pairs – x and x n , . . . , x k and x n +1 − k , . . . , x n and x n +1 . Only x n +1 – square of distance remains single. To calculate the angles,we use arrows directions (Figure 2) clockwise for vertexes – , . . . , n andcounterclockwise for vertexes n, . . . , n + 1 ; x = r + (cid:96) − r(cid:96) cos (cid:16) ◦ n + 1 − α (cid:17) and x n = r + (cid:96) − r(cid:96) cos (cid:16) ◦ n + 1 + α (cid:17) , ... k = r + (cid:96) − r(cid:96) cos (cid:16) k ◦ n + 1 − α (cid:17) and x n +1 − k = r + (cid:96) − r(cid:96) cos (cid:16) k ◦ n + 1 + α (cid:17) , ... x n = r + (cid:96) − r(cid:96) cos (cid:16) n ◦ n + 1 − α (cid:17) and x n +1 = r + (cid:96) − r(cid:96) cos (cid:16) n ◦ n + 1 + α (cid:17) ,x n +1 = r + (cid:96) − r(cid:96) cos α. There are n pairs, so total sum is S (2)2 n +1 = n +1 (cid:88) i =1 x i == (2 n + 1)( r + (cid:96) ) − r(cid:96) cos α −− r(cid:96) n (cid:88) k =1 (cid:18) cos (cid:16) k ◦ n + 1 − α (cid:17) + cos (cid:16) k ◦ n + 1 + α (cid:17)(cid:19) == (2 n + 1)( r + (cid:96) ) − r(cid:96) cos α −− r(cid:96) n (cid:88) k =1 α · cos (cid:16) k ◦ n + 1 (cid:17) == (2 n + 1)( r + (cid:96) ) −− r(cid:96) cos α (cid:18) n (cid:88) k =1 (cid:16) cos k ◦ n + 1 (cid:17)(cid:19) . n (cid:88) k =1 cos k ◦ n + 1 , multiply by ◦ n + 1 , and then use trigonometric identity C · cos D = sin( C + D ) − sin( D − C ) we get ◦ n + 1 (cid:16) cos 2 · ◦ n + 1 + cos 2 2 · ◦ n + 1 ++ cos 3 2 · ◦ n + 1 + · · · ++ cos( n −
1) 2 · ◦ n + 1 + cos n · ◦ n + 1 (cid:17) = sin 3 · ◦ n + 1 − sin 180 ◦ n + 1 ++ sin 5 · ◦ n + 1 − sin 3 · ◦ n + 1 ++ sin 7 · ◦ n + 1 − sin 5 · ◦ n + 1 +...+ sin (2 n − · ◦ n + 1 − sin (2 n − · ◦ n + 1 ++ sin (2 n + 1) · ◦ n + 1 − sin (2 n − · ◦ n + 1 == sin (2 n + 1) · ◦ n + 1 − sin 180 ◦ n + 1 = − sin 180 ◦ n + 1 . So, n (cid:88) k =1 cos k ◦ n + 1 = − and in odd case we obtain S (2)2 n +1 = n +1 (cid:88) i =1 x i = (2 n + 1)( r + (cid:96) ) . Summarize even and odd cases. For an arbitrary number n , we have: S (2) n = n ( r + (cid:96) ) , which finally proves the Theorem 2.1.12 Regular Triangle.Constant Sum of 4th Powers
The sum of the fourth powers of the distances to the vertexes is: S (4)3 = (cid:88) i =1 x i == ( r + (cid:96) − r(cid:96) cos α ) + (cid:0) r + (cid:96) − r(cid:96) cos(120 ◦ − α ) (cid:1) ++ (cid:0) r + (cid:96) − r(cid:96) cos(120 ◦ + α ) (cid:1) . Introduce new notations A = r + (cid:96) and B = 2 r(cid:96) ; S (4)3 = ( A − B cos α ) + (cid:0) A − B cos(120 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(120 ◦ + α ) (cid:1) == 3 A − AB (cid:0) cos α + cos(120 ◦ − α ) + cos(120 ◦ + α ) (cid:1) ++ B (cid:0) cos α + cos (120 ◦ − α ) + cos (120 ◦ + α ) (cid:1) ;cos α + cos(120 ◦ − α ) + cos(120 ◦ + α ) == cos α + 2 cos 120 ◦ cos α = 0 , cos α + cos (120 ◦ − α ) + cos (120 ◦ + α ) == 12 (cid:16) α + cos(240 ◦ − α ) + cos(240 ◦ + 2 α ) (cid:17) =
32 + 12 (cid:16) cos 2 α + cos(120 ◦ + 2 α ) + cos(120 ◦ − α ) (cid:17) = 32 . So, S (4)3 = 3 A + 32 B = 3( r + (cid:96) + 4 r (cid:96) ) . The sum S (4)3 does not contain α , i.e. this expression does not dependon direction OM and only depends on length – OM .Since the triangle is fixed r is constant and this expression is increasingunder (cid:96) ( (cid:96) > , so obtained relation is one to one: S (4)3 = const ⇐⇒ (cid:96) = const. We proved
Theorem 3.1.
The locus of points, such that the sum of thefourth powers S (4)3 of the distances to the vertexes of fixed regulartriangle is constant, is a circle if S (4)3 > r , where r is the distance from the center of the regular triangle tothe vertex. The center of the circle is at the center of the regulartriangle and the radius (cid:96) satisfies the following condition: S (4)3 = 3 (cid:104) ( r + (cid:96) ) + 2( r(cid:96) ) (cid:105) . If S (4)3 = 3 r the locus is a point – the center of the regulartriangle and if S (4)3 < r it is the empty set. ote. The sum of the sixth powers – S (6)3 contains α , so in this case thelocus is not a circle. S (4)4 = ( A − B cos α ) + (cid:0) A − B cos(90 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(180 ◦ − α ) (cid:1) + (cid:0) A − B cos(90 ◦ + α ) (cid:1) == 4 A − AB (cid:0) cos α + cos(90 ◦ − α )++ cos(180 ◦ − α ) + cos(90 ◦ + α ) (cid:1) ++ B (cid:0) cos α + cos (90 ◦ − α )++ cos (180 ◦ − α ) + cos (90 ◦ + α ) (cid:1) ;cos α + cos(90 ◦ − α ) + cos(180 ◦ − α ) + cos(90 ◦ + α ) == 2 cos 90 ◦ · cos(90 ◦ − α ) + 2 cos 90 ◦ · cos α = 0 , cos α + cos (90 ◦ − α ) + cos (180 ◦ − α ) + cos (90 ◦ + α ) == 12 (cid:16) α + cos(180 ◦ − α )++ cos(360 ◦ − α ) + cos(180 ◦ + 2 α ) (cid:17) = 2 ,S (4)4 = 4 A + 2 B = 4( r + (cid:96) + 4 r (cid:96) ) . heorem 4.1. The locus of points, such that the sum of thefourth powers S (4)4 of the distances to the vertexes of fixed squareis constant, is a circle if S (4)4 > r , where r is the distance from the center of the square to the vertex.The center of the circle is at the center of the square and theradius (cid:96) satisfying the following condition: S (4)4 = 4 (cid:104) ( r + (cid:96) ) + 2( r(cid:96) ) (cid:105) . If S (4)4 = 4 r the locus is a point – the center of the squareand if S (4)4 < r it is the empty set. The sum of sixth powers: S (6)4 = ( A − B cos α ) + (cid:0) A − B cos(90 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(180 ◦ − α ) (cid:1) + (cid:0) A − B cos(90 ◦ + α ) (cid:1) == 4 A − A B (cid:16) cos α + cos(90 ◦ − α )++ cos(180 ◦ − α ) + cos(90 ◦ + α ) (cid:17) ++ 3 AB (cid:16) cos α + cos (90 ◦ − α )++ cos (180 ◦ − α ) + cos (90 ◦ + α ) (cid:17) −− B (cid:16) cos α + cos (90 ◦ − α )++ cos (180 ◦ − α ) + cos (90 ◦ + α ) (cid:17) , os α + cos (90 ◦ − α ) + cos (180 ◦ − α ) + cos (90 ◦ + α ) == cos α + cos (90 ◦ − α ) − cos α − cos (90 ◦ − α ) = 0 . So, S (6)4 = 4 A + 3 AB · r + (cid:96) )( r + (cid:96) + 8 r (cid:96) ) . For fixed r , obtained expression is increasing under variable (cid:96) ( (cid:96) > , soit is one to one relation between the sum of the sixth powers S (6)4 and (cid:96) : S (6)4 = const ⇐⇒ (cid:96) = const. Theorem 4.2.
The locus of points, such that the sum of thesixth powers S (6)4 of the distances to the vertexes of fixed squareis constant, is a circle if S (6)4 > r , where r is the distance from the center of the square to the vertex.The center of the circle is at the center of the square and theradius (cid:96) satisfying the following condition: S (6)4 = 4 (cid:104) ( r + (cid:96) ) + 6( r + (cid:96) )( r(cid:96) ) (cid:105) . If S (6)4 = 4 r the locus is a point – the center of the squareand if S (6)4 < r it is the empty set. Note.
The sum of the eighth powers – S (8)4 contains α , so in this case thelocus is not a circle. 17 Regular Pentagon.Constant Sum of 4th, 6th and 8th Powers S (4)5 = ( r + (cid:96) − r(cid:96) cos α ) + (cid:0) r + (cid:96) − r(cid:96) cos(72 ◦ − α ) (cid:1) ++ (cid:0) r + (cid:96) − r(cid:96) cos(144 ◦ − α ) (cid:1) ++ (cid:0) r + (cid:96) − r(cid:96) cos(144 ◦ + α ) (cid:1) ++ (cid:0) r + (cid:96) − r(cid:96) cos(72 ◦ + α ) (cid:1) == ( A − B cos α ) + (cid:0) A − B cos(72 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(144 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(144 ◦ + α ) (cid:1) + (cid:0) A − B cos(72 ◦ + α ) (cid:1) == 5 A − AB (cid:16) cos α + cos(72 ◦ − α ) + cos(144 ◦ − α )++ cos(144 ◦ + α ) + cos(72 ◦ + α ) (cid:17) ++ B (cid:16) cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) . cos α + cos(72 ◦ − α ) + cos(144 ◦ − α ) + cos(144 ◦ + α )++ cos(72 ◦ + α ) = cos α + 2 cos α · ◦ + 2 cos α · cos 144 ◦ == cos α (cid:0) ◦ + cos 144 ◦ ) (cid:1) . Indeed, 18 in 18 ◦ = √ − , cos 72 ◦ + cos 144 ◦ = sin 18 ◦ − cos 36 ◦ == sin 18 ◦ − − ◦ ) == √ − − √ − − , i.e. cos α + cos(72 ◦ − α ) + cos(144 ◦ − α )++ cos(144 ◦ + α ) + cos(72 ◦ + α ) = 0 . Calculate the sum of the second powers: cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) == 12 (cid:16) α + cos(144 ◦ − α ) + cos(144 ◦ + 2 α )++ cos(288 ◦ − α ) + cos(288 ◦ + 2 α ) (cid:17) == 12 (cid:16) α + cos(144 ◦ − α ) + cos(144 ◦ + 2 α )++ cos(72 ◦ + 2 α ) + cos(72 ◦ − α ) (cid:17) = 52 . S (4)5 = 5 A + 52 B == 5( r + (cid:96) ) + 10 r (cid:96) = 5( r + (cid:96) + 4 r (cid:96) ) . Theorem 5.1.
The locus of points, such that the sum of thefourth powers S (4)5 of the distances to the vertexes of fixed regularpentagon is constant, is a circle if S (4)5 > r , where r is the distance from the center of the regular pentagon tothe vertex. The center of the circle is at the center of the regularpentagon and the radius (cid:96) satisfies the following condition: S (4)5 = 5 (cid:104) ( r + (cid:96) ) + 2( r(cid:96) ) (cid:105) . If S (4)5 = 5 r the locus is a point – the center of the pentagonand if S (4)5 < r it is the empty set. (6)5 = ( A − B cos α ) + (cid:0) A − B cos(72 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(144 ◦ − α ) (cid:1) + (cid:0) A − B cos(144 ◦ + α ) (cid:1) ++ (cid:0) A − B cos(72 ◦ + α ) (cid:1) == A − A B cos α + 3 AB cos α − B cos α ++ A − A B cos(72 ◦ − α ) + 3 AB cos (72 ◦ − α ) −− B cos (72 ◦ − α )++ A − A B cos(144 ◦ − α ) + 3 AB cos (144 ◦ − α ) −− B cos (144 ◦ − α )++ A − A B cos(144 ◦ + α ) + 3 AB cos (144 ◦ + α ) −− B cos (144 ◦ + α )++ A − A B cos(72 ◦ + α ) + 3 AB cos (72 ◦ + α ) −− B cos (72 ◦ + α ) == 5 A − A B (cid:16) cos α + cos(72 ◦ − α ) + cos(144 ◦ − α )++ cos(144 ◦ + α ) + cos(72 ◦ + α ) (cid:17) ++ 3 AB (cid:16) cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) −− B (cid:16) cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) . cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) == 14 (cid:16) α + cos 3 α ++ 3 cos(72 ◦ − α ) + cos(216 ◦ − α )++ 3 cos(144 ◦ − α ) + cos(432 ◦ − α )++ 3 cos(144 ◦ + α ) + cos(432 ◦ + 3 α )++ 3 cos(72 ◦ + α ) + cos(216 ◦ + 3 α ) (cid:17) == 14 (cid:16) (cid:16) cos α + cos(72 ◦ − α ) + cos(144 ◦ − α )++ cos(144 ◦ + α ) + cos(72 ◦ + α ) (cid:17) ++ cos 3 α + cos(216 ◦ − α ) + cos(432 ◦ − α )++ 3 cos(432 ◦ + 3 α ) + cos(216 ◦ + 3 α ) (cid:17) == 14 (cid:16) cos 3 α + cos(144 ◦ + 3 α ) + cos(72 ◦ − α )++ cos(72 ◦ + 3 α ) + cos(144 ◦ − α ) (cid:17) = 0 . So, 22 (6)5 = 5 A + 3 AB ·
52 == 5 A (cid:16) A + 3 B (cid:17) = 5( r + (cid:96) )( r + (cid:96) + 8 r (cid:96) ) . Theorem 5.2.
The locus of points, such that the sum of thesixth powers S (6)5 of the distances to the vertexes of fixed regularpentagon is constant, is a circle if S (6)5 > r , where r is the distance from the center of the regular pentagon tothe vertex. The center of the circle is at the center of the regularpentagon and the radius (cid:96) satisfies the following condition: S (6)5 = 5 (cid:104) ( r + (cid:96) ) + 6( r + (cid:96) )( r(cid:96) ) (cid:105) . If S (6)5 = 5 r the locus is a point – the center of the regularpentagon and if S (6)5 < r it is the empty set. S (8)5 = ( A − B cos α ) + (cid:0) A − B cos(72 ◦ − α ) (cid:1) ++ (cid:0) A − B cos(144 ◦ − α ) (cid:1) + (cid:0) A − B cos(144 ◦ + α ) (cid:1) ++ (cid:0) A − B cos(72 ◦ + α ) (cid:1) = A − A B cos α + 6 A B cos α −− AB cos α + B cos α + · · · == 5 A − A B (cid:16) cos α + cos(72 ◦ − α ) + cos(144 ◦ − α )++ cos(144 ◦ + α ) + cos(72 ◦ + α ) (cid:17) ++ 6 A B (cid:16) cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) −− AB (cid:16) cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) ++ B (cid:16) cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) . Calculate the sum of the fourth powers: cos α + cos (72 ◦ − α ) + cos (144 ◦ − α )++ cos (144 ◦ + α ) + cos (72 ◦ + α ) == 18 (cid:16) cos 4 α + 8 cos α − ◦ − α ) + 8 cos (72 ◦ − α ) − ◦ − α ) + 8 cos (144 ◦ − α ) − cos(576 ◦ + 4 α ) + 8 cos (144 ◦ + α ) − ◦ + 4 α ) + 8 cos (72 ◦ + α ) − (cid:17) == 18 (cid:16) − (cid:16) cos α + cos (72 ◦ − α )++ cos (144 ◦ − α ) + cos (144 ◦ + α ) + cos (72 ◦ + α ) (cid:17) ++ cos 4 α + cos(288 ◦ − α ) + cos(576 ◦ − α )++ cos(576 ◦ + 4 α ) + cos(288 ◦ + 4 α ) (cid:17) == 18 (cid:16) − ·
52 + cos 4 α + cos(72 ◦ + 4 α )++ cos(144 ◦ + 4 α ) + cos(144 ◦ − α ) + cos(72 ◦ − α ) (cid:17) == 158 ,S (8)5 = 5 A + 6 A B ·
52 + B ·
158 == 5( r + (cid:96) ) ( r + (cid:96) + 14 r (cid:96) ) + 30 r (cid:96) . Theorem 5.3.
The locus of points, such that the sum of theeighth powers S (8)5 of the distances to the vertexes of fixed regularpentagon is constant, is a circle if S (8)5 > r , here r is the distance from the center of the regular pentagon tothe vertex. The center of the circle is at the center of the regularpentagon and the radius (cid:96) satisfies the following condition: S (8)5 = 5 (cid:104) ( r + (cid:96) ) + 12( r + (cid:96) ) ( r(cid:96) ) + 6( r(cid:96) ) (cid:105) . If S (8)5 = 5 r the locus is a point – the center of the regularpentagon and if S (8)5 < r it is the empty set. Note.
The sum of the tenth powers contains α , so the locus is not a circle. Summarize obtained results. If d ( X, P i ) is the distance from X to thevertex P i of the regular n -sided polygon and r is the distance from thecenter of the regular polygon to the vertex, we have: (cid:110) X : n (cid:88) i =1 d ( X, P i ) = S (2) n = const (cid:111) == circle , if S (2) n > nr point , if S (2) n = nr ∅ , if S (2) n < nr , . (cid:110) X : (cid:88) i =1 d ( X, P i ) = S (4)3 = const (cid:111) == circle , if S (4)3 > r point , if S (4)3 = 3 r ∅ , if S (4)3 < r . (cid:110) X : (cid:88) i =1 d ( X, P i ) = S (4)4 = const (cid:111) == circle , if S (4)4 > r point , if S (4)4 = 4 r ∅ , if S (4)4 < r . (cid:110) X : (cid:88) i =1 d ( X, P i ) = S (6)4 = const (cid:111) == circle , if S (6)4 > r point , if S (6)4 = 4 r ∅ , if S (6)4 < r . . (cid:110) X : (cid:88) i =1 d ( X, P i ) = S (4)5 = const (cid:111) == circle , if S (4)5 > r point , if S (4)5 = 5 r ∅ , if S (4)5 < r . (cid:110) X : (cid:88) i =1 d ( X, P i ) = S (6)5 = const (cid:111) == circle , if S (6)5 > r point , if S (6)5 = 5 r ∅ , if S (6)5 < r . (cid:110) X : (cid:88) i =1 d ( X, P i ) = S (8)5 = const (cid:111) == circle , if S (8)5 > r point , if S (8)5 = 5 r ∅ , if S (8)5 < r . Now we are ready to state general hypothesis:28 ypothesis.
The locus of points, such that the sum of the (2 m ) -th powers S (2 m ) n of the distances to the vertexes of fixedregular n -sided polygon is constant, is a circle if S (2 m ) n > nr m , where m = 1 , , . . . , n − and r is the distance from the center of the regular polygon tothe vertex. The center of the circle is at the center of the regularpolygon.If S (2 m ) n = nr m the locus is a point – the center of the polygonand if S (2 m ) n < nr m it is the empty set. Thus, for each even (2 m ) -th power, so that m < n : (cid:110) X : n (cid:88) i =1 d m ( X, P i ) = S (2 m ) n = const (cid:111) == circle , if S (2 m ) n > nr m point , if S (2 m ) n = nr m ∅ , if S (2 m ) n < nr m . To prove the hypothesis, first of all two lemmas must be proved.29
Cosine Multiple Argumentsand Cosine Powers Sums
Lemma 7.1.
For arbitrary positive integers m and n , so that m < n , the following condition n (cid:88) k =1 cos (cid:18) m (cid:16) α − ( k −
1) 360 ◦ n (cid:17)(cid:19) = 0 is satisfied, where α is an arbitrary angle. Denote by P = e imα + e im ( α − ◦ n ) + e im ( α − ◦ n ) + · · · + e im ( α − ( n − ◦ n ) . By using Euler’s formula e iϕ = cos ϕ + i sin ϕ real part of P is: Re( P ) = n (cid:88) k =1 cos (cid:18) m (cid:16) α − ( k −
1) 360 ◦ n (cid:17)(cid:19) . The formula of the sum of geometric progression gives:30 = e imα (cid:18) e − im ◦ n + (cid:0) e − im ◦ n (cid:1) + · · · + (cid:0) e − im ◦ n (cid:1) n − (cid:19) == e imα − ( e − im ◦ n ) n − e − im ◦ n ,e − im ◦ = cos( − ◦ m ) + i ( − ◦ m ) = 1 . Since m < n , e − im ◦ n (cid:54) = 1 . So P = 0 , i.e. Re( P ) = 0 whichproves the Lemma 7.1. Note. If m = n , the sum equals n cos( nα ) . The sum is not zeroonly in cases when m is multiple of n . If m = pn , the sum is n cos( pnα ) . Lemma 7.2.
For arbitrary positive integers m and n , so that m < n and for an arbitrary angle α the following conditions aresatisfied:if m is odd n (cid:88) k =1 cos m (cid:16) α − ( k −
1) 360 ◦ n (cid:17) = 0; if m is even n (cid:88) k =1 cos m (cid:16) α − ( k −
1) 360 ◦ n (cid:17) = n (cid:16) m m (cid:17) m .
31y using power-reduction formula for cosine, when m is odd: cos m θ = 22 m m − (cid:88) k =0 (cid:18) mk (cid:19) cos (cid:0) ( m − k ) θ (cid:1) , we have n (cid:88) k =1 cos m (cid:16) α − ( k −
1) 360 ◦ n (cid:17) == cos m α + cos m (cid:16) α − ◦ n (cid:17) + · · · ++ cos m (cid:16) α − ( n −
1) 360 ◦ n (cid:17) == 22 m (cid:34) (cid:18) m (cid:19) cos mα + (cid:18) m (cid:19) cos( m − α + · · · + (cid:18) m m − (cid:19) cos α ++ (cid:18) m (cid:19) cos m (cid:16) α − ◦ n (cid:17) + (cid:18) m (cid:19) cos( m − (cid:16) α − ◦ n (cid:17) + · · · ++ (cid:18) m m − (cid:19) cos (cid:16) α − ◦ n (cid:17) + · · · ++ (cid:18) m (cid:19) cos m (cid:16) α − ( n −
1) 360 ◦ n (cid:17) + (cid:18) m (cid:19) cos( m − (cid:16) α − ( n −
1) 360 ◦ n (cid:17) ++ · · · + (cid:18) m m − (cid:19) cos (cid:16) α − ( n −
1) 360 ◦ n (cid:17)(cid:35) = m (cid:34) (cid:18) m (cid:19) (cid:18) cos mα + cos m (cid:16) α − ◦ n (cid:17) + · · · ++ cos m (cid:16) α − ( n −
1) 360 ◦ n (cid:17)(cid:19) ++ (cid:18) m (cid:19) (cid:18) cos( m − α +cos( m − (cid:16) α − ◦ n (cid:17) + · · · ++ cos( m − (cid:16) α − ( n −
1) 360 ◦ n (cid:17)(cid:19) +...+ (cid:18) m m − (cid:19) (cid:18) cos α + cos (cid:16) α − ◦ n (cid:17) + · · · ++ cos (cid:16) α − ( n −
1) 360 ◦ n (cid:17)(cid:19)(cid:35) . Since m < n , from the Lemma 7.1 follows – each sum equals zero, whichproves the first part of the Lemma 7.2.When m is even power-reduction formula for cosine is: cos m θ = 12 m (cid:18) m m (cid:19) + 22 m m − (cid:88) k =0 (cid:18) mk (cid:19) cos (cid:0) ( m − k ) θ (cid:1) . By using the same manner, as in preceding part the sum of the second ad-dends gives zero, and because the number of the first addends is n total sumequals: 33 (cid:16) m m (cid:17) m , which proves the Lemma 7.2. Theorem 8.1.
The locus of points, such that the sum of the (2 m ) -th powers S (2 m ) n of distances to the vertexes of fixed regular n -sided polygon is constant, is a circle if S (2 m ) n > nr m , where m = 1 , , . . . , n − and r is the distance from the center of the regular polygon tothe vertex. The center of the circle is at the center of the regularpolygon and the radius (cid:96) satisfies the following condition: S (2 m ) n = n (cid:34) ( r + (cid:96) ) m + [ m ] (cid:88) k =1 (cid:18) m k (cid:19) ( r + (cid:96) ) m − k ( r(cid:96) ) k (cid:18) kk (cid:19) (cid:35) . If S (2 m ) n = nr m the locus is a point – the center of the polygonand if S (2 m ) n < nr m the locus is the empty set. For (2 m ) -th power sum of the distances from an arbitrary point to thevertexes P i of regular n -sided polygon, we have:34 (2 m ) n = ( A − B cos α ) m + (cid:18) A − B cos (cid:16) ◦ n − α (cid:17)(cid:19) m ++ (cid:18) A − B cos (cid:16) · ◦ n − α (cid:17)(cid:19) m + · · · ++ (cid:18) A − B cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) m == nA m − (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) ◦ n − α (cid:17) + · · · ++ cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) ++ (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) ◦ n − α (cid:17) + · · · ++ cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) −− (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) ◦ n − α (cid:17) + · · · ++ cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) +... ± (cid:18) mm (cid:19) B m (cid:18) cos m α + cos m (cid:16) ◦ n − α (cid:17) + · · · ++ cos m (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) .
35y using the Lemma 7.2 each sum with negative sign “ − ” is zero, becausethey contain odd powers and only sum with even powers remains.If m is even the sum is: S (2 m ) n = nA m ++ (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) ◦ n − α (cid:17) + · · · ++ cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) +...+ (cid:18) mm (cid:19) B m (cid:18) cos m α + cos m (cid:16) ◦ n − α (cid:17) + · · · ++ cos m (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) == n (cid:32) A m + m (cid:88) k =1 (cid:18) m k (cid:19) A m − k B k k (cid:18) kk (cid:19) (cid:33) . If m is odd the sum is S (2 m ) n = nA m ++ (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) ◦ n − α (cid:17) + · · · ++ cos (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) + · · · + (cid:18) mm − (cid:19) AB m − (cid:18) cos m − α + cos m − (cid:16) ◦ n − α (cid:17) ++ · · · + cos m − (cid:16) ( n −
1) 360 ◦ n − α (cid:17)(cid:19) == n (cid:32) A m + m − (cid:88) k =1 (cid:18) m k (cid:19) A m − k B k k (cid:18) kk (cid:19) (cid:33) . Using integer part, it is possible to write obtained results in one formula: S (2 m ) n = n (cid:32) A m + [ m ] (cid:88) k =1 (cid:18) m k (cid:19) A m − k B k k (cid:18) kk (cid:19) (cid:33) , which proves the Theorem 8.1.Thus, for each even (2 m ) -th power m = 1 , , . . . , n − : (cid:110) X : n (cid:88) i =1 d m ( X, P i ) = S (2 m ) n = const (cid:111) == circle , if S (2 m ) n > nr m point , if S (2 m ) n = nr m ∅ , if S (2 m ) n < nr m . In case of circle, the radius (cid:96) satisfies:37 (2 m ) n = n (cid:34) ( r + (cid:96) ) m + [ m ] (cid:88) k =1 (cid:18) m k (cid:19) ( r + (cid:96) ) m − k ( r(cid:96) ) k (cid:18) kk (cid:19) (cid:35) . What happens if we consider the sum of even powers more than n − ?In these cases, cosine functions appear in the sum expressions (see note ofLemma 7.1), which proves that the locus is not a circle. But what kindof figures are there? The investigation of this subject is the aim of futurestudies. References [1] Prosolov Viktor,
Problems in Plane and Solid Geometry . Vol. 1.
PlaneGeometry . Translated and edit by Dimitry Leites, 2005.[2] Jones, Dustin. A Collection of Loci Using Two Fixed Points.
MissouriJournal of Mathematical Sciences , No. 19, 2007.
Author’s address:
Georgian-American High School, 18 Chkondideli Str., Tbilisi 0180, Georgia.
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