Cyclic Averages of Regular Polygons and Platonic Solids
aa r X i v : . [ m a t h . G M ] O c t Cyclic Averages of Regular Polygons andPlatonic Solids
Mamuka Meskhishvili
Abstract
The concept of the cyclic averages are introduced for a regular polygon P n and a Platonicsolid T n . It is shown that cyclic averages of equal powers are the same for various P n ( T n ),but their number is characteristic of P n ( T n ). Given the definition of a circle (sphere) bythe vertices of P n ( T n ) and on the base of the cyclic averages are established the commonmetrical relations of P n ( T n ). MSC.
Keywords and phrases.
Regular polygon, Platonic solid, circle, sphere, locus, sum of likepowers, rational distances problem
Consider a finite set of n points in the plane (space), then locus of points such that the sum ofthe squares of distances to the given points is constant, is a circle (sphere), whose center is atthe centroid of the given points [1, 2].Denote by M ( d , d , . . . , d n , L ) an arbitrary point in the plane (space) of a regular polygon(Platonic solid) of distances d , d , . . . , d n to the vertices A , A , . . . , A n , then: n X d i = n ( R + L ) , ( ∗ )where R is the radius of circumscribed circle (sphere) of the regular polygon (Platonic solid)and L is the distance between the point M and the centroid O .The symmetric equation exists for an equilateral triangle and an arbitrary point M ( d , d , d , L ) in the plane of the triangle3( d + d + d + a ) = ( d + d + d + a ) , ( ∗∗ )where a is the side of the triangle [8, 11]. 1he arbitrary point is always considered in the plane of the regular polygon, and in thespace of the Platonic solid, respectively.From relations ( ∗ ) and ( ∗∗ ) follows: X d i = 3( R + L ) , X d i = 3 (cid:0) ( R + L ) + 2 R L (cid:1) . For a given equilateral triangle, the side a as well as the circumradius R are fixed so that, Theorem 1.1.
The locus of points such that X d i = const is a circle, center of which is the centroid. As we see, the distances are considered to the second and the fourth powers. Naturally, weare interested to know what happens if we consider the distances of higher powers.
For an equilateral triangle the expression P d i , contains α – the angle between R and L , so thelocus is not a circle, but for a square case the answer is surprising: the locus of points such that X d i = const is a circle.Generally, the locus is a circle (sphere) if and only if the sum of power distances can beexpressed in terms of L and some fixed element (with length) of a given regular polygon (Platonicsolid). The fixed element is possible to express in terms of R , so we denote such sums by thesymbol P [ R,L ] , or P (2 m )[ R,L ] – to indicate the like powers of the distances.Denote by P n ( R ) and T n ( R ) a regular polygon and Platonic solid, respectively, with an n number of the vertices and circumscribed radius R . The value of the P [ R,L ] remains constantwhen the point M moves on the circle C ( O, L ) (sphere S ( O, L )). So,
Definition 2.1. P [ R,L ] – is the sum of like powers of the distances d , . . . , d n from an arbitrarypoint M ( d , . . . , d n , L ) to the vertices P n ( R ) ( T n ( R )) the value of which is constant for any pointof the C ( O, L ) ( S ( O, L )) .
2t is clear, the sum of odd power contains radicals and never will be P [ R,L ] .For establishing common properties of the P n and T n discussing average of P [ R,L ] is muchpreferred S (2 m ) n = 1 n X (2 m )[ R,L ] . Definition 2.2.
The cyclic averages S (2 m ) n ( S (2 m )[ n ] ) of a regular polygon (Platonic solid) is theaverage of the sum P (2 m )[ R,L ] . We call such averages the cyclic averages, because as we prove the cyclic averages of equalpowers of various P n ( T n ) for fixed R and L are the same (if they exist): S (2 m ) n = S (2 m ) n , if n ≤ n .On the other hand for any given P n the number of S (2 m ) n (as well as P (2 m )[ R,L ] ) is defineduniquely, so the number of the cyclic averages is characteristic of the regular polygon.For example, 2 cyclic averages exist for a regular 3-gon: S (2)3 and S (4)3 , while for a regular 4-gon – 3 cyclic averages: S (2)4 , S (4)4 and S (6)4 . They are in relations: S (2)3 = S (2)4 and S (4)3 = S (4)4 . To demonstrate the efficiency of cyclic averages the analogue of the relation ( ∗∗ ) will beobtained for the square. Firstly, we turn ( ∗∗ ) in terms of R and the cyclic averages – S (2)3 , S (4)3 : d + d + d R = (cid:16) d + d + d R (cid:17) , then replace with S (2)3 = S (2)4 , S (4)3 = S (4)4 and R = a √ Theorem 2.1.
For an arbitrary point M ( d , . . . , d , L ) in the plane of a square: d + d + d + d + 3 a ) = ( d + d + d + d + 2 a ) , where a is the side of the square. Circle as Locus of Constant P [ R,L ] Sums
Theorem 3.1.
For an arbitrary point M ( d , d , . . . , d n , L ) in the plane of regular polygon P n ( R ) : n X i =1 d mi = n (cid:20) ( R + L ) m + ⌊ m ⌋ X k =1 (cid:18) m k (cid:19)(cid:18) kk (cid:19) R k L k ( R + L ) m − k (cid:21) , where m = 1 , . . . , n − . First we need to prove two lemmas.
Lemma 3.1.
For arbitrary positive integers m and n , such that m < n , the following condition n X k =1 cos (cid:18) m (cid:16) α − ( k −
1) 2 πn (cid:17)(cid:19) = 0 is satisfied, where α is an arbitrary angle. Denote T = e imα + e im ( α − πn ) + e im ( α − πn ) + · · · + e im ( α − ( n − πn ) . The real part of T is Re( T ) = n X k =1 cos (cid:18) m (cid:16) α − ( k −
1) 2 πn (cid:17)(cid:19) . The formula of the sum of geometric progression gives T = e imα (cid:18) e − im πn + (cid:0) e − im πn (cid:1) + · · · + (cid:0) e − im πn (cid:1) n − (cid:19) = e imα − ( e − im πn ) n − e − im πn ,e − im π = cos( − πm ) + i sin( − πm ) = 1 . Since m < n , e − im πn = 1. So T = 0, i.e. Re( T ) = 0, which proves Lemma 3.1. Remark 3.1. If m ≥ n , the sum always contains α . Lemma 3.2.
For arbitrary positive integers m and n , such that m < n and for an arbitraryangle α the following conditions are satisfied:if m is odd n X k =1 cos m (cid:16) α − ( k −
1) 2 πn (cid:17) = 0;4 f m is even n X k =1 cos m (cid:16) α − ( k −
1) 2 πn (cid:17) = n (cid:0) m m (cid:1) m . When m is odd, using the power-reduction formula for cosinecos m θ = 22 m m − X k =0 (cid:18) mk (cid:19) cos (cid:0) ( m − k ) θ (cid:1) , we obtain n X k =1 cos m (cid:16) α − ( k −
1) 2 πn (cid:17) = cos m α + cos m (cid:16) α − πn (cid:17) + · · · + cos m (cid:16) α − ( n −
1) 2 πn (cid:17) = 22 m "(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 − πn (cid:17) + (cid:18) m (cid:19) cos( m − (cid:16) α − πn (cid:17) + · · · + (cid:18) m m − (cid:19) cos (cid:16) α − πn (cid:17) + · · · + (cid:18) m (cid:19) cos m (cid:16) α − ( n −
1) 2 πn (cid:17) + (cid:18) m (cid:19) cos( m − (cid:16) α − ( n −
1) 2 πn (cid:17) + · · · + (cid:18) m m − (cid:19) cos (cid:16) α − ( n −
1) 2 πn (cid:17) = 22 m "(cid:18) m (cid:19)(cid:18) cos mα + cos m (cid:16) α − πn (cid:17) + · · · + cos m (cid:16) α − ( n −
1) 2 πn (cid:17)(cid:19) + (cid:18) m (cid:19)(cid:18) cos( m − α +cos( m − (cid:16) α − πn (cid:17) + · · · + cos( m − (cid:16) α − ( n −
1) 2 πn (cid:17)(cid:19) + · · · + (cid:18) m m − (cid:19)(cid:18) cos α + cos (cid:16) α − πn (cid:17) + · · · + cos (cid:16) α − ( n −
1) 2 πn (cid:17)(cid:19) . Since m < n , from Lemma 3.1 it follows that each sum equals zero, which proves the first partof Lemma 3.2. 5hen m is even, the power-reduction formula for cosine iscos m θ = 12 m (cid:18) m m (cid:19) + 22 m m − X k =0 (cid:18) mk (cid:19) cos (cid:0) ( m − k ) θ (cid:1) . Analogously to the case with odd m , the sum of the second addenda vanishes, and since thenumber of the first addenda is n , the total sum equals n (cid:18) m m (cid:19) m , which proves Lemma 3.2. Proof of Theorem 3.1.
We introduce the new notations A = R + L and B = 2 RL.
Then n X i =1 d mi = ( A − B cos α ) m + (cid:18) A − B cos (cid:16) πn − α (cid:17)(cid:19) m + (cid:18) A − B cos (cid:16) · πn − α (cid:17)(cid:19) m + · · · + (cid:18) A − B cos (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) m . If m = 1, by Lemma 3.1 we have n X i =1 d mi = ( A − B cos α )+ (cid:18) A − B cos (cid:16) πn − α (cid:17)(cid:19) + · · · + (cid:18) A − B cos (cid:16) ( n − · πn − α (cid:17)(cid:19) = nA. Therefore S (2) n = R + L . If m >
1, we have n X i =1 d mi = nA m − (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) πn − α (cid:17) + · · · + cos (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) + (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) πn − α (cid:17) + · · · + cos (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) − (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) πn − α (cid:17) + · · · + cos (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) + · · ·± (cid:18) mm (cid:19) B m (cid:18) cos m α + cos m (cid:16) πn − α (cid:17) + · · · + cos m (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) . m is even n X i =1 d mi = nA m + (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) πn − α (cid:17) + · · · + cos (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) + · · · + (cid:18) mm (cid:19) B m (cid:18) cos m α + cos m (cid:16) πn − α (cid:17) + · · · + cos m (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) = n A m + m X k =1 (cid:18) m k (cid:19) A m − k B k k (cid:18) kk (cid:19)! . If m is odd n X i =1 d mi = nA m + (cid:18) m (cid:19) A m − B (cid:18) cos α + cos (cid:16) πn − α (cid:17) + · · · + cos (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) + · · · + (cid:18) mm − (cid:19) AB m − (cid:18) cos m − α + cos m − (cid:16) πn − α (cid:17) + · · · + cos m − (cid:16) ( n −
1) 2 πn − α (cid:17)(cid:19) = n A m + m − X k =1 (cid:18) m k (cid:19) A m − k B k k (cid:18) kk (cid:19)! . Using the floor function (the integer part), the obtained results can be combined into asingle formula as follows n X i =1 d mi = n (cid:18) A m + ⌊ m ⌋ X k =1 (cid:18) m k (cid:19) A m − k B k k (cid:18) kk (cid:19)(cid:19) , which proves the theorem.From Theorem 3.1 each sums n X i =1 d mi , where m = 1 , , . . . , n − P [ R,L ] sums. Beginning from the m ≥ n all sums of power distances contain α (Remark 3.1).For example for P the sums contain:- cos 3 α , if m = 3 , ,
5; 7 cos 3 α and cos 6 α , if m = 6 , , α , cos 6 α and cos 9 α , if m = 9 , , m ≥ n the sums n P i =1 d mi contain cosine of the multiples of nα . The study ofsuch sums is beyond the scope of this article.Therefore for P n exist an n − P [ R,L ] sums and if they are constant the locusfor each case is a circle: Theorem 3.2.
The locus of points such that the sum of the (2 m ) -th power of the distances tothe vertices of a given P n ( R ) is constant is a circle, if n X i =1 d mi > nR m , where m = 1 , , . . . , n − , whose center is the centroid of the P n ( R ) . Remark 3.2. - If n P i =1 d mi = nR m , the locus is the centroid of the polygon.- If n P i =1 d mi < nR m , the locus is the empty set. The properties of the cyclic average are as follows:
Property 4.1.
Each regular n -gon has an n − number of cyclic averages S (2) n , S (4) n , . . . , S (2 n − n . Property 4.2.
For fixed R and L , the cyclic averages of equal powers of various regular n -gonsare the same: S (2)3 = S (2)4 = S (2)5 = S (2)6 = · · · ,S (4)3 = S (4)4 = S (4)5 = S (4)6 = · · · ,S (6)4 = S (6)5 = S (6)6 = · · · ,S (8)5 = S (8)6 = · · · . roperty 4.3. Any relations in terms of the cyclic averages S (2 m ) n , the circumscibed radius R and the distance L , which are satisfied for a regular n -gon, are at the same time satisfied forany regular n -gon, where n ≤ n , i.e. S (2 m ) n can be replaced by S (2 m ) n . Eliminate L from the relations of Theorem 3.1 we obtain: Theorem 4.1.
For any regular n -gon: S (2 m ) n = ( S (2) n ) m + ⌊ m ⌋ X k =1 (cid:18) m k (cid:19)(cid:18) kk (cid:19) R k ( S (2) n − R ) k ( S (2) n ) m − k , where m = 2 , . . . , n − . In terms of S (2) n and S (4) n : Theorem 4.2.
For any regular n -gon: S (2 m ) n = ( S (2) n ) m + ⌊ m ⌋ X k =1 k (cid:18) m k (cid:19)(cid:18) kk (cid:19)(cid:0) S (4) n − ( S (2) n ) (cid:1) k ( S (2) n ) m − k , where m = 3 , . . . , n − . The first two relations of Theorem 3.1 imply:
Theorem 4.3.
For any regular n -gon: R = 12 (cid:16) S (2) n ± q S (2) n ) − S (4) n (cid:17) ,L = 12 (cid:16) S (2) n ∓ q S (2) n ) − S (4) n (cid:17) . The points on the circumscribed circle satisfy3( S (2) n ) = 2 S (4) n , so Theorem 4.4.
For any point on the circumscribed circle of the regular n -gon: (cid:16) n X i =1 d i (cid:17) = 2 n n X i =1 d i . .1 Equilateral triangle There are 2 cyclic averages: S (2)3 = 13 ( d + d + d ) = R + L ,S (4)3 = 13 ( d + d + d ) = ( R + L ) + 2 R L . In general case from Theorem 4.1, for n ≥ S (4) n + 3 R = ( S (2) n + R ) . Denote by the symbol – △ ( a,b,c ) the area of a triangle whose sides have lengths a , b , c . Thensolution of the system of the cyclic averages is: Theorem 4.5.
For any point M ( d , d , d , L ) and P ( R ) d = d ,d = 12 (cid:16) R + L ) − d ± √ △ ( R,L,d ) (cid:17) ,d = 12 (cid:16) R + L ) − d ∓ √ △ ( R,L,d ) (cid:17) . For P S (2) n ) − S (4) n = 13 (cid:16) ( d + d + d ) − d + d + d ) (cid:17) = 163 △ d ,d ,d ) , and R = 16 (cid:16) d + d + d ± √ △ ( d ,d ,d ) (cid:17) ,L = 16 (cid:16) d + d + d ∓ √ △ ( d ,d ,d ) (cid:17) . For any point on the circumscribed circle, follows the area – △ ( d ,d ,d ) should be zero.Indeed for the largest distance d = d + d holds. There are 3 cyclic averages: S (2)4 = 14 ( d + d + d + d ) = R + L ,S (4)4 = 14 ( d + d + d + d ) = ( R + L ) + 2 R L ,S (6)4 = 14 ( d + d + d + d ) = ( R + L ) + 6 R L ( R + L ) . From Theorems 4.1 and 4.2 10 heorem 4.6.
For any regular n -gon, where n ≥ : S (6) n = S (2) n (cid:0) ( S (2) n + 3 R ) − R (cid:1) ,S (6) n = S (2) n (cid:0) S (4) n − S (2) n ) (cid:1) . From Theorem 4.6 follows:8( d + d + d + d ) + ( d + d + d + d ) = 6( d + d + d + d )( d + d + d + d ) , which is equivalent to3( d + d − d − d )( d + d − d − d )( d + d − d − d ) = 0 . So d + d = d + d holds.Obtained relation has generalization for regular n -gon. If n is even for the diametricallyopposed vertices: Theorem 4.7.
For any regular n -gon, with even number of vertices n = 2 k : d + d k = d + d k = · · · = d k + d k = 2( R + L ) . Theorem 4.7 simplifies the system of the cyclic averages: S (4)4 + 3 R = ( S (2)4 + R ) ,d + d = d + d ;which is analogue to systems obtained in [7, 9]. Moreover, in terms of R and L , we get: d + d = d + d = 2( R + L ) ,d d + d d = 2( R + L ) . The solution of which is:
Theorem 4.8.
For any point M ( d , d , d , d , L ) and P ( R ) : d = d ,d = R + L ± △ ( R,L,d ) ,d = 2( R + L ) − d ,d = R + L ∓ △ ( R,L,d ) . P S (2) n ) − S (2) n = 116 h d + d + d + d ) − d + d + d + d ) i = 4 △ d , √ d ,d ) = 4 △ d , √ d ,d ) , and R = 14 ( d + d ) ± △ ( d , √ d ,d ) = 14 ( d + d ) ± △ ( d , √ d ,d ) ,L = 14 ( d + d ) ∓ △ ( d , √ d ,d ) = 14 ( d + d ) ∓ △ ( d , √ d ,d ) . For any point on the circumscribed circle the areas – △ ( d , √ d ,d ) and △ ( d , √ d ,d ) shouldbe zero. Indeed, if the point on the minor arc A A are satisfied d + √ d = d and d + d = √ d . There are 4, 5 and 6 cyclic averages for the P , P and P cases, respectively: S (2)5 = S (2)6 = S (2)7 = R + L ,S (4)5 = S (4)6 = S (4)7 = ( R + L ) + 2 R L ,S (6)5 = S (6)6 = S (6)7 = ( R + L ) + 6 R L ( R + L ) ,S (8)5 = S (8)6 = S (8)7 = ( R + L ) + 12 R L ( R + L ) + 6 R L ,S (10)6 = S (10)7 = ( R + L ) + 20 R L ( R + L ) + 30 R L ( R + L ) ,S (12)7 = ( R + L ) + 30 R L ( R + L ) + 90 R L ( R + L ) + 20 R L . These systems are simplified for the regular hexagon case only.The vertices A , A , A and A , A , A form two equilateral triangles, so they satisfy twocyclic relations for P . Generally for n -gon if n divisible by 3: Theorem 4.9.
For any regular n -gon, if n = 3 ℓd + d ℓ + d ℓ = · · · = d ℓ + d ℓ + d ℓ = 3( R + L ) ,d + d ℓ + d ℓ = · · · = d ℓ + d ℓ + d ℓ = 3 (cid:0) ( R + L ) + 2 R L (cid:1) . d + d = d + d = d + d = 2( R + L ) ,d + d + d = d + d + d = 3( R + L ) ,d + d + d = d + d + d = 3 (cid:0) ( R + L ) + 2 R L (cid:1) . By using these relations, we get explicit expressions for distances:
Theorem 4.10.
For any point M ( d , d , . . . , d , L ) and P ( R ) : d = d ,d = 12 (cid:16) R + L + d ± √ △ ( R,L,d ) (cid:17) ,d = 12 (cid:16) R + 3 L − d ± √ △ ( R,L,d ) (cid:17) ,d = 2( R + L ) − d ,d = 12 (cid:16) R + 3 L − d ∓ √ △ ( R,L,d ) (cid:17) ,d = 12 (cid:16) R + L + d ∓ √ △ ( R,L,d ) (cid:17) . For P : 3( S (2) n ) − S (4) n = 3 (cid:16) d + d + · · · + d (cid:17) − d + d + · · · + d
6= 13 (cid:0) ( d + d + d ) − d + d + d ) (cid:1) = 163 △ d ,d ,d ) = 163 △ d ,d ,d ) and R = 16 (cid:16) d + d + d ± √ △ ( d ,d ,d ) (cid:17) = 16 (cid:16) d + d + d ± √ △ ( d ,d ,d ) (cid:17) ,L = 16 (cid:16) d + d + d ∓ √ △ ( d ,d ,d ) (cid:17) = 16 (cid:16) d + d + d ∓ √ △ ( d ,d ,d ) (cid:17) . For any point on the circumscribed circle the area △ ( d ,d ,d ) as well as △ ( d ,d ,d ) vanishes.Indeed if the point on the minor arc A A : d + d = d and d + d = d . .4 Regular Octagon, Nonagon and Decagon There are 8, 9 and 10 cyclic averages for the P , P and P cases, respectively. The cyclicaverages from the second to the twelfth powers are the same as for regular heptagon, so we writeonly new ones: S (14)8 = S (14)9 = S (14)10 = ( R + L ) + 42 R L ( R + L ) + 210 R L ( R + L ) + 140 R L ( R + L ) ,S (16)9 = S (16)10 = ( R + L ) + 56 R L ( R + L ) + 420 R L ( R + L ) + 560 R L ( R + L ) + 70 R L ,S (18)10 = ( R + L ) + 72 R L ( R + L ) + 756 R L ( R + L ) + 1680 R L ( R + L ) + 630 R L ( R + L ) , All three cases n = 8 , ,
10 admit further simplifications.For P Theorem 4.7 gives: d + d = d + d = d + d = d + d = 2( R + L ) . The vertices A , A , A , A and A , A , A , A form two squares, so they satisfy “additional”cyclic relations for P .Generally, if n is divisible by 4: Theorem 4.11.
For any regular n -gon, if n = 4 p : d + d p + d p + d p = · · · = d p + d p + d p + d p = 4 (cid:0) ( R + L ) + 2 R L (cid:1) ,d + d p + d p + d p = · · · = d p + d p + d p + d p = 4 (cid:0) ( R + L ) + 6 R L ( R + L ) (cid:1) . For P Theorem 4.9 gives: d + d + d = d + d + d = d + d + d = 3( R + L ) ,d + d + d = d + d + d = d + d + d = 3 (cid:0) ( R + L ) + 2 R L (cid:1) . For P , from Theorem 4.7: d + d = d + d = d + d = d + d = d + d = 2( R + L ) . The vertices A , A , A , A , A and A , A , A , A , A form two regular pentagons, so theysatisfy “additional” cyclic relations for P .Generally, if n is divisible by 5: 14 heorem 4.12. For any regular n -gon, if n = 5 td + d t + d t + d t + d t = · · · = d t + d t + d t + d t + d t = 5( R + L ) ,d + d t + d t + d t + d t = · · · = d t + d t + d t + d t + d t = 5 (cid:0) ( R + L ) + 2 R L (cid:1) ,d + d t + d t + d t + d t = · · · = d t + d t + d t + d t + d t = 5 (cid:0) ( R + L ) + 6 R L ( R + L ) (cid:1) ,d + d t + d t + d t + d t = · · · = d t + d t + d t + d t + d t = 5 (cid:0) ( R + L ) + 12 R L ( R + L ) + 6 R L (cid:1) . To summarize the obtained results, we conclude: every regular n -gon has an n − n is the composite number we have “additional” relations for the distances,which are obtained from the cyclic averages of the n -gon, where n is divisible of n . n = 24 Is there a point all of whose distances to the vertices of the unit polygon are rational? Theproblem has a long history especially for the case of a square. An extensive historical review isgiven in [7, 9, 10]. For case of an equilateral triangle answer is positive [3]. According to [4] openproblems are in following cases n = 4 , , ,
12 and 24 . For n = 6 – only trivial point is known – the centroid of the unit hexagon.By Theorem 4.3 the side a n of the regular n -gon is: a n πn = S (2) n ± q S (2) n ) − S (4) n . For the unit icositetragon ( n = 24):sin π
24 = 12 vuuut S (2)24 ± q S (2)24 ) − S (4)24 S (4)24 − ( S (2)24 ) . The right side is the root of the fourth degree polynomial equation with rational coefficients:8 (cid:0) S (4) n − ( S (2) n ) (cid:1) x − S (2) n x + 1 = 0 , thus it is the algebraic number of degree ≤
4. On the other handsin π
24 = 12 r − q √ , is the algebraic number of degree > heorem 5.1. There is not a point in the plane that is at rational distances from the verticesof the unit regular -gon. For positive answers for the P and P cases the necessary conditions are the rationalitiesof the equal areas:- △ ( d , √ d ,d ) = △ ( d , √ d ,d ) , if n = 4;- √ △ ( d ,d ,d ) = √ △ ( d ,d ,d ) , if n = 6. P [ R,L ] Sums
For regular polygons with different vertices the number of the P [ R,L ] sums are different too. Aswe see, unlike the plane case, dual Platonic solids have the same number of the P [ R,L ] sums:regular tetrahedron – X (2)[ R,L ] , X (4)[ R,L ] ;octahedron and cube – X (2)[ R,L ] , X (4)[ R,L ] , X (6)[ R,L ] ;icosahedron and dodecahedron – X (2)[ R,L ] , X (4)[ R,L ] , X (6)[ R,L ] , X (8)[ R,L ] , X (10)[ R,L ] . To prove these, we consider each Platonic solid separately. In all cases, we consider solidscentered at the origin and use simple Cartesian coordinates.
The coordinates of the vertices T ( R ): A , ( c, ± c, ± c ) , A , ( − c, ± c, ∓ c ) and R = √ c. Consider an arbitrary point in space M ( d , d , d , d , L ) with the coordinates: ( x, y, z ). Thedistance between M and the centroid O of the tetrahedron: L = x + y + z . Then, d , = ( x − c ) + ( y ∓ c ) + ( z ∓ c ) = R + L + 2 c ( − x ∓ y ∓ z ) ,d , = ( x + c ) + ( y ∓ c ) + ( z ± c ) = R + L + 2 c ( x ∓ y ± z ) , X d i = (cid:0) R + L + 2 c ( − x ∓ y ∓ z ) (cid:1) + (cid:0) R + L + 2 c ( x ∓ y ± z ) (cid:1) = 4( R + L ) + 4 c (cid:0) ( x + y + z ) + ( − x + y + z ) + ( x − y + z ) + ( x + y − z ) (cid:1) = 4 (cid:16) ( R + L ) + 43 R L (cid:17) . If for T ( R ): X d i > R , then Theorem 6.1.
The locus of points in the space such that the sum of the fourth power of thedistances to the vertices of a given regular tetrahedron is constant is a sphere whose center isthe centroid of the tetrahedron.
Remark 6.1. - If P d i = 4 R the locus is the centroid.- If P d i < R the locus is the empty set. The sums of the distances of the power more than 4 contain x , y and z (like α for the planecase), so for T only the sums of the second and fourth powers are P [ R,L ] sums. The coordinates of the vertices of the octahedron T ( R ): A , ( ± c, , , A , (0 , ± c, , A , (0 , , ± c ) and R = c. For an arbitrary point P ( d , d , . . . , d , L ): d , = R + L ± Rx,d , = R + L ± Ry,d , = R + L ± Rz.
Beginning from T each Platonic solid (except tetrahedron) has diametrically opposed17ertices, so for them Theorem 4.7 is satisfied. For the sums of the fourth and sixth powers: X d i = ( R + L ± Rx ) + ( R + L ± Ry ) + ( R + L ± Rz ) = 6 (cid:16) ( R + L ) + 43 R L (cid:17) , X d i = 6( R + L ) + 24( R + L ) R ( x + y + z )= 6 (cid:0) ( R + L ) + 4 R L ( R + L ) (cid:1) . For the cube T ( R ): A , ( ∓ c, ∓ c, ∓ c ) , A , ( ± c, ± c, ∓ c ) ,A , ( ± c, ∓ c, ± c ) , A , ( ∓ c, ± c, ± c )and R = √ c .The distances from the P ( d , d , . . . , d , L ): d , = R + L ± c ( x + y + z ) , d , = R + L ∓ c ( x + y − z ) ,d , = R + L ∓ c ( x + z − y ) , d , = R + L ± c ( x − y − z ) . The vertices A , A , A , A and A , A , A , A form two regular tetrahedrons, so theysatisfy the regular tetrahedron relations. Theorem 6.2.
For an arbitrary point in the space, the sum of the quadruple of the distancesto the vertices of the cube which lie on parallel faces and are endpoints of skew face diagonals,satisfies X d k − = X d k = 4( R + L ) , X d k − = X d k = 4 (cid:16) ( R + L ) + 43 R L (cid:17) . Remark 6.2.
These quadruples do not contain the distances to diametrically opposed vertices.
Thus, X d i = 8 (cid:16) ( R + L ) + 43 R L (cid:17) . X d i = (cid:0) R + L ± c ( x + y + z ) (cid:1) + (cid:0) R + L ∓ c ( x + y − z ) (cid:1) + (cid:0) R + L ∓ c ( x + z − y ) (cid:1) + (cid:0) R + L ∓ c ( y + z − x ) (cid:1) = 8( R + L ) + 24( R + L ) c (cid:16) ( x + y + z ) + ( x + y − z ) + ( x − y + z ) + ( x − y − z ) (cid:17) = 8 (cid:0) ( R + L ) + 4 R L ( R + L ) (cid:1) . If for T ( R ) and T ( R ) is satisfied n X i =1 d mi > nR m , n = 6 , Theorem 6.3.
The locus of points in the space such that the sum of the sixth (fourth) power ofdistances to the vertices of a given octahedron (cube) is constant is a sphere whose center is thecentroid of the octahedron (cube).
Remark 6.3. - If n P d mi = nR m the locus is the centroid.- If n P d mi < nR m the locus is the empty set. The coordinates of the vertices of icosahedron T ( R ): A , (0 , ± c, ± cϕ ) , A , (0 , ∓ c, ± cϕ ) ,A , ( ± c, ± cϕ, , A , ( ± c, ∓ cϕ, ,A , ( ± cϕ, , ± c ) , A , ( ± cϕ, , ∓ c ) , where ϕ is the golden ratio ϕ = √ and R = c p ϕ . For an arbitrary point P ( d , d , . . . , d , L ): d , = R + L ∓ c ( y + zϕ ) , d , = R + L ± c ( y − zϕ ) ,d , = R + L ∓ c ( x + yϕ ) , d , = R + L ∓ c ( x − yϕ ) ,d , = R + L ∓ c ( z + xϕ ) , d , = R + L ± c ( z − xϕ ) . X d i = X d i + X d i + X d i = 4( R + L ) + 16 c ( y + z ϕ ) + 4( R + L ) + 16 c ( x + y ϕ )+ 4( R + L ) + 16 c ( z + x ϕ )= 12( R + L ) + 16 c (1 + ϕ )( x + y + z )= 12 (cid:16) ( R + L ) + 43 R L (cid:17) . X d i = X d i + X d i + X d i = 4( R + L ) + 48 c ( R + L )( y + z ϕ )+ 4( R + L ) + 48 c ( R + L )( x + y ϕ )+ 4( R + L ) + 48 c ( R + L )( z + x ϕ )= 12 (cid:0) ( R + L ) + 4 R L ( R + L ) (cid:1) . For the sum of the eighth power X d i = 4( R + L ) + 96 c ( R + L ) ( y + z ϕ ) + 64 c ( y + z ϕ + 6 y z ϕ ) , X d i = 12( R + L ) + 96 c ( R + L ) ( x + y + z )(1 + ϕ )+ 64 c (cid:0) ( x + y + z )(1 + ϕ ) + 6( x y + x z + y z ) ϕ (cid:1) , Because 1 + ϕ = 3 ϕ and ϕ = 15 (1 + ϕ ) , X d i = 12 (cid:16) ( R + L ) + 8 R L ( R + L ) + 165 R L (cid:17) . X d i = 4( R + L ) + 80( R + L ) c (cid:0) ( y + zϕ ) + ( y − zϕ ) (cid:1) + 160( R + L ) c (cid:0) ( y + zϕ ) + ( y − zϕ ) (cid:1)
20 4( R + L ) + 160( R + L ) c ( y + z ϕ )+ 320( R + L ) c ( y + z ϕ + 6 y z ϕ ) . X d i = 12( R + L ) + 160( R + L ) c ( x + y + z )(1 + ϕ )+ 320( R + L ) c (cid:0) (1 + ϕ )( x + y + z ) + 3 ϕ (2 x y + 2 x z + 2 y z ) (cid:1) = 12 (cid:16) ( R + L ) + 403 R L ( R + L ) + 16 R L ( R + L ) (cid:17) . Divide the vertices of the dodecahedron – T ( R ) into two groups, the vertices A , A , . . . , A which form a cube and other vertices – A , A , . . . , A . Then the coordinates: A , ( ∓ c, ∓ c, ∓ c ) , A , ( ± c, ± c, ∓ c ) ,A , ( ± c, ∓ c, ± c ) , A , ( ∓ c, ± c, ± c ) ,A , (cid:16) , ± cϕ , ± cϕ (cid:17) , A , (cid:16) , ∓ cϕ , ± cϕ (cid:17) ,A , (cid:16) ± cϕ , ± cϕ, (cid:17) , A , (cid:16) ∓ cϕ , ± cϕ, (cid:17) ,A , (cid:16) ± cϕ, , ± cϕ (cid:17) , A , (cid:16) ± cϕ, , ∓ cϕ (cid:17) . and R = √ c .Consider an arbitrary point P ( d , d , . . . , d , L ). For the distances d , d , . . . , d we use therespective distances of the cube, and for others: d , = R + L ∓ c (cid:16) yϕ + zϕ (cid:17) , d , = R + L ± c (cid:16) yϕ − zϕ (cid:17) ,d , = R + L ∓ c (cid:16) xϕ + yϕ (cid:17) , d , = R + L ± c (cid:16) xϕ − yϕ (cid:17) ,d , = R + L ∓ c (cid:16) zϕ + xϕ (cid:17) , d , = R + L ± c (cid:16) zϕ − xϕ (cid:17) , X d i = 8( R + L ) + 323 R L + X d i = 8( R + L ) + 323 R L + 12( R + L ) + 16 c ( x + y + z ) (cid:16) ϕ + ϕ (cid:17) = 20 (cid:16) ( R + L ) + 43 R L (cid:17) . X d i = 8( R + L ) + 32 R L ( R + L ) + 12( R + L ) + 3( R + L )16 c ( x + y + z ) (cid:16) ϕ + ϕ (cid:17) = 20 (cid:0) ( R + L ) + 4 R L ( R + L ) (cid:1) . X d i = (cid:0) R + L ± c ( x + y + z ) (cid:1) + (cid:0) R + L ∓ c ( x + y − z ) (cid:1) + (cid:0) R + L ∓ c ( x + z − y ) (cid:1) + (cid:0) R + L ± c ( x − y − z ) (cid:1) = 8( R + L ) + 64 R L ( R + L ) + 649 R (2 L + 8 x y + 8 x z + 8 y z ) , X d i = 12( R + L ) + 96( R + L ) L c + 64 c (cid:16) ( x + y + z ) (cid:16) ϕ + 1 ϕ (cid:17) + 6 x y + 6 x z + 6 y z (cid:17) = 12( R + L ) + 96( R + L ) R L + 649 R (cid:0) x + y + z ) + 6 x y + 6 x z + 6 y z (cid:1) , X d i = 20 (cid:16) ( R + L ) + 8 R L ( R + L ) + 165 R L (cid:17) . Like T , maximal power for T which depends on R and L only is 10. Indeed, X d i = (cid:0) R + L ± c ( x + y + z ) (cid:1) + (cid:0) R + L ∓ c ( x + y − z ) (cid:1) + (cid:0) R + L ∓ c ( x + z − y ) (cid:1) + (cid:0) R + L ± c ( x − y − z ) (cid:1) = 8( R + L ) + 320( R + L ) c L + 320( R + L ) c (cid:0) x + y + z ) + 12( x y + x z + y z ) (cid:1) , X d i = 12( R + L ) + 160( R + L ) c L (cid:16) ϕ + ϕ (cid:17) + 320( R + L ) c (cid:16)(cid:16) ϕ + ϕ (cid:17) ( x + y + z ) + 6( x y + x z + y z ) (cid:17) , X d i = 20 (cid:16) ( R + L ) + 403 R L ( R + L ) + 16 R L ( R + L ) (cid:17) . If for T ( R ) and T ( R ) is satisfied n X i =1 d mi > nR m , n = 12 , , then Theorem 6.4.
The locus of points in the space such that the sum of the m -th power of distancesto the vertices of a given icosahedron (dodecahedron) is constant is a sphere, when m = 1 , , , and . The center of the sphere is the centroid of the icosahedron (dodecahedron).
Remark 6.4. - If n P i =1 d mi = nR m the locus is the centroid.- If n P i =1 d mi < nR m the locus is the empty set. Summarize the obtained results, in terms of the cyclic averages:
Theorem 7.1.
The cyclic averages of the Platonic solids are the following: S (2)[4] = S (2)[6] = S (2)[8] = S (2)[12] = S (2)[20] = R + L ,S (4)[4] = S (4)[6] = S (4)[8] = S (4)[12] = S (4)[20] = ( R + L ) + 43 R L ,S (6)[6] = S (6)[8] = S (6)[12] = S (6)[20] = ( R + L ) + 4 R L ( R + L ) ,S (8)[12] = S (8)[20] = ( R + L ) + 8 R L ( R + L ) + 165 R L ,S (10)[12] = S (10)[20] = ( R + L ) + 403 R L ( R + L ) + 16 R L ( R + L ) . Eliminate L and R from the relations, we obtain direct relations among the cyclic averagesof the Platonic solids. 23 heorem 7.2. For each Platonic solid ( n = 4 , , , , : S (4)[ n ] + 169 R = (cid:16) S (2)[ n ] + 23 R (cid:17) . This result for regular simplicial and regular polytopic distances is obtained in [5] and [12],respectively.
Theorem 7.3.
For each Platonic solid, except the tetrahedron ( n = 6 , , , : S (6)[ n ] = S (2)[ n ] (cid:0) ( S (2)[ n ] + 2 R ) − R (cid:1) ,S (6)[ n ] = S (2)[ n ] (cid:0) S (4)[ n ] − S (2)[ n ] ) (cid:1) . Theorem 7.4.
For the icosahedron and the dodecahedron ( n = 12 , : S (8)[ n ] − ( S (2)[ n ] ) = 8 R ( S (2)[ n ] − R ) (cid:16) ( S (2)[ n ] ) + 25 R ( S (2)[ n ] − R ) (cid:17) ,S (10)[ n ] − ( S (2)[ n ] ) = 8 R S (2)[ n ] ( S (2)[ n ] − R ) (cid:16)
53 ( S (2)[ n ] ) + 2 R ( S (2)[ n ] − R ) (cid:17) ,S (8)[ n ] = 15 (cid:16) S (4)[ n ] ) + 12 S (4)[ n ] ( S (2)[ n ] ) − S (2)[ n ] ) (cid:17) ,S (10)[ n ] = S (2)[ n ] S (4)[ n ] (cid:0) S (4)[ n ] − S (2)[ n ] ) (cid:1) . Like the plane cases, in some space cases we have “additional” relations. Each Platonicsolid, except the tetrahedron satisfies Theorem 4.7 and for the cube and the dodecahedronTheorem 6.2.For the radius of the circumscribed sphere and the distance between the point and thecentroid:
Theorem 7.5.
For each Platonic solid ( n = 4 , , , , : R = 12 (cid:16) S (2)[ n ] ± r S (2)[ n ] ) − S (4)[ n ] (cid:17) ,L = 12 (cid:16) S (2)[ n ] ∓ r S (2)[ n ] ) − S (4)[ n ] (cid:17) . The points on the circumscribed sphere satisfy4( S (2)[ n ] ) = 3 S (4)[ n ] , so Theorem 7.6.
For any point on the circumscribed sphere of each Platonic solid ( n = 4 , , , , : (cid:16) n X i =1 d i (cid:17) = 3 n n X i =1 d i . Conclusion
In the present paper, we introduce the P [ R,L ] sums and define the cyclic averages of the regularpolygons and the Platonic solids. We prove the main property of the cyclic averages – the equalityof them for various regular polygons and Platonic solids. By means of the cyclic averages thedistances of an arbitrary point to the vertices of the regular polygons (the plane case) and thePlatonic solids (the space case) are investigated. All cases of constant sum of like powers of thedistances, when the locus is a circle (a sphere), are found. General metrical relations for regularpolygons (Platonic solids), which were known in special cases only, are established. Rationaldistances problem solved for the n = 24 case. Acknowledgement
The author would like to thank Georgia Young Scientists Union (GYSU). This research wasfunded by GYSU. Contract no. 2506-19.
References [1] T. M. Apostol and M. A. Mnatsakanian, Sums of squares of distances in m -space. AmericanMathematical Monthly (2003), no. 6, 516 – 526; DOI: 10.2307/3647907.[2] T. M. Apostol and M. A. Mnatsakanian, Sums of squares of distances.
Math Horizons (2001), 21 – 22; DOI: 10.1080/10724117.2001.12021858.[3] R. Barbara and A. Karam, The rational distance problem for equilateral triangles.
Communications in Mathematics and Applications (2018), no. 2, 139 – 145; DOI:10.26713/cma.v9i2.659.[4] R. Barbara, Points at rational distance from the vertices of a unit polygon. Bulletin of theIranian Mathematical Society (2009), no. 2, 209 – 215.[5] J. Bentin, 79.15 Regular simplicial distances. The Mathematical Gazette (1995),p. 106; DOI: 10.2307/3620008.[6] J. Bentin, 81.32 Regular polygonal distances.
The Mathematical Gazette
81 (1997), no. 491,277 – 279; DOI: 10.2307/3619212.[7] T. G. Berry, Points at rational distance from the corners of a unit square.
Annali della ScuolaNormale Superiore di Pisa. Classe di Scienze. Serie IV (1990), no. 4, 505 – 529.[8] M. Gardner, Mathematical circus. More puzzles, games, paradoxes, and other mathematicalentertainments from Scientific American . Revised reprint of the 1981 edition. With a prefaceby Donald Knuth. MAA Spectrum. Mathematical Association of America, Washington, DC,1992.[9] R. K. Guy,
Unsolved Problems in Number Theory . Springer-Verlag, New York–Berlin, 1981;DOI: 10.1007/978-1-4757-1738-9. 2510] R. K. Guy, Tiling the square with rational triangles.
Number Theory and Applications , RAMollin (ed.),
NATO Adv. Study Inst. Ser. C (1989), 45 – 101.[11] B. J. McCartin,
Mysteries of the Equilateral Triangle
Forum Geometricorum (2016), 227 – 232; URL:http://forumgeom.fau.edu/FG2016volume16/FG201627.pdf.[13] P. Tangsupphathawat, Algebraic trigonometric values at rational multipliers of π . Acta etCommentationes Universitatis Tartuensis de Mathematica (2014), no. 1, 9 – 18; DOI:10.12697/ACUTM.2014.18.02. Author’s address:
Department of Mathematics, Georgian-American High School, 18 Chkondideli Str., Tbilisi 0180,Georgia.E-mail: [email protected]@gmail.com