Formulas Among Diagonals in the Regular Polygon and the Catalan Numbers

Forum Geometricorum b Volume 6 (2006) 255–262. bbb FORUM GEOM ISSN 1534-1178

Matthew Hudelson

Abstract. We look at relationships among the lengths of diagonals in the regular polygon. Speciﬁcally, we derive formulas for all diagonals in terms of the shortest diagonals and other formulas in terms of the next-to-shortest diagonals, assuming unit side length. These formulas are independent of the number of sides of the regular polygon. We also show that the formulas in terms of the shortest diagonals involve the famous Catalan numbers.

1. Motivation In [1], Fontaine and Hurley develop formulas that relate the diagonal lengths of a regular n-gon. Specifically, given a regular convex n-gon whose vertices are P0,P1,...,Pn−1, define dk as the distance between P0 and Pk. Then the law of sines yields d sin kπ k = n . d jπ j sin n Defining kπ sin n rk = π , sin n the formulas given in [1] are min{k,h,n−k,n−h} rhrk = r|k−h|+2i−1 i=1 and 1 s = r r k(2j−1) k j=1 where s =min{j>0:jk ≡±1 mod n}. dk 1 ≤ k ≤ n − 1 rk = , Notice that for , d1 but there is no a priori restriction on k in the definition of rk. Thus, it would make perfect sense to consider r0 =0and r−k = −rk not to mention rk for non-integer values of k as well. Also, the only restriction on n in the definition of rn is that n not be zero or the reciprocal of an integer.

Publication Date: October 23, 2006. Communicating Editor: Paul Yiu. 256 M. Hudelson

2. Short proofs of rk formulas 1 Using the identity sin α sin β = 2 (cos(α − β) − cos(α + β)), we can provide some short proofs of formulas equivalent, and perhaps simpler, to those in [1]: Proposition 1. For integers h and k, k−1 rhrk = rh−k+2j+1. j=0 Proof. Letting h and k be integers, we have

π −2 hπ kπ rhrk =(sin ) sin sin n n n π 1 (h − k)π (h + k)π =(sin )−2 cos − cos n 2 n n   π k−1 1 (h − k +2j)π (h − k +2j +2)π =(sin )−2  cos − cos  n 2 n n j=0 π k−1 (h − k +2j +1)π π =(sin )−2 sin sin n n n j=0 k−1 = rh−k+2j+1. j=0 The third equality holds since the sum telescopes. Note that we can switch the roles of h and k to arrive at the formula h−1 rkrh = rk−h+2j+1. j=0 To illustrate that this is not contradictory, consider the example when k =2and h =5. From Proposition 1, we have

r5r2 = r4 + r6. Reversing the roles of h and k,wehave

r2r5 = r−2 + r0 + r2 + r4 + r6.

Recalling that r0 =0and r−j = rj, we see that these two sums are in fact equal. The reciprocal formula in [1] is proven almost as easily: Proposition 2. Given an integer k relatively prime to n, 1 s = r r k(2j−1) k j=1 where s is any (not necessarily the smallest) positive integer such that ks ≡ ±1 mod n. Formulas among diagonals in the regular polygon and the Catalan numbers 257

Proof. Starting at the right-hand side, we have s π kπ −1 s k(2j − 1)π kπ r = sin sin sin sin k(2j−1) n n n n j=1 j=1 π kπ −1 s 1 k(2j − 2)π 2jkπ = sin sin cos − cos n n 2 n n j=1 π kπ −1 1 2skπ = sin sin cos 0 − cos n n 2 n π kπ −1 skπ = sin sin sin2 n n n π kπ −1 π = sin sin sin2 n n n kπ −1 π = sin sin n n 1 = . rk

Here, the third equality follows from the telescoping sum, and the ﬁfth follows from the deﬁnition of s.

3. From powers of r2 to Catalan numbers We use the special case of Proposition 1 when h =2, namely

rkr2 = rk−1 + rk+1, to develop formulas for powers of r2.

Proposition 3. For nonnegative integers m,

m m r m = r . 2 i 1−m+2i i=0

Proof. We proceed by induction on m. When we have m =0, the formula reduces 0 to r2 = r1 and both sides equal 1. This establishes the basis step. For the inductive step, we assume the result for m = n and begin with the sum on the right-hand side for m = n +1. 258 M. Hudelson

n+1 n+1 n +1 n n r = + r i −n+2i i i − 1 −n+2i i=0 i=0 n+1 n+1 n n = r + r i − 1 −n+2i i −n+2i i=0 i=0   n n n n =  r  + r j 2−n+2j i −n+2i j=0 i=0 n n = (r + r ) i −n+2i 2−n+2i i=0 n n = r r i 2 1−n+2i i=0 n = r2r2 n+1 = r2 which completes the induction. The first equality uses the standard identity for n +1 n n = + . binomial coefficients i i i − 1 The third equality is by means of the change of index j = i − 1 and the fact that n =0 c<0 c>n c if or . The fifth equality is from Proposition 1 and the sixth is from the induction hy- pothesis.

Now, we use Proposition 3 and the identity r−k = −rk to consider an expression m for r2 as a linear combination of rk’s where k>0, i.e., we wish to determine the coefﬁcients αk in the sum m+1 m r2 = αm,krk. k=1 From Proposition 2, the sum is known to end at k = m +1. In fact, we can determine αk directly. One contribution occurs when k =1− m +2i,ori = 1 2 (m + k − 1). A second contribution occurs when −k =1− m +2i,ori = 1 2 (m − k − 1). Notice that if m and k have the same parity, there is no contribution to αm,k. Piecing this information together, we ﬁnd   m m − , m − k α = 1 (m + k − 1) 1 (m − k − 1) if is odd; m,k  2 2 0, if m − k is even. Notice that if m − k is odd, then m m 1 = 1 . 2 (m − k +1) 2 (m + k − 1) Formulas among diagonals in the regular polygon and the Catalan numbers 259

Therefore,   m m − , m + k α = 1 (m − k +1) 1 (m − k − 1) if is odd; m,k  2 2 0, if m + k is even. Intuitively, if we arrange the coefﬁcients of the original formula in a table, in- dexed horizontally by k ∈ Z and vertically by m ∈ N, then we obtain Pascal’s triangle (with an extra row corresponding to m =0attached to the top):

−2 −10123456 0 0 0 0100000 1 0 0 0010000 2 0 0 0101000 3 0 0 1020100 4 0 1 0303010 5 1 0 4060401 Next, if we subtract the column corresponding to s = −k from that corresponding to s = k, we obtain the αm,k array:

123456 0 100000 1 010000 2 101000 3 020100 4 203010 5 050401 As a special case of this formula, consider what happens when m =2p and k =1.Wehave 2p 2p α = − 1 p p − 1 (2p)! (2p)! = − p!p! (p − 1)!(p +1)! (2p)! p = 1 − p!p! p +1 1 2p = p +1 p which is the closed form for the pth Catalan number.

4. Inverse formulas, polynomials, and binomial coefﬁcients

We have a formula for the powers of r2 as linear combinations of the rk values. We now derive the inverse relationship, writing the rk values as linear combinations of powers of r2. We start with Proposition 1: r2rk = rk−1 + rk+1. 260 M. Hudelson

We demonstrate that for natural numbers k, there exist polynomials Pk(t) such that rk = Pk(r2). We note that P0(t)=0and P1(t)=1. Now assume the existence of Pk−1(t) and Pk(t). Then from the identity

rk+1 = r2rk − rk−1 we have Pk+1(t)=tPk(t) − Pk−1(t) which establishes a second-order recurrence for the polynomials Pk(t). Armed k ≥ 0 with this, we show for , k − 1 − i P (t)= (−1)itk−1−2i. k i i By inspection, this holds for k =0as the binomial coefficients are all zero in the sum. Also, when k =1, the i =0term is the only nonzero contributor to the sum. Therefore, it is immediate that this formula holds for k =0, 1. Now, we use the recurrence to establish the induction. Given k ≥ 1, Pk+1(t)=tPk(t) − Pk−1(t) k − 1 − i k − 2 − j = t (−1)itk−1−2i − (−1)jtk−2−2j i j i j k − 1 − i k − 1 − i = (−1)itk−2i − (−1)i−1tk−2i i i − 1 i i k − i = (−1)itk−2i i i as desired. The third equality is obtained by replacing j with i − 1. As a result, we obtain the desired formula for rk in terms of powers of r2: Proposition 4. k − 1 − i r = (−1)ir k−1−2i. k i 2 i Assembling these coefficients into an array similar to the αm,k array in the pre- vious section, we have 0123456 1 1000000 2 0100000 3 −1010000 4 0 −201000 5 10−30 100 6 030−4010 7 −1060−501 This array displays the coefficient β(k, m) in the formula m rk = β(k, m)r2 , m Formulas among diagonals in the regular polygon and the Catalan numbers 261 where k is the column number and m is the row number. An interesting obser- vation is that this array and the α(m, k) array are inverses in the sense of ”array multiplication”: 1,m= p α(m, i)β(i, p)= ; 0, otherwise. i

Also, the k(th) column of the β(k, m) array can be generated using the generating function xk(1 + x2)−1−k. Using machinery in §5.1 of [2], this leads to the con- clusion that the columns of the original α(m, k) array can be generated using an inverse function; in this case, the function that generates the m(th) column of the α(m, k) array is √ m 1 − 1 − 4x2 xm−1 . 2x2

This game is similar but slightly more complicated in the case of r3. Here, we use

r3rk = rk+2 + rk + rk−2 which leads to

rk+2 =(r3 − 1)rk + rk−2.

With this, we show that for k ≥ 0, there are functions, but not necessarily polynomials, Qk(t) such that rk = Qk(r3). From the above identity, we have

Qk+2(t)=(t − 1)Qk(t) − Qk−2(t).

This establishes a fourth-order recurrence relation for the functions Qk(t) so de- termining the four functions Q0,Q1,Q2, and Q3 will establish the recurrence. By inspection, Q0(t)=0, Q1(t)=1, and Q3(t)=t so all that remains is√ to deter- 2 mine Q2(t).Wehave√ r3 = r2 − 1 from Proposition 4. Therefore, r2 = r3 +1 and so Q2(t)= t +1. We now claim

Proposition 5. For all natural numbers k, √ k − 1 − i Q (t)= t +1 (−1)i(t − 1)k−1−2i 2k i i k − i k − 1 − i Q (t)= (−1)i(t − 1)k−2i + (−1)i(t − 1)k−1−2i. 2k+1 i i i i 262 M. Hudelson

Proof. We proceed by induction on k. These are easily checked to match the functions Q0,Q1,Q2, and Q3 for k =0, 1.Fork ≥ 2,wehave Q2k(t)=(t − 1)Q2(k−1)(t) − Q2(k−2)(t) √ k − 2 − i = t +1 (t − 1) (−1)i(t − 1)k−2−2i i i  k − 3 − j − (−1)j (t − 1)k−3−2j j j √ k − 2 − i k − 2 − i = t +1 + (−1)i(t − 1)k−1−2i i i − 1 i √ k − 1 − i = t +1 (−1)i(t − 1)k−1−2i i i as desired. The third equality is obtained by replacing j with i − 1. The proof for Q2k+1 is similar, treating each sum separately, and is omitted. As a corollary, we have Corollary 6. √ k − 1 − i r = r +1 (−1)i(r − 1)k−1−2i 2k 3 i 3 i k − i k − 1 − i r = (−1)i(r − 1)k−2i + (−1)i(r − 1)k−1−2i. 2k+1 i 3 i 3 i i References [1] A. Fontaine and S. Hurley, Proof by picture: Products and reciprocals of diagonal length ratios in the regular polygon, Forum Geom., 6 (2006) 97–101. [2] H. Wilf, Generatingfunctionology, Academic Press, 1994.

Department of Mathematics, Washington State University, Pullman, WA 99164-3113 E-mail address: [email protected]