The Inner Radius of n/m Stars, in Surds

Julian D. A. Wiseman

Contents: if the boundary of an n/m star is drawn, there are n points on the exterior bounding circle, and another n points on an interior circle (see diagram). The table below shows, in surds, the radius of this interior circle, for several values of n and m: n/m general case, 5/2, 6/2, 8/2, 8/3, 10/2, 10/3, 10/4, 12/2, 12/3, 12/4, 12/5, 15/2, 15/3, 15/4, 15/5, 15/6, 15/7, 16/2, 16/3, 16/4, 16/5, 16/6, 16/7, 20/2, 20/3, 20/4, 20/5, 20/6, 20/7, 20/8, and 20/9.

Publication history: only here. Usual disclaimer and copyright terms apply. Also see the values of Sin[] and Cos[] in surds, and the values of Cosecant[] = Cosec[] = Csc[] = 1/Sin[] and Secant[] = Sec[] = 1/Cos[] in surds.

If the boundary of an n/m star is drawn, there are n points on the exterior bounding circle, and another n points on an interior circle (see diagram on right). The table below shows, in surds, the radius of this interior circle, for several values of n and m. The surds are are shown in several formats.

• Graphical formula: a .png, derived from…

• LaTex: a LaTeX expression.

• Excel: copy-pasteable into Excel, which will automatically convert the “Sqrt” into an upper-case “SQRT”.

• CalcCenter: if one enters Sqrt[2] directly into Mathematica CalcCenter (the budget version of Mathematica), it automatically evaluates the expression numerically, frustrating an attempt to work with surds. To prevent this integers inside the inner-most Sqrts have been replaced with the likes of Int5, which CalcCenter treats as a variable. If using the full-expense Mathematica instances of “Int” may be removed by a preprocessor or by setting Int2=2; Int3=3; Int5=5;.

• Postscript: using 3 5 sqrt sub 2 div rather than 180 5 div dup 2 mul cos exch cos div is about as efficient computationally, but is, in some sense, far more elegant.

Help! Endeavours have been made to represent these values as simply as possible. But further simplifications would be welcomed, credit being given.

Trivia: as m→+∞ with i fixed, the inner radius of the (2m+i)/m star tends downwards to i/(2+i) + m^–2×π²i(i+1)/(24(i+2)) + O(m^–4), and hence the inner radii are bounded below by ⅓.

Open question: stars (6i–2)/i and (18i–6)/(6i–2) have the same inner radius, as do (6i–4)/i and (18i–12)/(6i–3), for integer i≥2 (PDF of overlapping 8/2 and 24/9 stars, and of 10/2 and 30/10 and 14/3 and 42/15 stars). This is easily shown using the double-angle formulae. But are all pairs of stars with matching inner radii in one of these two forms? A proof (or, less probably, a counterexample) would be welcomed. (Also observe that the inner radius of both forms tends to unity, being >0.99 when i≥31, and >0.99975 when i≥1210.)

Errors: whilst the outputs have been tested, it is possible that errors remain. Please do test things before embedding them somewhere important—and if errors or possible improvements are found, tell the author.

Star	n/m	Approximate inner radius	Graphical formula	LaTeX	Excel	CalcCenter	PostScript
	n/m			\frac{\cos{\left(\frac{\pi m}{n}\right)}}{\cos{\left(\frac{\pi (m-1)}{n}\right)}}	=Cos(Pi()*m/n) / Cos(Pi()*(m-1)/n)	Cos[π m / n] / Cos[π (m-1) / n]	180 n div dup m mul cos exch m 1 sub mul cos div
	5/2	≈ 0.38196601125		\frac{1}{2} \left(3 - \sqrt{5}\right)	=(3-Sqrt(5))/2	(3-Sqrt[Int5])/2	3 5 sqrt sub 2 div
	6/2	≈ 0.57735026919		\frac{\sqrt{3}}{3}	=Sqrt(3)/3	Sqrt[Int3]/3	3 sqrt 3 div
	8/2	≈ 0.76536686473		\sqrt{2 - \sqrt{2}}	=Sqrt(2-Sqrt(2))	Sqrt[2-Sqrt[Int2]]	2 2 sqrt sub sqrt
	8/3	≈ 0.541196100146		\frac{1}{2} \sqrt{4 - 2 \sqrt{2}}	=Sqrt(4-2*Sqrt(2))/2	Sqrt[4-2 Sqrt[Int2]]/2	4 2 sqrt 2 mul sub sqrt 2 div
	10/2	≈ 0.850650808352		\sqrt{\frac{1}{10} \left(5 + \sqrt{5}\right)}	=Sqrt( (5+Sqrt(5)) / 10 )	Sqrt[ (5+Sqrt[Int5]) / 10 ]	5 5 sqrt add 10 div sqrt
	10/3	≈ 0.726542528005		\sqrt{5 - 2 \sqrt{5}}	=Sqrt(5-2*Sqrt(5))	Sqrt[5-2 Sqrt[Int5]]	5 5 sqrt 2 mul sub sqrt
	10/4	≈ 0.525731112119		\frac{1}{10} \sqrt{50 - 10 \sqrt{5}}	=Sqrt(50-10*Sqrt(5))/10	Sqrt[50-10 Sqrt[Int5]]/10	50 5 sqrt 10 mul sub sqrt 10 div
	12/2	≈ 0.896575472168		\frac{\sqrt{2} }{2}\left(3 - \sqrt{3}\right)	=Sqrt(2) * (3-Sqrt(3)) / 2	Sqrt[Int2] (3-Sqrt[Int3]) / 2	3 3 sqrt sub 2 sqrt mul 2 div
	12/3	≈ 0.816496580928		\frac{\sqrt{6}}{3}	=Sqrt(6)/3	Sqrt[Int2 Int3]/3	6 sqrt 3 div
	12/4	≈ 0.707106781187		\frac{\sqrt{2}}{2}	=Sqrt(2)/2	Sqrt[Int2]/2	2 sqrt 2 div
	12/5	≈ 0.517638090205		\frac{\sqrt{2} }{2}\left(\sqrt{3} - 1\right)	=Sqrt(2) * (Sqrt(3)-1) / 2	Sqrt[Int2] (Sqrt[Int3]-1) / 2	3 sqrt 1 sub 2 sqrt mul 2 div
	15/2	≈ 0.933954606603		\frac{1}{4} \left(5 \sqrt{5} + \sqrt{150 - 66 \sqrt{5}} - 9\right)	=( 5*Sqrt(5) + Sqrt(150-66*Sqrt(5)) - 9 ) / 4	( 5 Sqrt[Int5] + Sqrt[150-66 Sqrt[Int5]] - 9 ) / 4	5 sqrt dup -66 mul 150 add sqrt exch 5 mul add 9 sub 4 div
	15/3	≈ 0.885579351971		\frac{1}{4} \left(\sqrt{150 + 66 \sqrt{5}} - 3 \sqrt{5} - 7\right)	=( Sqrt(150+66*Sqrt(5)) - 3*Sqrt(5) - 7 ) / 4	( Sqrt[150+66 Sqrt[Int5]] - 3 Sqrt[Int5] - 7 ) / 4	5 sqrt dup 66 mul 150 add sqrt exch 3 mul sub 7 sub 4 div
	15/4	≈ 0.827090915285		\frac{1}{4} \left(\sqrt{30 - 6 \sqrt{5}} + \sqrt{5} - 3\right)	=( Sqrt(30-6*Sqrt(5)) + Sqrt(5) - 3 ) / 4	( Sqrt[30-6 Sqrt[Int5]] + Sqrt[Int5] - 3 ) / 4	5 sqrt dup -6 mul 30 add sqrt add 3 sub 4 div
	15/5	≈ 0.747238274932		\frac{1}{2} \left(\sqrt{15 - 6 \sqrt{5}} + \sqrt{5} - 2\right)	=( Sqrt(15-6*Sqrt(5)) + Sqrt(5) - 2 ) / 2	( Sqrt[15-6 Sqrt[Int5]] + Sqrt[Int5] - 2 ) / 2	5 sqrt dup -6 mul 15 add sqrt add 2 sub 2 div
	15/6	≈ 0.61803398875		\frac{1}{2} \left(\sqrt{5} - 1\right)	=(Sqrt(5)-1)/2	(Sqrt[Int5]-1)/2	5 sqrt 1 sub 2 div
	15/7	≈ 0.338261212718		\frac{1}{4} \left(\sqrt{30 + 6 \sqrt{5}} - \sqrt{5} - 3\right)	=( Sqrt(30+6*Sqrt(5)) - Sqrt(5) - 3 ) / 4	( Sqrt[30+6 Sqrt[Int5]] - Sqrt[Int5] - 3 ) / 4	5 sqrt dup 6 mul 30 add sqrt sub neg 3 sub 4 div
	16/2	≈ 0.941979402598		\sqrt{\left(3 + 2 \sqrt{2}\right) \left(2 - \sqrt{2 + \sqrt{2}}\right)}	=Sqrt( (3+2*Sqrt(2)) * (2-Sqrt(2+Sqrt(2))) )	Sqrt[ (3+2 Sqrt[Int2]) (2-Sqrt[2+Sqrt[Int2]]) ]	2 sqrt dup 2 mul 3 add exch 2 add sqrt 2 sub neg mul sqrt
	16/3	≈ 0.899976223136		\sqrt{\frac{1}{2} \sqrt{20 - 14 \sqrt{2}} - \sqrt{2} + 2}	=Sqrt( Sqrt(20-14*Sqrt(2))/2 - Sqrt(2) + 2 )	Sqrt[ Sqrt[20-14 Sqrt[Int2]]/2 - Sqrt[Int2] + 2 ]	2 sqrt neg dup 14 mul 20 add sqrt 2 div add 2 add sqrt
	16/4	≈ 0.850430094767		\sqrt{\left(2 - \sqrt{2}\right) \left(2 - \sqrt{2 - \sqrt{2}}\right)}	=Sqrt( (2-Sqrt(2)) * (2-Sqrt(2-Sqrt(2))) )	Sqrt[ (2-Sqrt[Int2]) (2-Sqrt[2-Sqrt[Int2]]) ]	2 2 sqrt sub dup sqrt 2 sub neg mul sqrt
	16/5	≈ 0.785694958387		\frac{\sqrt{2} }{2}\sqrt{\left(2 - \sqrt{2 - \sqrt{2}}\right)}	=Sqrt(2) * Sqrt( (2-Sqrt(2-Sqrt(2))) ) / 2	Sqrt[Int2] Sqrt[ (2-Sqrt[2-Sqrt[Int2]]) ] / 2	2 sqrt dup 2 sub neg sqrt 2 sub neg sqrt mul 2 div
	16/6	≈ 0.688811980234		\sqrt{\left(3 - 2 \sqrt{2}\right) \left(2 + \sqrt{2 - \sqrt{2}}\right)}	=Sqrt( (3-2*Sqrt(2)) * (2+Sqrt(2-Sqrt(2))) )	Sqrt[ (3-2 Sqrt[Int2]) (2+Sqrt[2-Sqrt[Int2]]) ]	2 sqrt neg dup 2 add sqrt 2 add exch 2 mul 3 add mul sqrt
	16/7	≈ 0.509795579104		\frac{1}{2} \sqrt{2 \left(2 + \sqrt{2}\right) \left(2 - \sqrt{2 + \sqrt{2}}\right)}	=Sqrt( 2 * (2+Sqrt(2)) * (2-Sqrt(2+Sqrt(2))) ) / 2	Sqrt[ 2 (2+Sqrt[Int2]) (2-Sqrt[2+Sqrt[Int2]]) ] / 2	2 sqrt 2 add dup sqrt 2 sub neg mul 2 mul sqrt 2 div
	20/2	≈ 0.962911555402		\frac{1}{2} \sqrt{25 + 11 \sqrt{5}} - \frac{\sqrt{2}}{4}\left(\sqrt{5} + 5\right)	=Sqrt(25+11*Sqrt(5))/2 - Sqrt(2)*(Sqrt(5)+5)/4	Sqrt[25+11 Sqrt[Int5]]/2 - Sqrt[Int2] (Sqrt[Int5]+5)/4	5 sqrt dup 11 mul 25 add sqrt 2 div exch 5 add 2 sqrt mul 4 div sub
	20/3	≈ 0.936859701734		\frac{\sqrt{2} }{10}\left(\sqrt{25 - 10 \sqrt{5}} + 5\right)	=Sqrt(2) * ( Sqrt(25-10*Sqrt(5)) + 5 ) / 10	Sqrt[Int2] ( Sqrt[25-10 Sqrt[Int5]] + 5 ) / 10	25 5 sqrt 10 mul sub sqrt 5 add 2 sqrt mul 10 div
	20/4	≈ 0.907980999479		\frac{1}{4} \left(2 \sqrt{\sqrt{5} + 5} - \sqrt{2} \left(\sqrt{5} - 1\right)\right)	=( 2*Sqrt(Sqrt(5)+5) - Sqrt(2)*(Sqrt(5)-1) ) / 4	( 2 Sqrt[Sqrt[Int5]+5] - Sqrt[Int2] (Sqrt[Int5]-1) ) / 4	5 sqrt dup 5 add sqrt 2 mul exch 1 sub 2 sqrt mul sub 4 div
	20/5	≈ 0.874032048898		\frac{\sqrt{2} }{2}\left(\sqrt{5} - 1\right)	=Sqrt(2) * (Sqrt(5)-1) / 2	Sqrt[Int2] (Sqrt[Int5]-1) / 2	5 sqrt 1 sub 2 sqrt mul 2 div
	20/6	≈ 0.831253875555		\frac{1}{2} \sqrt{5 - \sqrt{5}}	=Sqrt(5-Sqrt(5))/2	Sqrt[5-Sqrt[Int5]]/2	5 5 sqrt sub sqrt 2 div
	20/7	≈ 0.772374771175		\frac{\sqrt{2}}{4}\left(\sqrt{5} + 1\right) - \frac{1}{10} \sqrt{25 - 5 \sqrt{5}}	=Sqrt(2)*(Sqrt(5)+1)/4 - Sqrt(25-5*Sqrt(5))/10	Sqrt[Int2] (Sqrt[Int5]+1)/4 - Sqrt[25-5 Sqrt[Int5]]/10	5 sqrt dup 1 add 2 sqrt mul 4 div exch -5 mul 25 add sqrt 10 div sub
	20/8	≈ 0.680668416084		\frac{1}{2} \left(\sqrt{10 - 4 \sqrt{5}} + \sqrt{2} \left(\sqrt{5} - 2\right)\right)	=( Sqrt(10-4*Sqrt(5)) + Sqrt(2)*(Sqrt(5)-2) ) / 2	( Sqrt[10-4 Sqrt[Int5]] + Sqrt[Int2] (Sqrt[Int5]-2) ) / 2	5 sqrt dup -4 mul 10 add sqrt exch 2 sub 2 sqrt mul add 2 div
	20/9	≈ 0.506232562894		\frac{\sqrt{2}}{4}\left(\sqrt{5} + 3\right) - \frac{1}{2} \sqrt{\sqrt{5} + 5}	=Sqrt(2)*(Sqrt(5)+3)/4 - Sqrt(Sqrt(5)+5)/2	Sqrt[Int2] (Sqrt[Int5]+3)/4 - Sqrt[Sqrt[Int5]+5]/2	5 sqrt dup 3 add 2 sqrt mul 4 div exch 5 add sqrt 2 div sub

Julian D. A. Wiseman, June 2008