Main index Other Papers index About author

Valid HTML 4.01 Transitional

The inner radius of a 5/2 star is (3–√5)/2

The horizontal line that divides the 5/2 star into two equal areas is at y = ¼(5√5–11) ≈ 0.04508497, which touches the star at x = ±½√(½(5–√5)) ≈ ±0.587785.

Excel formulae for plotting a unit 5/2 star;
range names having a two-pixel black border.

Sqrt5
= Sqrt(5)
  
Star_5_2_XStar_5_2_Y
= 0= 1
= Sqrt(50 - 22*Sqrt5) / 4= (Sqrt5 - 1) / 4
= Sqrt( (5+Sqrt5) / 8 )= (Sqrt5 - 1) / 4
= Sqrt(5 - 2*Sqrt5) / 2= 1 - Sqrt5/2
= Sqrt( (5-Sqrt5) / 8 )= -(1 + Sqrt5) / 4
= 0= (Sqrt5 - 3) / 2
= -Sqrt( (5-Sqrt5) / 8 )= -(1 + Sqrt5) / 4
= -Sqrt(5 - 2*Sqrt5) / 2= 1 - Sqrt5/2
= -Sqrt( (5+Sqrt5) / 8 )= (Sqrt5 - 1) / 4
= -Sqrt(50 - 22*Sqrt5) / 4= (Sqrt5 - 1) / 4
= 0= 1

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, the values of Cosecant[] = Cosec[] = Csc[] = 1/Sin[] and Secant[] = Sec[] = 1/Cos[] in surds, and the values of Tan[] 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.

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 k fixed, the inner radius of the (2m+k)/m star tends downwards to k/(2+k) + m–2×π2k(k+1)/(24(k+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), as is easily shown using the product-to-sum formulae. But are all pairs of stars with matching inner radii in one of these two forms?—a question discussed in greater detail in Open Mathematical Questions at jdawiseman.com. 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.

Starn/mApproximate
inner radius
Graphical formulaLaTeXExcelCalcCenterPostScript
n/mInner radius of a n/m star\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]m n 180 exch div 2 copy mul cos 3 1 roll exch 1 sub mul cos div
5/2 star5/2≈ 0.38196601125Inner radius of a 5/2 star\frac{1}{2} \left(3 - \sqrt{5}\right)=(3-Sqrt(5))/2(3-Sqrt[Int5])/23 5 sqrt sub 2 div
6/2 star6/2≈ 0.57735026919Inner radius of a 6/2 star\frac{\sqrt{3}}{3}=Sqrt(3)/3Sqrt[Int3]/33 sqrt 3 div
8/2 star8/2≈ 0.76536686473Inner radius of a 8/2 star\sqrt{2 - \sqrt{2}}=Sqrt(2-Sqrt(2))Sqrt[2-Sqrt[Int2]]2 2 sqrt sub sqrt
8/3 star8/3≈ 0.541196100146Inner radius of a 8/3 star\frac{1}{2} \sqrt{4 - 2 \sqrt{2}}=Sqrt(4-2*Sqrt(2))/2Sqrt[4-2 Sqrt[Int2]]/24 2 sqrt 2 mul sub sqrt 2 div
10/2 star10/2≈ 0.850650808352Inner radius of a 10/2 star\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 star10/3≈ 0.726542528005Inner radius of a 10/3 star\sqrt{5 - 2 \sqrt{5}}=Sqrt(5-2*Sqrt(5))Sqrt[5-2 Sqrt[Int5]]5 5 sqrt 2 mul sub sqrt
10/4 star10/4≈ 0.525731112119Inner radius of a 10/4 star\frac{1}{10} \sqrt{50 - 10 \sqrt{5}}=Sqrt(50-10*Sqrt(5))/10Sqrt[50-10 Sqrt[Int5]]/1050 5 sqrt 10 mul sub sqrt 10 div
12/2 star12/2≈ 0.896575472168Inner radius of a 12/2 star\frac{\sqrt{2} }{2}\left(3 - \sqrt{3}\right)=Sqrt(2) * (3-Sqrt(3)) / 2Sqrt[Int2] (3-Sqrt[Int3]) / 23 3 sqrt sub 2 sqrt mul 2 div
12/3 star12/3≈ 0.816496580928Inner radius of a 12/3 star\frac{\sqrt{6}}{3}=Sqrt(6)/3Sqrt[Int2 Int3]/36 sqrt 3 div
12/4 star12/4≈ 0.707106781187Inner radius of a 12/4 star\frac{\sqrt{2}}{2}=Sqrt(2)/2Sqrt[Int2]/22 sqrt 2 div
12/5 star12/5≈ 0.517638090205Inner radius of a 12/5 star\frac{\sqrt{2} }{2}\left(\sqrt{3} - 1\right)=Sqrt(2) * (Sqrt(3)-1) / 2Sqrt[Int2] (Sqrt[Int3]-1) / 23 sqrt 1 sub 2 sqrt mul 2 div
15/2 star15/2≈ 0.933954606603Inner radius of a 15/2 star\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 ) / 45 sqrt dup -66 mul 150 add sqrt exch 5 mul add 9 sub 4 div
15/3 star15/3≈ 0.885579351971Inner radius of a 15/3 star\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 ) / 45 sqrt dup 66 mul 150 add sqrt exch 3 mul sub 7 sub 4 div
15/4 star15/4≈ 0.827090915285Inner radius of a 15/4 star\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 ) / 45 sqrt dup -6 mul 30 add sqrt add 3 sub 4 div
15/5 star15/5≈ 0.747238274932Inner radius of a 15/5 star\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 ) / 25 sqrt dup -6 mul 15 add sqrt add 2 sub 2 div
15/6 star15/6≈ 0.61803398875Inner radius of a 15/6 star\frac{1}{2} \left(\sqrt{5} - 1\right)=(Sqrt(5)-1)/2(Sqrt[Int5]-1)/25 sqrt 1 sub 2 div
15/7 star15/7≈ 0.338261212718Inner radius of a 15/7 star\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 ) / 45 sqrt dup 6 mul 30 add sqrt sub neg 3 sub 4 div
16/2 star16/2≈ 0.941979402598Inner radius of a 16/2 star\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 star16/3≈ 0.899976223136Inner radius of a 16/3 star\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 star16/4≈ 0.850430094767Inner radius of a 16/4 star\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 star16/5≈ 0.785694958387Inner radius of a 16/5 star\frac{\sqrt{2} }{2}\sqrt{\left(2 - \sqrt{2 - \sqrt{2}}\right)}=Sqrt(2) * Sqrt( (2-Sqrt(2-Sqrt(2))) ) / 2Sqrt[Int2] Sqrt[ (2-Sqrt[2-Sqrt[Int2]]) ] / 22 sqrt dup 2 sub neg sqrt 2 sub neg sqrt mul 2 div
16/6 star16/6≈ 0.688811980234Inner radius of a 16/6 star\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 star16/7≈ 0.509795579104Inner radius of a 16/7 star\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))) ) / 2Sqrt[ 2 (2+Sqrt[Int2]) (2-Sqrt[2+Sqrt[Int2]]) ] / 22 sqrt 2 add dup sqrt 2 sub neg mul 2 mul sqrt 2 div
20/2 star20/2≈ 0.962911555402Inner radius of a 20/2 star\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)/4Sqrt[25+11 Sqrt[Int5]]/2 - Sqrt[Int2] (Sqrt[Int5]+5)/45 sqrt dup 11 mul 25 add sqrt 2 div exch 5 add 2 sqrt mul 4 div sub
20/3 star20/3≈ 0.936859701734Inner radius of a 20/3 star\frac{\sqrt{2} }{10}\left(\sqrt{25 - 10 \sqrt{5}} + 5\right)=Sqrt(2) * ( Sqrt(25-10*Sqrt(5)) + 5 ) / 10Sqrt[Int2] ( Sqrt[25-10 Sqrt[Int5]] + 5 ) / 1025 5 sqrt 10 mul sub sqrt 5 add 2 sqrt mul 10 div
20/4 star20/4≈ 0.907980999479Inner radius of a 20/4 star\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) ) / 45 sqrt dup 5 add sqrt 2 mul exch 1 sub 2 sqrt mul sub 4 div
20/5 star20/5≈ 0.874032048898Inner radius of a 20/5 star\frac{\sqrt{2} }{2}\left(\sqrt{5} - 1\right)=Sqrt(2) * (Sqrt(5)-1) / 2Sqrt[Int2] (Sqrt[Int5]-1) / 25 sqrt 1 sub 2 sqrt mul 2 div
20/6 star20/6≈ 0.831253875555Inner radius of a 20/6 star\frac{1}{2} \sqrt{5 - \sqrt{5}}=Sqrt(5-Sqrt(5))/2Sqrt[5-Sqrt[Int5]]/25 5 sqrt sub sqrt 2 div
20/7 star20/7≈ 0.772374771175Inner radius of a 20/7 star\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))/10Sqrt[Int2] (Sqrt[Int5]+1)/4 - Sqrt[25-5 Sqrt[Int5]]/105 sqrt dup 1 add 2 sqrt mul 4 div exch -5 mul 25 add sqrt 10 div sub
20/8 star20/8≈ 0.680668416084Inner radius of a 20/8 star\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) ) / 25 sqrt dup -4 mul 10 add sqrt exch 2 sub 2 sqrt mul add 2 div
20/9 star20/9≈ 0.506232562894Inner radius of a 20/9 star\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)/2Sqrt[Int2] (Sqrt[Int5]+3)/4 - Sqrt[Sqrt[Int5]+5]/25 sqrt dup 3 add 2 sqrt mul 4 div exch 5 add sqrt 2 div sub

Julian D. A. Wiseman, June 2008


Main index Top About author