Regular polygon inside another? Inspired by this question, I was wondering if one can generalize to the case of an $n$-gon. 

For example, when $n=5$ we have this picture:



where $ABCDE$ is a regular pentagon, $AA_1=BB_1=\cdots=EE_1$. Also $A,A_1,B_1$ are colinear in that order, and so on.

Can we conclude that $A_1B_1C_1D_1E_1$ is regular? Or is there a counterexample for that?
Update

Won't include the trig as rather lengthy, but included as visual complement to give genreal idea:






Note that a shutter on a traditional camera works on this same principal:



Note also that Narasimham's observation of the paths taken offer visual proof of the conjecture, and for greater $n$, are quite visually compelling:






Related



Code at Narasimham's request:

Manipulate[
Show[\[Theta] = LambertW[1]//N; length = -10; g = 0.2; 
tt = {1.337, 1.236, 1.18, 1.144, 1.121, 1.104, 1.091, 1.081, 1.073, 
1.066, 1.061, 1.056, 1.052, 1.049, 1.046, 1.043, 1.041, 1.039};
r = c*(rr = 
  If[n &lt; 2*Pi, 
   Pi + n*(-1 + 2*Pi) + 
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] - 
       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] + 
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]], 
   Pi + n*(-1 + 2*Pi) + 
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] + 
       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] - 
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]]]);
 aa = Table[{Cos[1 + (2*k*Pi)/n]*Cos[\[Theta]] - 
  Sin[1 + (2*k*Pi)/n]*Sin[\[Theta]], 
 Cos[\[Theta]]*Sin[1 + (2*k*Pi)/n] + 
  Cos[1 + (2*k*Pi)/n]*Sin[\[Theta]]}, {k, Join[{n}, Range[n]]}]; 
 li = Table[{(2*
    Sin[r/n]*(-Sin[1 + ((2*s + 1)*Pi)/n + \[Theta]] + 
      Cos[Pi/n]*Sin[(n + (2*s + 1)*Pi + r + n*\[Theta])/n]))/(3 + 
    Cos[Pi/n]^2 - 4*Cos[Pi/n]*Cos[r/n] - 
    Sin[Pi/n]^2), (2*(Cos[1 + ((2*s + 1)*Pi)/n + \[Theta]] - 
      Cos[Pi/n]*Cos[(n + (2*s + 1)*Pi + r + n*\[Theta])/n])*
    Sin[r/n])/(3 + Cos[Pi/n]^2 - 4*Cos[Pi/n]*Cos[r/n] - 
    Sin[Pi/n]^2)}, {s, 1, n}]; 
 list = Join[Drop[li, n - 2], Take[li, n - 2]];
 Graphics[{Opacity[If[TrueQ[JamesBond] == True, 1, 0]], Black, 
 Polygon[{{-(1 + g), -(1 + g)}, {1 + g, -(1 + g)}, {1 + g, 
   1 + g}, {-(1 + g), 1 + g}}], {Red, Thick, 
 Opacity[If[TrueQ[JamesBond] == True, 0, 
   If[TrueQ[circles] == True, 1, 0]]], Circle[]}, {Red, Thick, 
 Opacity[If[TrueQ[JamesBond] == True, 0, 
   If[TrueQ[circles] == True, 1, 0]]], Circle[{0, 0}, Cos[Pi/n]]},
 Opacity[If[TrueQ[JamesBond] == True, 0, 1]], Thick, Blue, 
Line[aa], Opacity[If[TrueQ[JamesBond] == True, 1, 0]], White, 
Polygon[list], Opacity[1], 
If[TrueQ[JamesBond] == True, White, Blue], Thick, 
bb = Table[Line[{aa[[u]], list[[u]]}], {u, 1, n}], 
Opacity[If[TrueQ[JamesBond] == True, 0, 1]], Red, 
PointSize[Large], Point[list], 
If[TrueQ[JamesBond] == True, White, Red], 
Opacity[If[TrueQ[JamesBond] == True, 1, 
  If[TrueQ[lines] == True, 1, 0]]], 
Table[Line[{bb[[u, 1, 
    1]], {bb[[u, 1, 2, 
      1]] + ((bb[[u, 1, 2, 1]] - bb[[u, 1, 1, 1]])*length)/
      Sqrt[Abs[(bb[[u, 1, 1, 1]] - 
            bb[[u, 1, 1, 2]])*2 + (bb[[u, 1, 1, 2]] - 
            bb[[u, 1, 2, 2]])*2]], 
    bb[[u, 1, 2, 
      2]] + ((bb[[u, 1, 2, 2]] - bb[[u, 1, 1, 2]])*length)/
      Sqrt[Abs[(bb[[u, 1, 1, 1]] - 
            bb[[u, 1, 1, 2]])*2 + (bb[[u, 1, 1, 2]] - 
            bb[[u, 1, 2, 2]])*2]]}}], {u, 1, n}]}], 
 PlotRange -&gt; {{-(1 + g), 1 + g}, {-(1 + g), 1 + g}}], {{c, 
 If[TrueQ[min] == True, 1.18, 1]}, 
If[TrueQ[min] == True, Evaluate[tt[[n - 2]]], 1],
Dynamic[
1*n*(Pi/If[n &lt; 2*Pi, 
   Pi + n*(-1 + 2*Pi) + 
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] - 

       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] + 
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]], 
   Pi + n*(-1 + 2*Pi) + 
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] + 
       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] - 
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]]])]}, {{n, 5}, 3, 20, 
 1}, {{circles, False}, {True, False}}, {{lines, False}, {True, 
 False}}, {{JamesBond, False}, {True, False}}, {{min, False}, {True,  False}}]


paths calculated separately with

n = 10;
Show[\[Theta] = LambertW[1]//N;
aa = Table[{Cos[1 + (2 k \[Pi])/n] Cos[\[Theta]] - 
 Sin[1 + (2 k \[Pi])/n] Sin[\[Theta]], 
Cos[\[Theta]] Sin[1 + (2 k \[Pi])/n] + 
 Cos[1 + (2 k \[Pi])/n] Sin[\[Theta]]}, {k, Join[{n}, Range[n]]}];

Graphics[{Thick, Blue, Line[aa]}], Show[\[Theta] = LambertW[1];

 Graphics[Thick, Blue, Line[aa]], 
  r = c*(Pi + 20*(-1 + 2*Pi) + 
  20*ArcTan[(4*Cos[Pi/20]^2*Sin[1 - Pi/10] - 
       2*Sqrt[2*(3 + Sqrt[5/8 + Sqrt[5]/8])]*Cos[1 - Pi/10]*
        Sin[Pi/20])/(Cos[1] + Cos[1 - Pi/5] + 2*Cos[1 - Pi/10] + 
       2*Sqrt[2*(3 + Sqrt[5/8 + Sqrt[5]/8])]*Sin[1 - Pi/10]*
        Sin[Pi/20])]);

  Table[ParametricPlot[\[Theta] = LambertW[1]//N; {(
  2 Sin[r/n] (-Sin[1 + ((2 s + 1) \[Pi])/n + \[Theta]] + 
    Cos[\[Pi]/n] Sin[(n + (2 s + 1) \[Pi] + r + n \[Theta])/n]))/(
  3 + Cos[\[Pi]/n]^2 - 4 Cos[\[Pi]/n] Cos[r/n] - Sin[\[Pi]/n]^2), (
  2 (Cos[1 + ((2 s + 1) \[Pi])/n + \[Theta]] - 
    Cos[\[Pi]/n] Cos[(n + (2 s + 1) \[Pi] + r + n \[Theta])/
      n]) Sin[r/n])/(
  3 + Cos[\[Pi]/n]^2 - 4 Cos[\[Pi]/n] Cos[r/n] - 
  Sin[\[Pi]/n]^2)}, {c, 0, 0.5}, Axes -&gt; False, PlotRange -&gt; All, 
  PlotStyle -&gt; {Red, Thick}], {s, 1, n}]]]


just change n as desired.



Original answer




Idea

Since all regular polygons can be inscribed in the unit circle, a circle can then be inscribed in the polygon (dashed) and a further polygon inside that. If lines then join the vertices of the outer polygon to the corresponding ones of the inner polygon, the outer polygon can be fixed while the inner is rotated. The adjoining lines will then form a third polygon that decreases in size to $0$ when the inner polygon is rotated a half turn. Since the outer is fixed, the relation follows.

Details

A polygon inscribed in a unit circle has vertices at $$e^{i(1 + 2\pi k/n)}$$ for all $k=1$ to $n$, where $n$ is the number of sides of the polygon. This can be described in cartesian coordinates as $$\{\Re\  e^{i(1 + 2\pi k/n)},\Im\  e^{i(1 + 2\pi k/n)}\}.$$ The polygon can then be rotated about the origin so that the uppermost vertex lies on the $y$-axis by multiplying by matrix

\begin{pmatrix}
\Re \ e^{i\Omega} &amp; -\Im \ e^{i\Omega}\\
\Im \ e^{i\Omega}&amp; \Re \ e^{i\Omega}\\
\end{pmatrix}

where $\Omega$ is the Omega constant.

A smaller circle, with radius $\Re \ e^{i\pi/n}$, where $n$ is the number of sides of the polygon, can then be inscribed inside it.

The polygon to be inscribed and rotated within that will then have coordinates

\begin{align}
\Re \ e^{i\pi/n}
\begin{pmatrix}
\Re \ e^{i\Omega_{1}} &amp; -\Im \ e^{i\Omega_{1}}\\
\Im \ e^{i\Omega_{1}}&amp; \Re \ e^{i\Omega_{1}}\\
\end{pmatrix}
\cdot \{\Re\  e^{i(1 + 2\pi k/n)},\Im\  e^{i(1 + 2\pi k/n)}\}
\end{align}

where $\Omega_{1}=\dfrac{r}{n}+\Omega$, with $r$ ranging from $0$ to $n\pi.$



Mathematica code to play with:

Manipulate[ Show[\[Theta] = LambertW[1] // N; 
c = Re[E^( I Pi/n)]; \[CapitalTheta] = r/n + \[Theta]; 
Graphics[{(*Circle[]*){Circle[{0, 0}, c]}, 
Line[aa = (Table[{{Re[E^( I \[Theta])], -Im[E^( I \[Theta])]}, 
{Im[E^( I \[Theta])], Re[E^( I \[Theta])]}}.{Re[ E^(I  (1 + 2 \[Pi] k/n))], 
Im[E^(I (1 + 2 \[Pi] k/n))]}, {k, Join[{n}, Range[n]]}])],
Line[bb = (c Table[{{Re[E^( I \[CapitalTheta])], -Im[E^( I \[CapitalTheta])]}, 
{Im[E^( I \[CapitalTheta])], 
Re[E^( I \[CapitalTheta])]}}.{ Re[E^(I ( 1 + 2 \[Pi]  k/n))], 
Im[E^(I ( 1 + 2 \[Pi]  k/n))]}, {k,Join[{n}, Range[n]]}])], 
Line[Table[{aa[[q]], bb[[q]]}, {q, 1, n}]] }], Axes -&gt; False, 
PlotRange -&gt; {{-1, 1}, {-1, 1}}], {{r, 0}, 0, n Pi}, {{n, 3}, 3, 20, 1}]
I think it is true. 

The construction (and hence the proof) will be easier to start with inclined angle instead of length.



Starting from $A$, the ray $AB_1$ is drawn inclining at angle $x_1$ to the adjacent side $AB$. Then the ray $BC_1$ with $x_2 = x_1$. The process is continued until $EA_1$ is formed.

Note that all x’s are equal. Therefore all y’s are also equal because the polygon is regular.

Then, $⊿ABB_1$ is congruent to $⊿BCC_1$ …… (ASA)

Therefore, $BB_1 = CC_1 = …$. 

This further means $A_1B_1 = B_1C_1 = ….$, implying that the constructed polygon is also regular.