The Butterfly Theorem

Fig. 1, C is the midpoint of a chord EF. PQ and UV are two chords passing through C. PV meets EF at X. UQ meets EF at Y. The butterfly theorem says, CX = CY.    

The Butterfly Theorem is a classical result in Euclidean geometry. It gives a beautiful shape of butterfly (Fig. 2) along with two equal length segments.

Fig. 2, the butterfly in the butterfly theorem

Fig. 3, proof of the theorem. O is the center of the circle. M, N are midpoint of the chord PV and UQ respectively. 

Poof:

Let M, N be the midpoint of chord PV and UQ respectively. O is the center of the circle.

Points P, V, Q, U are on the same circle. So $\angle VPQ = \angle VUQ$ and $\angle PVU = \angle PQU$. So  $\triangle CPV \simeq \triangle CUQ$. So $\frac{MV}{NQ} = \frac{PV/2}{UQ/2} = \frac{PV}{UQ} = \frac{VC}{QC}$.

$\frac{MV}{NQ} = \frac{VC}{QC}$ plus $\angle PVU  = \angle PQU$ implies $\triangle MVC \simeq \triangle CQN$. Thus $\angle VMC = \angle CNQ$.

OM is perpendicular to PV. ON is perpendicular to UQ. OC is perpendicular to EF (CX, CY). So O, M, X, C are on the same circle. O, C, Y, N are on the same circle. Thus, $\angle XOC = \angle XMC = \angle YNC = \angle YOC$. Note that OC is perpendicular to EF. Therefore, CX = CY.

The proof of the theorem gives us another butterfly (Fig. 4), which also consists of a pair of similar triangles.

Fig. 4. 

Frenet Tube

Given a 3D curve $\vec{r}(t) = (x(t), y(t), z(t)), t \in I=[0,1]$,  Tubify[ {x, y, z},  a] turns it into a tube around the curve. The Mathematica Tube[] has similar function (works on lines). The essential thing is to find the unit circle in the normal plane for each point (Shown in Fig. 1). The tangential vector is simply $\vec{\alpha} = \vec{r}'/|\vec{r}'|$, where $\vec{r}' = \mathrm{d} \vec{r}/\mathrm{d}t$. The normal vector is $\vec \beta = \vec{\alpha}' / |\vec{\alpha}'| = \frac{\vec{r}'' |\vec{r}'|^2 - \vec{r}' \vec{r}'\cdot\vec{r}''}{|\vec{r}'| \sqrt{|\vec{r}''|^2 |\vec{r}'|^2 - (\vec{r}'\cdot\vec{r}'')^2} }$. The binormal vector is $\vec{\gamma} = \vec{\alpha} \times \vec{\beta} =  \frac{\vec{r}' \times \vec{r}''  }{\sqrt{|\vec{r}''|^2 |\vec{r}'|^2 - (\vec{r}'\cdot\vec{r}'')^2} } $.

Fig. 1: black, $\vec{\alpha}$, green $\vec{\beta}$, blue $\vec{\gamma}$. 

Tubify[{x_, y_, z_}, r_] :=
 ({x[#1], y[#1], z[#1]}
    + (r[#1] Cos[#2])/
      Norm[(Norm[{x'[#1], y'[#1], z'[#1]}]^2 {x''[#1], y''[#1], 
           z''[#1]} - {x'[#1], y'[#1], z'[#1]}.{x''[#1], y''[#1], 
            z''[#1]} {x'[#1], y'[#1], z'[#1]})] (Norm[{x'[#1], y'[#1],
            z'[#1]}]^2 {x''[#1], y''[#1], 
         z''[#1]} - {x'[#1], y'[#1], z'[#1]}.{x''[#1], y''[#1], 
          z''[#1]} {x'[#1], y'[#1], z'[#1]})
    + (r [#1] Sin[#2] )/
      Norm[{-z'[#1]*y''[#1] + z''[#1]*y'[#1], 
        z'[#1]*x''[#1] - z''[#1]*x'[#1], -y'[#1]*x''[#1] + 
         x'[#1]*y''[#1]}] {-z'[#1]*y''[#1] + z''[#1]*y'[#1], 
      z'[#1]*x''[#1] - z''[#1]*x'[#1], -y'[#1]*x''[#1] + 
       x'[#1]*y''[#1]}) &

All the arguments are functions.

Example 1:

Show[ParametricPlot3D[
  Tubify[{(2 + Cos[3 #]) Cos[ #] &, (2 + Cos[3 #]) Sin[#] &, 
     Sin[3 #] &}, 0.25 &][u, v],
  {u, 0, 2 Pi}, {v, 0, 2 Pi}, PlotStyle -> Opacity[0.2],
  Mesh -> None, Boxed -> True, Axes -> False, BoxRatios -> Automatic, 
  PerformanceGoal -> "Quality", PlotPoints -> 80, MaxRecursion -> 0, 
  PlotLabel -> 
   Style[TraditionalForm /@ {(2 + Cos[3 t]) Cos[ 
        t], (2 + Cos[3 t]) Sin[t], Sin[3 t]}, 20]
  ],
 ParametricPlot3D[{(2 + Cos[3 u]) Cos[ u], (2 + Cos[3 u]) Sin[u], 
   Sin[3 u]}, {u, 0, 2 Pi}, 
  PlotStyle -> Directive[{Red, Opacity[1], Thick}]], ImageSize -> 600
 ]

A similar code using Tube[]:

ParametricPlot3D[
  {(2 + Cos[3 u]) Cos[ u], (2 + Cos[3 u]) Sin[u], 
   Sin[3 u]}, {u, 0, 2 Pi}, PlotStyle -> Opacity[0.2],
  Mesh -> None, Boxed -> True, Axes -> False, BoxRatios -> Automatic, 
  PerformanceGoal -> "Quality", PlotPoints -> 80, MaxRecursion -> 0, 
  PlotLabel -> 
   Style[TraditionalForm /@ {(2 + Cos[3 t]) Cos[ 
        t], (2 + Cos[3 t]) Sin[t], Sin[3 t]}, 20]
  ]/.Line[pts_, rest___] :> Tube[pts, 0.1, rest]

Fig. 2

Example 2, trefoil knot:

Show[ParametricPlot3D[
  Tubify[{Sin[#] + 2 Sin[2 #] &, Cos[#] - 2 Cos[2 #] &, -Sin[3 #] &}, 
    0.25 &][u, v],
  {u, 0, 2 Pi}, {v, 0, 2 Pi}, PlotStyle -> Opacity[0.3],
  Mesh -> None, Boxed -> True, Axes -> False, BoxRatios -> Automatic, 
  PerformanceGoal -> "Quality", PlotPoints -> 100, MaxRecursion -> 0, 
  PlotLabel -> Style[TraditionalForm /@ trefoil[t], 20]
  ], ImageSize -> 600
 ]

Fig. 3 Trefoil Knot

Similarly, we can also define the Frenet's Ribbon:

Ribbonize[{x_, y_, z_}, r_] :=
 ({x[#1], y[#1], z[#1]}
    + r [#1] #2 /
      Norm[{-z'[#1]*y''[#1] + z''[#1]*y'[#1], 
        z'[#1]*x''[#1] - z''[#1]*x'[#1], -y'[#1]*x''[#1] + 
         x'[#1]*y''[#1]}] {-z'[#1]*y''[#1] + z''[#1]*y'[#1], 
      z'[#1]*x''[#1] - z''[#1]*x'[#1], -y'[#1]*x''[#1] + 
       x'[#1]*y''[#1]}) &

Fig. 4, Trefoil Ribbon

Fig. 5, Helix Ribbon

more examples:

Fig. 6, A Seashell Surface

Fig. 7, Double Helix (cf. DNA)

Fig. 8, A Figure-eight Knot

ParametricPlot3D[
 Tubify[{#/Pi Cos[#] &, #/Pi Sin[#] &, #/Pi &}, Sin[#/4] &][u, u*80],
 {u, 0, 4 Pi}, PlotStyle -> Directive[{Red, Thick, Opacity[0.8]}],
 Mesh -> None, Boxed -> True, Axes -> False, BoxRatios -> Automatic,
 PlotRange -> All, PerformanceGoal -> "Quality", MaxRecursion -> 0, 
 PlotPoints -> 2560, PlotLabel -> Style["", 20], 
 ImageSize -> 500
 ]

Fig. 9, a coilfied 3D curve

Quiz: area of the shaded part

Inside the above rectangle, the area of the light-blue triangle is 2 and the area of the light-yellow triangle is 3. Question: what is the area of the gray polyon?