3-D Analytic Geometry

This is half of a regular dodecahedron , so a trip to Wikipedia gives that the dihedral angle (angle between two adjacent sides) is $2 \arctan \phi \approx 116.565^\circ$ . Thus the pentagons must be rotated by $180^\circ - 2 \arctan \phi \approx \boxed{63.435^\circ}$ .

Consider the orthogonal projection of the 5 pentagons onto the XY plane, after the rotation. At this point, the line segments joining the vertices that unite, at the end of the rotation, form a larger pentagon.

image1

image2

A right triangle (shown in red, in the second image), can be drawn with a hypotenuse (first image, the initial frame) equal to

$h = e \sin(108^{\circ})$

where $e$ is the length of the edge.

Back to the first image, we note that (the black segment) is the apothem $a$ of the pentagon (the perpendicular distance from the center of the pentagon to any of its sides). We have

$a = e / (2 \tan(72^{\circ}/2) ) = e /(2 \tan(36^{\circ}))$

Now considering the larger pentagon, it edge length is given by

$e' = 2 e \sin(108^{\circ}/2) = 2 e \sin(54^{\circ})$

Hence by similarity, the apothem of the larger pentagon is scaled by the same ratio as the edges of the pentagon, and therefore,

$r = a' / a = e' / e = 2 \sin(54^{\circ}) )$

When the rotation is complete, the difference between the two apothems is the adjacent side to the sought angle of rotation.

Hence,

$x = a' - a = a ( a' / a - 1) = a ( r - 1)$

Finally, we can write,

$\cos \theta = x / h = (r - 1) ( e / (2 \tan(36^{\circ}) ) / (e \sin(108^{\circ}))$

$\cos \theta = ( 2 \sin(54^{\circ}) ) - 1) /( 2 \sin(108^{\circ} ) \tan(36^{\circ} ) )$

$\cos \theta = 0.4472135954999$

Hence,

$\theta = 63.435^{\circ}$

This can be solved simply in 3 steps by taking components of sides.

Step1: Realize that the edges meet in the plane of symmetry as shown. Let the angle by which pentagon is rotated (the one we have to find) be α. And let the edge make angle β in the plane of symmetry.

Plane of symmetry

Step2: Refer to the image below. We know the angles 54 and 18. Lengths of parts highlighted in blue should be equal.

First condition Hence

L cosβ cos54 = L sin18

Or β = 58.16

Step3: The vertical components as shown below should be equal.

Second condition Hence

L cos18 sin α = L sin β

Substituting β from step 2,

α = 63.435

@Hosam Hajjir can we generalise this for n-sided polygon? like for squares its 90 degrees, for triangle its 120.

I would be easier to explain if I know how to post images here. Help me if you know how.

Let a vertex of the central pentagon on the x-y plane be A. The movable pentagon on the right be ABCDE and the left, AB'C'D'E'. The points are assigned symmetrically at A, so that B and B', C and C', D and D', and E and E' are equidistant from A. So that what is true on ABCDE is also true on AB'C'D'E'. B and B' are the nearest vertices on the movable pentagons to A. BD is parallel to AE and likeswise B'D' and AE'. Let the line from A perpendicular to BD meet BD at F. Now if we look directly on top, when the pentagons rotate up, we note that BD is always parallel to AE, the length BF remains the same, and the projection of AF is always perpendicular to BD and AE. The projection of of AF just get shorter as the angle of rotation increases to $90^o$ . It is the same for B'D', B'F' and AF'. The distance between B and B' becomes shorter as the angle of rotation increases until B and B' coincide.

Let BF = B'F' = $a$ , AF = AF' = $b$ and the angle of rotation be $\theta$ . Then the projection of AF, $x=b \cos {\theta}$ . Since BD and B'D' are always parallel to AE and AE' respectively. When B and B' coincide, $\angle DBD' = \angle EAE' = 108^o$ and $\angle ABF = \frac {1}{2} \angle DBD' = 54^o$ . It is noted that the projection of AF, $x = a \tan {54^o}$ , Therefore,

$\Rightarrow x=b \cos {\theta} = a \tan {54^o}$

$\Rightarrow \cos {\theta} = \dfrac {a \tan {54^o} } {b}$

We note that $\angle CBD = \frac {1}{2}(180^o - \angle BCD) = \frac {1}{2} (180^o - 108^o) = 36^o$ .

Therefore, $\angle DBA = \angle CBA - \angle CBD = 108^o - 36^o = 72^o$ and $\angle BAF = 90^o - 72^o = 18^o$ .

$\Rightarrow \tan {18^o} = \frac {a}{b}$

$\Rightarrow \cos {\theta} = \dfrac {a \tan {54^o} } {b} = tan {18^o} \tan {54^o} = 0.324919696 \times 1.37638192 = 0.447213595$

$\Rightarrow \theta = \cos^{-1} {0.447213595} = \boxed {63.435^o}$

solved by planar line symmetry.