Pent-hexagon & Hex-pentagon

Label some of the polygon vertices as seen in the figure.

All angles which are inscribed in the circumcircles of the regular polygons have measures which are multiples of $\dfrac{1}{2}\times \dfrac{360{}^\circ }{5}=36{}^\circ$ in the case of the regular pentagons and multiples of $\dfrac{1}{2}\times \dfrac{360{}^\circ }{6}=30{}^\circ$ , in the case of regular hexagons.

Moreover, $\angle DAE=360{}^\circ -\left( 120{}^\circ +2\times 108{}^\circ \right)=24{}^\circ$ and $\angle SPT=360{}^\circ -\left( 108{}^\circ +2\times 120{}^\circ \right)=12{}^\circ$

Furthermore, $\triangle ABC$ and $\triangle PQR$ are isoscele triangles, thus $\angle CBA=\frac{1}{2}\left\{ 180{}^\circ -\left( 72{}^\circ +24{}^\circ +36{}^\circ \right) \right\}=24{}^\circ$ and $\angle RQP=\dfrac{1}{2}\left\{ 180{}^\circ -\left( 90{}^\circ +12{}^\circ +30{}^\circ \right) \right\}=24{}^\circ$

Combining all the above, we find that ${\color{#3D99F6}{Blue}} \ angle=36{}^\circ +24{}^\circ =60{}^\circ \ \ \ \ \ and \ \ \ \ {\color{#D61F06}{Red}} \ angle=30{}^\circ +24{}^\circ =54{}^\circ$ Hence, $\boxed{{\color{#3D99F6}{Blue}} > {\color{#D61F06}{Red}}}$ .

The following is the image by Chew Seong Cheong

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For the solution, define the points as follows:

Note that $\angle BAD + \angle BAE = \angle FAC + \angle CAG = 108^{\circ}$ , which is the interior angle of a regular pentagon. Since $\bigtriangleup BAD$ and $\bigtriangleup CAF$ are both isosceles, $\angle BAD = \angle GAF = 36^{\circ}$ , so $\angle CAG = \angle BAE = 72^{\circ}$ . So

$\angle BAC = 360^{\circ} - (\angle DAG + \angle BAD + \angle CAG) = 360^{\circ} - (120^{\circ} + 108\circ) = 132^{\circ}$

Since $\bigtriangleup BAC$ is also isosceles,

$\angle CBA = \angle CAB = \dfrac{1}{2}\left(180^{\circ} - \angle BAC\right) = 24^{\circ}$

which tells us that

${\color{#3D99F6}\text{Blue}} = \angle CBA + \angle ABD = 60^{\circ}$

We can follow the similar steps for the pentagon tangent to two hexagons. Angle-chasing, we see that since $\angle LHJ = \angle JHN = 90^{\circ}$ and $\angle HIK = \angle JHM = 30^{\circ}$ , then $\angle IHK + \angle JHN = 120^{\circ}$ , which also makes up the interior angle of a regular hexagon. Therefore, since

$\angle JHI = 360^{\circ} - (\angle KHN + 120^{\circ}) = 132^{\circ} \rightarrow \angle JIH = \angle IJH = 24^{\circ}$

It tells us that

${\color{#D61F06}\text{Red}} = \angle JIH + \angle HIK = 54^{\circ}$

For this problem, $\boxed{{\color{#3D99F6}\text{Blue}} > {\color{#D61F06}\text{Red}}}$