A Circle and a Centroid

Let $DF$ be the height of the pentagon, $O$ the center of the inscribed circle, $OM\bot CD$ , $GK\bot CD$ and $OL\bot GK$ . Denote the radius of the circle by $R$ .
Simple angle chasing gives that $\angle DOM=\angle DGK=36{}^\circ$ .
For the angles $36{}^\circ$ and $72{}^\circ$ we use the following trigonometric values: $\cos 36{}^\circ =\dfrac{\varphi }{2} \ \ \ \ \ \tan 36{}^\circ =\dfrac{\sqrt{3-\varphi }}{\varphi } \ \ \ \ \ \tan 72{}^\circ =\dfrac{\sqrt{2+\varphi }}{\varphi -1}$ where $\varphi$ is the golden ratio $\dfrac{1+\sqrt{5}}{2}$ .

First,

$DG=\dfrac{2}{3}DF=\dfrac{2}{3}\cdot FB\cdot \tan 72{}^\circ =\dfrac{2}{3}\cdot \dfrac{1}{2}\cdot \dfrac{\sqrt{2+\varphi }}{\varphi -1}=\dfrac{1}{3}\cdot \dfrac{\sqrt{2+\varphi }}{\varphi -1}$ Hence,

$GK=DG\cdot \cos 36{}^\circ =\dfrac{1}{3}\cdot \dfrac{\sqrt{2+\varphi }}{\varphi -1}\cdot \dfrac{\varphi }{2} \ \ \ \ \ (1)$ On the other hand,

$GK=GL+LK=R\cdot \cos 36{}^\circ +R=R\cdot \left( \dfrac{\varphi }{2}+1 \right) \ \ \ \ \ (2)$ Combining $(1)$ and $(2)$ ,

$\begin{aligned} R\cdot \left( \dfrac{\varphi }{2}+1 \right)=\dfrac{1}{3}\cdot \dfrac{\sqrt{2+\varphi }}{\varphi -1}\cdot \dfrac{\varphi }{2}\Rightarrow R & =\dfrac{1}{3}\cdot \dfrac{\varphi \sqrt{2+\varphi }}{\left( \varphi -1 \right)\left( \varphi +2 \right)} \\ & =\dfrac{1}{3}\cdot \dfrac{\varphi \sqrt{2+\varphi }}{{{\varphi }^{2}}+\varphi -2} \\ & =\dfrac{1}{3}\cdot \dfrac{\varphi \sqrt{2+\varphi }}{2\varphi -1} \ \ \ \ \ \text{since }{{\varphi }^{2}}=\varphi +1 \\ & =\dfrac{1}{3}\cdot \dfrac{\sqrt{2+\varphi }}{2-\dfrac{1}{\varphi }} \\ & \text{ } \\ \Rightarrow \boxed{R=\dfrac{1}{3}\cdot \dfrac{\sqrt{2+\varphi }}{3-\varphi }} \ \ \ \ \ \text{since }\frac{1}{\varphi }=\varphi -1 \\ \end{aligned}$

Finally, $\begin{aligned} DM & =OM\cdot \tan 36{}^\circ \\ & =R\cdot \tan 36{}^\circ \\ & =\dfrac{1}{3}\cdot \dfrac{\sqrt{2+\varphi }}{3-\varphi }\cdot \dfrac{\sqrt{3-\varphi }}{\varphi } \\ & =\dfrac{1}{3}\cdot \dfrac{\sqrt{6+\varphi -{{\varphi }^{2}}}}{3\varphi -{{\varphi }^{2}}} \\ & =\dfrac{1}{3}\cdot \dfrac{\sqrt{5}}{2\varphi -1} \ \ \ \ \ \text{since }{{\varphi }^{2}}=\varphi +1 \\ & =\dfrac{1}{3} \\ \end{aligned}$

For the answer, $p=1$ , $q=3$ , thus, $p+q=\boxed{4}$ .

---

2nd proof (using homothecy) The circle we are interested in is the image of the incircle of the regular pentagon, with homothetic center point $D$ and scale factor $r=\dfrac{DG}{DF}=\dfrac{2}{3}$ .
Since the point of tangency of the incircle with side $DC$ is the midpoint $N$ of $DC$ and its image is $M$ it holds that

$\dfrac{DM}{DN}=r\Rightarrow DM=r\cdot DN=\dfrac{2}{3}\cdot \dfrac{1}{2}DC=\boxed{\dfrac{1}{3}}$ Likewise, the radius $R$ of the circle is $\dfrac{2}{3}$ that of the radius of the incircle, thus,

$R=\dfrac{2}{3}\cdot \dfrac{1}{10}\sqrt{25+10\sqrt{5}}=\dfrac{\sqrt{25+10\sqrt{5}}}{15}$

Point G as a centroid is always at a height 1/3 of the apex from the base, and since ABD is an isosceles triangle, DG is perpendicular to AB, with twice the length of G to AB distance. If we draw a smaller regular pentagon with D being one of the vertices and the circle as its incircle, it's easy to see that one of its other vertices would be the midpoint of MC. It's equivalent to enlarging the circle to be ABCDE's incircle so that G' touches the AB base and M' becomes the midpoint of DC.

DM / DM' = DG / DG'
DM / (1/2) = (2h/3) / (h)
DM = (1/2) × (2/3) = 1/3
Answer = 1 + 3 = 4

Bonus : the r would also be 2/3 of a regular unit pentagon's incircle, so
r = (2/3) × ( √(25 + 10√5) / 10 )
= √(25 + 10√5) / 15

Let $DN$ be perpendicular to $AB$ ; and the circle has a center at $O$ and intersects $DN$ at $F$ and $G$ . We note that $\angle DAB = 72^\circ$ and $\angle DOM = 36^\circ$ . The the altitude $DN = \dfrac 12 \tan 72^\circ$ . Since $G$ is the centroid of $\triangle ABD$ , $DG = \dfrac 23 \cdot DN = \dfrac {\tan 72^\circ}3$ .

Let the radius of the circle be $r$ , $DF = k$ , and $DM = x$ . Then $DG = k + 2r \implies k = \dfrac {\tan 72^\circ}3 - 2r$ and $x = r \tan 36^\circ$ . By tangent-secant theorem ,

$\begin{aligned} x^2 & = k(k+2r) \\ r^2 \tan^2 36^\circ & = \left(\frac {\tan 72^\circ}3 - 2r \right) \frac {\tan 72^\circ}3 \\ r^2 \tan^2 36^\circ & = \left(\frac {\tan 72^\circ}3 - r \right)^2 - r^2 \\ r^2 (1+ \tan^2 36^\circ) & = \left(\frac {\tan 72^\circ}3 - r \right)^2 \\ r^2 \sec^2 36^\circ & = \left(\frac {\tan 72^\circ}3 - r \right)^2 \\ \implies r & = \frac {\tan 72^\circ}{3(1+\sec 36^\circ)} \\ x & = r \tan 36^\circ = \frac {\tan 72^\circ \tan 36^\circ}{3(1+\sec 36^\circ)} \\ & = \frac {2\sin^2 36^\circ \cos 36^\circ}{3\cos 72^\circ(\cos 36^\circ+1)} \\ & = \frac {2(1-\cos 36^\circ)\cos 36^\circ}{3(2\cos^2 36^\circ - 1)} & \small \blue{\text{Note that }\cos 36^\circ = \frac \varphi 2 \text{, where}} \\ & = \frac {2\varphi - \varphi^2}{3(\varphi^2-2)} = \frac {2\varphi - \varphi -1}{3(\varphi + 1 -2)} & \small \blue{\varphi = \frac {1+\sqrt 5}2 \text{ is the golden ratio.}} \\ & = \frac 13 \end{aligned}$

Therefore $p+q = 1 + 3 = \boxed 4$ .