Decagon in a Pentagon

The decagon and the pentagon share the same apothem, and the area of a regular polygon with $n$ sides and an apothem $a$ is $A = na^2 \tan \big(\frac{180°}{n}\big)$ , so $A_d = 10a^2 \tan \big(\frac{180°}{10}\big) = 10a^2 \sqrt{\frac{5 - 2\sqrt{5}}{5}}$ and $A_p = 5a^2 \tan \big(\frac{180°}{5}\big) = 5a^2 \sqrt{5 - 2\sqrt{5}}$ .

Therefore, $\big(\frac{A_d}{A_p}\big)^2 = \Bigg(\frac{10a^2 \sqrt{\frac{5 - 2\sqrt{5}}{5}}}{5a^2 \sqrt{5 - 2\sqrt{5}}}\Bigg)^2 = \frac{4}{5}$ , so $a = 4$ , $b = 5$ , and $a + b = \boxed{9}$ .

In the diagram below, $O$ is the centre of the decagon (and of the pentagon) and $A$ is the midpoint of one of the sides of the pentagon. $B$ and $C$ are vertices of the decagon and pentagon.

The required ratio of the areas is $\frac{A_d}{A_p}=\frac{\triangle OAB + \triangle OA'B}{\triangle OAC}=\frac{2\triangle OAB}{\triangle OAC}=\frac{2AB}{AC}$ , where the last relation holds as triangles $OAB$ and $OAC$ share the side $OA$ , and $B$ is on $AC$ .

But $AB=OA \cdot \tan{18^\circ}$ and $AC=OA \cdot \tan{36^\circ}$ ; so the ratio we want is just $\frac{2\tan{18^\circ}}{\tan{36^\circ}} = \frac{2}{\sqrt{5}}$ (using standard trig results for angles in a pentagon) giving the answer $2^2+(\sqrt5)^2=\boxed9$ .

The function inRadius computes the incircle of a n-sided regular polygon giving that the vertices are on a circle of radius of 1. The function uses the area of a triangle formula where the length of a single edge is the base and the desired inRadius is the altitude of the triangle.

$\text{inRadius}=\text{FullSimplify}\left[\frac{2 \text{Area}\left[\text{Triangle}\left[\text{Join}\left[\text{Take}[\text{CirclePoints}[\{0,0\},\{1,0\},\text{\$\#\$1}],2],\left( \begin{array}{cc} 0 & 0 \\ \end{array} \right)\right]\right]\right]}{\text{EuclideanDistance}\text{@@}\text{Take}[\text{CirclePoints}[\{0,0\},\{1,0\},\text{\$\#\$1}],2]}\right]\&;$

The inRadii are used to scale the two polygons so that they have the same inRadii. The ratio of their areas is computed and the expression is simplified.

$\text{FullSimplify}\left[\left(\frac{\text{Area}[\text{Polygon}[\text{CirclePoints}[\{0,0\},\{\text{inRadius}(5),0\},10]]]}{\text{Area}\left[\text{Polygon}\left[\text{CirclePoints}\left[\{0,0\},\left\{\text{inRadius}(10),\frac{\pi }{10}\right\},5\right]\right]\right]}\right)^2\right] \Rightarrow \frac45$

$\begin{aligned} &\text{In the above figure ABCDEFGHIJ is a decagon inscribed in the pentagon PQRST,O is their common center.}\\\\ \text{We have,}\\ \large \text{Area of the decagon}&=10\times \text{Area}({\Delta OIH})\\ \large \text{Area of the pentagon}&=5\times \text{Area}({\Delta POT})\\\\ \text{Now,} \text{ Area}(\Delta OIH)&=\dfrac{1}{2} d h \hspace{4mm}\color{#3D99F6}\small h \text{ is the height}\\ \text{ Area}(\Delta POT)&=\dfrac{1}{2} (d+2x) h\\\\ \text{Consider } \Delta PWD\\ \sin{\dfrac{\alpha}{2}}&=\dfrac{d}{2x}\hspace{4mm}\color{#3D99F6}\small\text{Where } \alpha \text{ is the interior angle of a regular pentagon}=108^{\small\circ}\\\\ \implies2x&=\dfrac{d}{\sin{\dfrac{\alpha}{2}}}=\dfrac{d}{\sin{54^{\small\circ}}}\\\\ &=\dfrac{d}{\dfrac{\sqrt{5}+1}{4}}\hspace{4mm}\color{#3D99F6}\small,\sin{54^{\small\circ}}=\dfrac{\sqrt{5}+1}{4}\\ \implies (d+2x)&=\left(\dfrac{4}{\sqrt{5}+1}+1\right)d\\ &=((\sqrt{5}-1)+1)d\\\\ &=\sqrt{5}d\\\\ \left(\dfrac{\text{Area of the decagon}}{\text{Area of the pentagon}}\right)^2&=\left(\dfrac{10\times \text{Area}({\Delta OIH})}{5\times \text{Area}({\Delta POT})}\right)^2\\ &=\left(\dfrac{10\times\dfrac{1}{2} d h}{5\times\dfrac{1}{2} (d+2x) h}\right)^2\\ &=\left(\dfrac{2d h}{\sqrt{5}d h}\right)^2\\ &=\left(\dfrac{2}{\sqrt5}\right)^2\\ &=\dfrac{4}{5}\\ \implies a+b&=\color{#EC7300} \boxed{\color{#333333}9}\end{aligned}$

Since we are dealing with ratio, we need only consider one-fifth of the two regular polygons as shown in the figure.

$\begin{aligned} \frac {A_d}{A_p} & = \frac {[OABC]}{[OADC]} & \small \color{#3D99F6} \text{where }[\ \cdot\ ] \text{ denotes the area of }\cdot, \\ & = \frac {[OBC]}{[OBC]+[CDE]} & \small \color{#3D99F6} \text{Let }OB=OC = 1 \\ & = \frac {\frac 12 \sin 36^\circ}{\frac 12 \sin 36^\circ + \frac 12 CE \times DE} \\ & = \frac {\sin 36^\circ}{\sin 36^\circ + \sin^2 18^\circ \tan 36^\circ} \\ & = \frac 1{1+ \frac {\sin^2 18^\circ}{\cos 36^\circ}} \\ & = \frac {\cos 36^\circ}{\cos 36^\circ+ \frac 12(1-\cos 36^\circ)} \\ & = \frac {2 \cos 36^\circ}{1+ \cos 36^\circ} & \small \color{#3D99F6} \text{See note: }\cos 36^\circ = \frac {1+\sqrt 5}4 \\ & = \frac {2(1+\sqrt 5)}{5+\sqrt 5} = \frac 2{\sqrt 5} \end{aligned}$

Therefore, $\left(\dfrac {A_d}{A_p}\right)^2 = \left(\dfrac 2{\sqrt 5}\right)^2 = \dfrac 45$ $\implies a+b = 4+5 = \boxed 9$ .

---

Note:

$\begin{aligned} \cos \frac \pi 5 + \cos \frac {3\pi}5 & = \frac 12 \\ \cos \frac \pi 5 - \cos \frac {2\pi}5 & = \frac 12 \\ 2\cos \frac \pi 5 - 2 \left( 2 \cos^2 \frac \pi5 - 1\right) & = 1 \\ 4 \cos^2 \frac \pi 5 - 2\cos \frac \pi 5 - 1& = 0 \\ \implies \cos \frac \pi 5 = \frac {1+\sqrt 5}4 \end{aligned}$