One Odd Octadecagon

Consider a quarter of the cyan polygon. The angles between the sides and the horizontal are as shown. Split the area into $a$ to $g$ , then we have:

$\begin{aligned} a & = \frac 12 \left(\cos \frac \pi{14} + \cos \frac {5\pi}{14} + \cos \frac \pi{14} + \cos \frac {3\pi}{14} \right) & \small \blue{\text{Note that }\sin \theta = \cos \left(\frac \pi 2 - \theta\right)} \\ & = \frac 12 \sin \frac \pi 7 + \frac 12 \sin \frac {2\pi}7 + \sin \frac {3\pi}7 \\ b & = \sin \frac \pi{14} \left(\cos \frac {5\pi}{14} + \cos \frac \pi{14} + \cos \frac {3\pi}{14} \right) \\ & = \blue{\cos \frac {6\pi}{14} \cos \frac {5\pi}{14}} + \red{\sin \frac \pi{14} \cos \frac \pi{14}} + \blue{\cos \frac {6\pi}{14} \cos \frac {3\pi}{14}} & \small \blue{\text{and }\cos A \cos B = \frac 12 (\cos (A-B) + \cos (A+B))} \\ & = \blue{\frac 12 \left(\cos \frac \pi{14} + \cos \frac {11\pi}{14} \right)} + \red{\frac 12 \sin \frac \pi 7} + \blue{\frac 12 \left(\cos \frac {3\pi}{14} + \cos \frac {9\pi}{14} \right)} & \small \red{\text{also }\sin (2\theta) = 2\sin \theta \cos \theta} \\ & = \frac 12 \sin \frac {3\pi}7 - \frac 12 \sin \frac {2\pi}7 + \frac 12 \sin \frac \pi 7 + \frac 12 \sin \frac {2\pi}7 - \frac 12 \sin \frac \pi 7 \\ & = \frac 12 \sin \frac {3\pi}7 \end{aligned}$

$\begin{aligned} c & = \left(\sin \frac {5\pi}{14} - \sin \frac \pi{14} \right)\left(\cos \frac \pi{14} + \cos \frac {3\pi}{14}\right) \\ & = \left(\cos \frac {2\pi}{14} - \cos \frac {6\pi}{14} \right)\left(\cos \frac \pi{14} + \cos \frac {3\pi}{14}\right) \\ & = \frac 12 \left(\cos \frac \pi{14} + \cancel{\cos \frac {3\pi}{14}} + \cos \frac \pi{14} + \cancel{\cos \frac {5\pi}{14}} - \cancel{\cos \frac {5\pi}{14}} - \cancel{\cos \frac {7\pi}{14}}^0 - \cancel{\cos \frac {3\pi}{14}} - \cos \frac {9\pi}{14} \right) \\ & = \frac 12 \sin \frac \pi 7 + \sin \frac {3\pi}7 \end{aligned}$

$d = \dfrac 12 \sin \dfrac {3\pi}{14} \cos \dfrac {3\pi}{14} = \dfrac 14 \sin \dfrac {3\pi}7$ , $e=g = \dfrac 14 \sin \dfrac \pi 7$ , and $f = \dfrac 14 \sin \dfrac {5\pi}7 = \dfrac 14 \sin \left(\pi - \dfrac {5\pi}7 \right) = \dfrac 14 \sin \dfrac {2\pi}7$ . Therefore the area of the polygon:

$\begin{aligned} A_{\rm polygon} & = 4(a+b+c+d+e+f+g) \\ & = 4\left(\left(\frac 12 + \frac 12 + \frac 14 + \frac 14 \right) \sin \frac \pi 7 + \left(\frac 12 + \frac 14 \right) \sin \frac {2\pi} 7 + \left(1 + \frac 12 + 1 + \frac 14 \right) \sin \frac {3\pi} 7 \right) \\ & = 4\left(\frac 32 \sin \frac \pi 7 + \frac 34 \sin \frac {2\pi} 7 + \frac {11}4 \sin \frac {3\pi} 7 \right) \\ & = 6\sin \frac \pi 7 + 3\sin \frac {2\pi} 7 + 11 \sin \frac {3\pi} 7 \end{aligned}$

Therefore $ABC = 6 \times 3 \times 11 = \boxed{198}$ .

The cyan polygon can be tiled with $6$ purple unit rhombi (with interior angles of $\cfrac{\pi}{7}$ and $\cfrac{6\pi}{7}$ ), $3$ green unit rhombi (with interior angles of $\cfrac{2\pi}{7}$ and $\cfrac{5\pi}{7}$ ), and $11$ blue unit rhombi (with interior angles of $\cfrac{3\pi}{7}$ and $\cfrac{4\pi}{7}$ ), as follows:

Since each purple unit rhombi has an area of $2 \cdot \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin \cfrac{\pi}{7} = \sin \cfrac{\pi}{7}$ , each green unit rhombi has an area of $2 \cdot \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin \cfrac{2\pi}{7} = \sin \cfrac{2\pi}{7}$ , and each blue unit rhombi has an area of $2 \cdot \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin \cfrac{3\pi}{7} = \sin \cfrac{3\pi}{7}$ , the total area is:

$6 \sin \cfrac{\pi}{7} + 3 \sin \cfrac{2\pi}{7} + 11 \sin \cfrac{3\pi}{7}$ .

Therefore, $A = 6$ , $B = 3$ , $C = 11$ , and $ABC = \boxed{198}$ .

Note that we can cut the shape into smaller regions area as shown below where congruent shapes are colored alike. Also note that the internal angle of a regular heptagon is $\theta=\frac{5\pi}{7}$ , which can be easily derived if we draw segments from the center of the heptagon to each vertex and find the angles of one of the resulting seven congruent isosceles triangles. The identities $\textcolor{#3D99F6}{\text{sin}(\pi-x)=\text{sin}(x)}$ , $\textcolor{#20A900}{\text{sin}x.\text{sin}y=\frac{\text{cos}(x-y)-\text{cos}(x+y)}{2}}$ , $\textcolor{#20A900}{\text{cos}x.\text{cos}y=\frac{\text{cos}(x-y)+\text{cos}{x+y}}{2}}$ and $\textcolor{#20A900}{\text{sin}x.\text{cos}y=\frac{\text{sin}(x-y)+\text{sin}(x+y)}{2}}$ will be used a lot in the course of our work so whenever the first one is used, it will be colored in blue, and whenever one of the other three is used, it will be colored in green. Sometimes the green formulae will be used for $x=y$ and will be equivalent to $\textcolor{#20A900}{2\text{sin}x\text{cos}x=\text{sin}2x}$ , $\textcolor{#20A900}{2\text{sin}^2x=1-\text{cos}2x}$ and $\textcolor{#20A900}{2\text{cos}^2x=\text{cos}2x+1}$ .

We'll address each of the four shapes individually. $$ $1.$ Cyan Triangle $(\triangle BEQ)$ .

It is an isosceles triangle with equal side-lengths of $1$ and $\angle E=2\pi-2\theta=\frac{4\pi}{7}$ . Also $\angle B=\angle Q = \frac{3\pi}{14}$ . Its area is: $S_{\textcolor{cyan}{\text{cyan}}}=\frac{BE.EQ.\text{sin}\angle BEQ}{2}=\frac{\textcolor{#3D99F6}{\text{sin}\frac{4\pi}{7}}}{2}= \frac{\text{sin}\frac{3\pi}{7}}{2}$ Drawing the height from $\angle E$ we get two congruent right triangles with which it's easily found that $BQ=2\text{sin}\frac{2\pi}{7}=QP$ since these two segments are bases of the congruent triangles. $$ $2.$ Yellow Trapezoid $(ABCD)$ .

It is an isosceles trapezoid. Draw the heights from $C$ and $D$ to the base and label the intersection points with $C_1$ and $D_1$ . The two heights split the trapezoid to two congruent right triangles and a rectangle. $\angle C=2\pi-2\theta=\frac{4\pi}{7}\Rightarrow \angle B=\frac{3\pi}{7}$ . Then: $AD_1=BC_1=\text{cos}\frac{3\pi}{7}; CC_1=h=\text{sin}\frac{3\pi}{7}; AB=AD_1+D_1C_1+C_1B=2\text{cos}\frac{3\pi}{7}+1$ $S_{\textcolor{#CEBB00}{\text{yellow}}}=\frac{(AB+CD)h}{2}={\textcolor{#20A900}{\text{sin}\frac{3\pi}{7}\text{cos}\frac{3\pi}{7}}+\text{sin}\frac{3\pi}{7}=\frac{\textcolor{#3D99F6}{\text{sin}\frac{6\pi}{7}}}{2}+\text{sin}\frac{3\pi}{7}=\frac{\text{sin}\frac{\pi}{7}}{2}+\text{sin}\frac{3\pi}{7}}$ $3.$ Red Trapezoid $(MQBA)$ .

It is again an isosceles trapezoid but its legs are $AM=BQ=2\text{sin}\frac{2\pi}{7}$ (which we know from above). Again, draw the heights $AA_1$ and $BB_1$ and split the trapezoid to a rectangle and and two congruent right triangles. First: $\angle QBA=2\pi-(\angle ABC+\theta+\angle QBE)=\frac{9\pi}{14} \Rightarrow \angle MQB=\frac{5\pi}{14}$ Now from $\triangle BB_1Q\cong\triangle AA_1M$ we have: $MA_1=QB_1=BQ\text{cos}\frac{5\pi}{14}=2\text{sin}\frac{2\pi}{7}\textcolor{#EC7300}{\text{cos}\frac{5\pi}{14}}\ \ \ \ \ \textcolor{#EC7300}{\text{cos}\left(\frac{\pi}{2}-x\right)=\text{sin}x}$ $MA_1=QB_1=2\text{sin}\frac{2\pi}{7}\text{sin}\frac{\pi}{7}$ $BB_1=h_1=BQ\text{sin}\frac{5\pi}{14}=2\text{sin}\frac{2\pi}{7}\textcolor{#EC7300}{\text{sin}\frac{5\pi}{14}}\ \ \ \ \ \textcolor{#EC7300}{\text{sin}\left(\frac{\pi}{2}-x\right)=\text{cos}x}$ $BB_1=h_1=2\text{sin}\frac{2\pi}{7}\text{cos}\frac{\pi}{7}$ $MQ=MA_1+A_1B_1+B_1Q=2MA_1+AB=4\text{sin}\frac{2\pi}{7}\text{sin}\frac{\pi}{7}+2\text{cos}\frac{3\pi}{7}+1$ Using the formula for trapezoid's area: $S_{\textcolor{#D61F06}{\text{red}}}=\frac{(MQ+AB)h_1}{2}=\left(4\text{sin}\frac{2\pi}{7}\text{sin}\frac{\pi}{7}+4\text{cos}\frac{3\pi}{7}+2\right)\text{sin}\frac{2\pi}{7}\text{cos}\frac{\pi}{7}$ $S_{\textcolor{#D61F06}{\text{red}}}=2\textcolor{#20A900}{\text{sin}^2\frac{2\pi}{7}}.2\textcolor{#20A900}{\text{sin}\frac{\pi}{7}\text{cos}\frac{\pi}{7}}+2\text{sin}\frac{2\pi}{7}\left(2\textcolor{#20A900}{\text{cos}\frac{\pi}{7}\text{cos}\frac{3\pi}{7}}\right)+2\textcolor{#20A900}{\text{sin}\frac{2\pi}{7}\text{cos}\frac{\pi}{7}}$ $S_{\textcolor{#D61F06}{\text{red}}}=\left(1-\text{cos}\frac{4\pi}{7}\right)\text{sin}\frac{2\pi}{7}+2\textcolor{#20A900}{\text{sin}\frac{2\pi}{7}\text{cos}\frac{2\pi}{7}}+2\textcolor{#20A900}{\text{sin}\frac{2\pi}{7}\text{cos}\frac{4\pi}{7}}+\text{sin}\frac{\pi}{7}+\text{sin}\frac{3\pi}{7}$ $S_{\textcolor{#D61F06}{\text{red}}}=\text{sin}\frac{2\pi}{7}-\textcolor{#20A900}{\text{sin}\frac{2\pi}{7}\text{cos}\frac{4\pi}{7}}+\textcolor{#3D99F6}{\text{sin}\frac{4\pi}{7}}+\text{sin}\frac{-2\pi}{7}+\textcolor{#3D99F6}{\text{sin}\frac{6\pi}{7}}+\text{sin}\frac{\pi}{7}+\text{sin}\frac{3\pi}{7}$ $S_{\textcolor{#D61F06}{\text{red}}}=\text{sin}\frac{2\pi}{7}-\frac{1}{2}\text{sin}\frac{-2\pi}{7}-\frac{1}{2}\textcolor{#3D99F6}{\text{sin}\frac{6\pi}{7}}+\text{sin}\frac{3\pi}{7}-\text{sin}\frac{2\pi}{7}+\text{sin}\frac{\pi}{7}+\text{sin}\frac{\pi}{7}+\text{sin}\frac{3\pi}{7}$ $S_{\textcolor{#D61F06}{\text{red}}}=\text{sin}\frac{2\pi}{7}+\frac{1}{2}\text{sin}\frac{2\pi}{7}-\frac{1}{2}\text{sin}\frac{\pi}{7}+\text{sin}\frac{3\pi}{7}-\text{sin}\frac{2\pi}{7}+\text{sin}\frac{\pi}{7}+\text{sin}\frac{\pi}{7}+\text{sin}\frac{3\pi}{7}$ Collecting like terms results in: $S_{\textcolor{#D61F06}{\text{red}}}=\frac{3}{2}\text{sin}\frac{\pi}{7}+\frac{1}{2}\text{sin}\frac{2\pi}{7}+2\text{sin}\frac{3\pi}{7}$ $4.$ Green Rectangle $(MNPQ)$ .

From above we have the side-lengths of the rectangle so evaluating its area is just a matter of calculations: $S_{\textcolor{#20A900}{\text{green}}}=PQ.QM=BQ.QM=2\text{sin}\frac{2\pi}{7}\left(4\text{sin}\frac{2\pi}{7}\text{sin}\frac{\pi}{7}+2\text{cos}\frac{3\pi}{7}+1\right)$ $S_{\textcolor{#20A900}{\text{green}}}=8\textcolor{#20A900}{\text{sin}^2\frac{2\pi}{7}}\text{sin}\frac{\pi}{7}+4\textcolor{#20A900}{\text{sin}\frac{2\pi}{7}\text{cos}\frac{3\pi}{7}}+2\text{sin}\frac{2\pi}{7}$ $S_{\textcolor{#20A900}{\text{green}}}=4\text{sin}\frac{\pi}{7}\left(1-\text{cos}\frac{4\pi}{7}\right)+2\text{sin}\frac{-\pi}{7}+2\textcolor{#3D99F6}{\text{sin}\frac{5\pi}{7}}+2\text{sin}\frac{2\pi}{7}$ $S_{\textcolor{#20A900}{\text{green}}}=4\text{sin}\frac{\pi}{7}-4\textcolor{#20A900}{\text{sin}\frac{\pi}{7}\text{cos}\frac{4\pi}{7}}-2\text{sin}\frac{\pi}{7}+2\text{sin}\frac{2\pi}{7}+2\text{sin}\frac{2\pi}{7}$ $S_{\textcolor{#20A900}{\text{green}}}=2\text{sin}\frac{\pi}{7}-2\text{sin}\frac{-3\pi}{7}-2\textcolor{#3D99F6}{\text{sin}\frac{5\pi}{7}}+4\text{sin}\frac{2\pi}{7}$ $S_{\textcolor{#20A900}{\text{green}}}=2\text{sin}\frac{\pi}{7}+2\text{sin}\frac{3\pi}{7}-2\text{sin}\frac{2\pi}{7}+4\text{sin}\frac{2\pi}{7}$ $S_{\textcolor{#20A900}{\text{green}}}=2\text{sin}\frac{\pi}{7}+2\text{sin}\frac{2\pi}{7}+2\text{sin}\frac{3\pi}{7}$ Finally we have to evaluate the area of the original octadecagon $S$ : $S=6S_{\textcolor{cyan}{\text{cyan}}}+2S_{\textcolor{#CEBB00}{\text{yellow}}}+2S_{\textcolor{#D61F06}{\text{red}}}+S_{\textcolor{#20A900}{\text{green}}}$ Substituting and simplifying gives: $S=6\text{sin}\frac{\pi}{7}+3\text{sin}\frac{2\pi}{7}+11\text{sin}\frac{3\pi}{7}$ Therefore the answer is $6*3*11=\fbox{198}$ .