The Grand Ennui

It can be easily deduced by rotating the nonagon such that the bottom side becomes the side nearest to the original red-shaded triangle, that the two red-shaded triangles are congruent.

Next, we show that the triangle in yellow is also an equilateral triangle. To do this we trace the line segment of one of its sides all the way to the vertex of the nonagon it touches. Drawing a right-triangle taking two of its vertices as these vertices of the nonagon (shown in the above figure in red), we find that

$\text{Base} = a + a \cos 40^{\circ} + a \cos 80^{\circ} + a \cos 120^{\circ} = a (1 + \cos 40^{\circ} + \cos 80^{\circ} + \cos 120^{\circ} ) \\ \text{Height} = a \sin 40^{\circ} + a \sin 80^{\circ} + a \sin 120^{\circ} = a ( \sin 40^{\circ} + \sin 80^{\circ} + \sin 120^{\circ} )$

where $a$ is the side length of nonagon and the angles (in degrees) are in the arithmetic progression of $40^{\circ}$ and come from the fact that an exterior angle of nonagon is equal to $\dfrac{360^{\circ}}{9} = 40^{\circ}$ .

The angle made by one side of yellow-triangle is thus given by

$\begin{aligned} \tan^{-1} \left( \dfrac{\text{Height}}{\text{Base}} \right) &= \tan^{-1} \left( \dfrac{a ( \sin 40^{\circ} + \sin 80^{\circ} + \sin 120^{\circ} )}{a (1 + \cos 40^{\circ} + \cos 80^{\circ} + \cos 120^{\circ} )} \right) \\ &= \tan^{-1} \left( \dfrac{(\sin 40^{\circ} + \sin 80^{\circ}) + \sin 120^{\circ}}{(\cos 40^{\circ} + \cos 80^{\circ}) + (1+\cos 120^{\circ} )} \right) \\ &= \tan^{-1} \left( \dfrac{2 \sin 60^{\circ} \cos 20^{\circ} + \sin 60^{\circ}}{2 \cos 60^{\circ} \cos 20^{\circ} + (1-\cos 60^{\circ} )} \right) \\ &= \tan^{-1} \left( \dfrac{2 \sin 60^{\circ} \cos 20^{\circ} + \sin 60^{\circ}}{2 \cos 60^{\circ} \cos 20^{\circ} + \cos 60^{\circ} } \right) \\ &= \tan^{-1} \left( \dfrac{\sin 60^{\circ}}{\cos 60^{\circ}} \cdot \left( \dfrac{2 \cos 20^{\circ} + 1}{2 \cos 20^{\circ} + 1} \right) \right) \\ &= \tan^{-1} \left( \tan 60^{\circ} \right) \\ &= 60^{\circ} \end{aligned}$

By the vertical symmetry of nonagon about the centre of its side, the other base angle of yellow triangle is also $60^{\circ}$ . Hence, the yellow triangle is equilateral.

Next, note that the diagonal of the nonagon which contains the bottom side of the smaller red-shaded triangle is parallel to the bottom side of the nonagon (due to both vertices being at a height of $a \sin 40^{\circ}$ above the bottom side of nonagon). This deduction also proves that the red-shaded triangle is indeed equilateral.

Now, we use the fact that the yellow triangle and the red-shaded triangle are similar and use the relationship that the ratio of their heights is the required ratio of their sides.

• Height of yellow triangle : $a \sin 60^{\circ}$ .
• Height of red-shaded triangle : $a (\sin 60^{\circ} - \sin 40^{\circ})$ .

Thus, the required ratio is

$\begin{aligned} \dfrac{a (\sin 60^{\circ} - \sin 40^{\circ})}{a \sin 60^{\circ}} &= 1 - \dfrac{\sin 40^{\circ}}{\sin 60^{\circ}} \\ &= 1 - \dfrac{2}{\sqrt{3}} \sin (30^{\circ} + 10^{\circ} ) \\ &= 1 - \dfrac{2}{\sqrt{3}} \left( \sin 30^{\circ} \cos 10^{\circ} + \cos 30^{\circ} \sin 10^{\circ} \right) \\ &= 1 - \dfrac{2}{\sqrt{3}} \left( \dfrac{\cos 10^{\circ}}{2} + \frac{\sqrt{3}}{2} \sin 10^{\circ} \right) \\ &= 1 - \dfrac{1}{\sqrt{3}} \cos 10^{\circ} - \sin 10^{\circ} \\ &= 1 - \frac{1}{\sqrt{3}} \cos \left( \dfrac{\pi}{18} \right) - \sin \left( \dfrac{\pi}{18} \right) \end{aligned}$

Giving $a = 1, b =3, c= 18 \implies a+b+c = \boxed{22}$ .