Heptagonal Triangle

Draw a circle circumscribing the regular heptagon with center $O$ .

As a central angle, $\angle AOB = \frac{2\pi}{7}$ , and by the inscribed angle theorem $\angle ACB = \frac{\pi}{7}$ . Similarly, $\angle BOC = \frac{4\pi}{7}$ so $\angle BAC = \frac{2\pi}{7}$ , and $\angle AOC = \frac{8\pi}{7}$ so $\angle ABC = \frac{4\pi}{7}$ . Therefore, the angles of $\triangle ABC$ are $\angle A = \frac{2\pi}{7}$ , $\angle B = \frac{4\pi}{7}$ , and $\angle C = \frac{\pi}{7}$ .

Let $x = \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{4\pi}{7}$ and $y = \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$ . Then

$8xy = 8 \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{4\pi}{7} \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$

$8xy = (2 \sin \frac{\pi}{7} \cos \frac{\pi}{7}) (2 \sin \frac{2\pi}{7} \cos \frac{2\pi}{7}) (2 \sin \frac{4\pi}{7} \cos \frac{4\pi}{7})$

$8xy = \sin \frac{2\pi}{7} \sin \frac{4\pi}{7} \sin \frac{8\pi}{7}$

$8xy = \sin \frac{2\pi}{7} \sin \frac{4\pi}{7} (-\sin \frac{\pi}{7})$

$8xy = -\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{4\pi}{7}$

$8xy = -x$

$y = -\frac{1}{8}$

Therefore, $y = \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = -\frac{1}{8}$ . Also,

$8x^2 = 8 \sin^2 \frac{\pi}{7} \sin^2 \frac{2\pi}{7} \sin^2 \frac{4\pi}{7}$

$8x^2 = (2 \sin^2 \frac{\pi}{7}) (2 \sin^2 \frac{2\pi}{7}) (2 \sin^2 \frac{4\pi}{7})$

$8x^2 = (1 - \cos \frac{2\pi}{7}) (1 - \cos \frac{4\pi}{7}) (1 - \cos \frac{8\pi}{7})$

$8x^2 = (1 - \cos \frac{2\pi}{7}) (1 - \cos \frac{4\pi}{7}) (1 + \cos \frac{\pi}{7})$

$8x^2 = (1 + \cos \frac{\pi}{7}) (1 - \cos \frac{2\pi}{7}) (1 - \cos \frac{4\pi}{7})$

$8x^2 = 1 + \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7} - \cos \frac{\pi}{7}\cos \frac{2\pi}{7} - \cos \frac{\pi}{7}\cos \frac{4\pi}{7} + \cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + \cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$

$8x^2 = 1 + \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7} - \cos \frac{\pi}{7}\cos \frac{2\pi}{7} - \cos \frac{\pi}{7}\cos \frac{4\pi}{7} + \cos \frac{2\pi}{7}\cos \frac{4\pi}{7} - \frac{1}{8}$

$8x^2 = \frac{7}{8} + \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7} - \cos \frac{\pi}{7}\cos \frac{2\pi}{7} - \cos \frac{\pi}{7}\cos \frac{4\pi}{7} + \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$

$8x^2 = \frac{7}{8} + (\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7}) - \frac{1}{2}(2 \cos \frac{\pi}{7}\cos \frac{2\pi}{7} + 2 \cos \frac{\pi}{7}\cos \frac{4\pi}{7} - 2 \cos \frac{2\pi}{7}\cos \frac{4\pi}{7})$

$8x^2 = \frac{7}{8} + (\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7}) - \frac{1}{2}((\cos \frac{3\pi}{7} + \cos \frac{\pi}{7}) + (\cos \frac{5\pi}{7} + \cos \frac{3\pi}{7}) - (\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}))$

$8x^2 = \frac{7}{8} + (\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7}) - \frac{1}{2}((-\cos \frac{4\pi}{7} + \cos \frac{\pi}{7}) + (-\cos \frac{2\pi}{7} - \cos \frac{4\pi}{7}) - (-\cos \frac{\pi}{7} + \cos \frac{2\pi}{7}))$

$8x^2 = \frac{7}{8} + (\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7}) - \frac{1}{2}(2 \cos \frac{\pi}{7} - 2 \cos \frac{2\pi}{7} - 2 \cos \frac{4\pi}{7})$

$8x^2 = \frac{7}{8} + (\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7}) - (\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7})$

$8x^2 = \frac{7}{8}$

$x = \frac{\sqrt{7}}{8}$

Therefore, $x = \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{4\pi}{7} = \frac{\sqrt{7}}{8}$ .

For any triangle, $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ , so for this triangle,

$\tan A + \tan B + \tan C = \tan \frac{\pi}{7} \tan \frac{2\pi}{7} \tan \frac{4\pi}{7} = \frac{\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{4\pi}{7}}{\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}} = \frac{\frac{\sqrt{7}}{8}}{-\frac{1}{8}} = \boxed{-\sqrt{7}}$ .

If we write $F(t) \; = \; \frac{2t}{1-t^2}$ then $F(F(F(t))) - t \; = \; -\frac{t (1 + t^2) (t^6 - 21t^4 + 35t^2 - 7)}{ t^8 - 28t^6 + 70t^4 - 28t^2 + 1}$ and hence the roots of the polynomial $G(t) \; = \; t^6 - 21t^4 + 35t^2 - 7$ are $\tan\tfrac{k\pi}{7}$ for $1 \le k \le 6$ . Moreover we can write $G(t) \; = \; (t^3 - 7t)^2 - 7(t^2+1)^2 \; = \; (t^3 - \sqrt{7}t^2 - 7t - \sqrt{7}) (t^3 + \sqrt{7}t^2 - 7t + \sqrt{7}) \; = \; G_1(t)G_2(t)$ and we can show that $G_1(t)$ divides the polynomial $(1-t^2)^3G_1(F(t))$ , and that $G_2(t)$ divides $(1-t^2)^3G_2(F(t))$ .

Thus the roots of $G(t)$ split into two groups, $\tan\tfrac{\pi}{7}\,,\,\tan\tfrac{2\pi}{7}\,,\, \tan\tfrac{4\pi}{7}$ and $\tan\tfrac{3\pi}{7}\,,\,\tan\tfrac {6\pi}{7}\,,\,\tan\tfrac{12\pi}{7} = \tan\tfrac{5\pi}{7}$ . One set gives the zeroes of $G_1(t)$ , and the other set gives the zeroes of $G_2(t)$ . Thus it follows that $\tan A + \tan B + \tan C = \tan\tfrac{\pi}{7} + \tan\tfrac{2\pi}{7} + \tan\tfrac{4\pi}{7}$ must be either $\sqrt{7}$ or $-\sqrt{7}$ . Since $\tan\tfrac{3\pi}{7} \; = \; \frac{\tan\frac{2\pi}{7} + \tan\frac{\pi}{7}}{1 - \tan\frac{2\pi}{7}\tan\frac{\pi}{7}} \; > \; \tan\tfrac{\pi}{7} + \tan\tfrac{2\pi}{7}$ it follows that $\tan A + \tan B + \tan C = \tan\tfrac{\pi}{7} + \tan\tfrac{2\pi}{7} - \tan\tfrac{3\pi}{7} < 0$ , so we deduce that the answer is $\boxed{-\sqrt{7}}$ .

Angle $A$ is $\dfrac{360}{7} degrees$ , $B$ is $\dfrac{720}{7} degrees$ and $C$ is $\dfrac{180}{7} degrees$ . So $tanA+tanB+tanC=-√ 7$