Number Theory in Trigonometry!

First, note that $\tan(x)$ is an increasing function (in the interval $(0,\frac{\pi}{2})$ ), with $\tan(\frac{\pi}{4}) = 1$ and $\tan(\frac{\pi}{3}) = \sqrt3 < 2$ .

Clearly $C < \frac{\pi}{3}$ , so must equal $\frac{\pi}{4}$ , with $\tan(C) = 1$ .

Hence, $\frac{\pi}{3} < B < \frac{3\pi}{8} < A < \frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12} < \frac{4\pi}{9}$ .

This straightaway confirms 3 of the inequalities.

$A > \frac{2\pi}{5}$ must be false.

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Bonus 1:

$A > \frac{2\pi}{5} \implies \frac{\pi}{3} < B < \frac{7\pi}{20}$ .

$\sqrt3 < \tan(B) < \tan(\frac{7\pi}{20}) < 2$ . So this is false.

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Bonus 2:

One can further note that $\tan( \frac{3\pi}{8}) = 1 + \sqrt2 < 3$ , so $\tan(B) = 2$ , and $\tan (A+B) = \frac{\tan(A) + 2}{1-2\tan(A)} = \tan(\frac{3\pi}{4}) = -1$

So, $\tan(A) + 2 = 2 \tan(A) - 1$ , and $\tan(A) = 3$ . So a solution does exist, and is unique.

Since $A$ , $B$ , and $C$ are angles in $\triangle ABC$ , $A + B + C = \pi$ . Then:

$A + B = \pi - C$

$\tan(A + B) = \tan(\pi - C)$

$\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\tan \pi - \tan C}{1 + \tan \pi \tan C}$

$\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$

$\tan A + \tan B = -\tan C + \tan A \tan B \tan C$

$\tan A + \tan B + \tan C = \tan A \tan B \tan C$

Let $x = \tan C$ , $y = \tan B$ , and $z = \tan A$ . Since $\tan A$ , $\tan B$ , and $\tan C$ are all integers such that $A > B > C$ , then $x$ , $y$ , and $z$ are nonzero integers such that $|x| \leq |y| \leq |z|$ , and from the above identity,

$x + y + z = xyz$

Then $|xyz| = |x + y + z| \leq |x| + |y| + |z| \leq 3|z|$ , which means that $|xy| \leq 3$ , meaning $(|x|, |y|)$ is either $(1, 1)$ , $(1, 2)$ , or $(1, 3)$ .

Testing all the values with the given restrictions gives two solutions for $(x, y, z)$ which are $(-3, -2, -1)$ and $(1, 2, 3)$ , but $(-3, -2, -1)$ gives $A + B + C \neq \pi$ , so the only solution is $(x, y, z) = (1, 2, 3)$ , which gives $(A, B, C) \approx (0.3976\pi, 0.3524\pi, 0.25\pi)$ .

Therefore, $A < \frac{2}{5}\pi$ , so $A > \frac{2}{5}\pi$ is false.

For a triangle $\tan A \tan B \tan C = \tan A + \tan B + \tan C$ . For $\tan A$ , $\tan B$ , and $\tan C$ to be integers and $A > B > C$ , the only possibility is $\tan A=3$ , $\tan B=2$ , and $\tan C=1$ so that $3\times 2 \times 1 = 3+2+1 = 6$ .

$\implies \begin{cases} A \approx 1.249045772 \approx 0.3976 \pi \begin{cases} \color{#D61F06} \not > \dfrac {2\pi}5 \\ \color{#3D99F6} < \dfrac {4\pi}9 \end{cases} \\ B \approx 1.107148718 \approx 0.3524\pi \begin{cases} \color{#3D99F6} > \dfrac \pi 3 \\ \color{#3D99F6} < \dfrac {5\pi}{12}\end{cases} \end{cases}$

Therefore, $\boxed{A > \dfrac {2\pi}5}$ is false .