Question

Let $A,B,C$ are three angles of a triangle such that $A=\frac{\pi }{4}$ and $\tan B\tan C=p$. Then the minimum positive value of $\left[p\right]$ is
(where $[.]$ is the greatest integer function)

Answer 1

$A+B+C=\pi \Rightarrow B+C=\frac{3\pi }{4}$
So, $0<B,C<\frac{3\pi }{4}$

Now, $\tan B\tan C=p$
$\Rightarrow \frac{\sin B\sin C}{\cos B\cos C}=p\Rightarrow \frac{\cos B\cos C-\sin B\sin C}{\cos B\cos C+\sin B\sin C}=\frac{1-p}{1+p}\Rightarrow \frac{\cos (B+C)}{\cos (B-C)}=\frac{1-p}{1+p}\Rightarrow \cos (B-C)=-\frac{1+p}{\sqrt{2}(1-p)}$

We know that,
$0\le B-C<\frac{3\pi }{4}\Rightarrow -\frac{1}{\sqrt{2}}<\cos (B-C)\le 1\Rightarrow -\frac{1}{\sqrt{2}}<-\frac{1+p}{\sqrt{2}(1-p)}\le 1\Rightarrow -\sqrt{2}\le \frac{1+p}{(1-p)}<1$

$\frac{1+p}{(1-p)}\ge -\sqrt{2}\Rightarrow \frac{p+1}{p-1}-\sqrt{2}\le 0\Rightarrow \frac{p-(\sqrt{2}+1{)}^2}{p-1}\ge 0\Rightarrow p<1\text{or}p\ge (\sqrt{2}+1{)}^2\cdots (1)$

$\frac{1+p}{(1-p)}<1\Rightarrow \frac{2p}{p-1}>0\Rightarrow p<0\text{or}p>1\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$p<0\text{or}p\ge (\sqrt{2}+1{)}^2$
Hence, the minimum positive value of $\left[p\right]$ is $5$.