Find the interval of $k$ for given conditions In a triangle $PQR$,  $\tan(P)+\tan(Q)+\tan(R)=k$. Find the interval in which $k$ should lie so that: 
$(A)$ there exists only one isosceles triangle $PQR$. 
$(B)$ there exists exactly two isosceles triangle $PQR$. 
$(C)$ Can there exist three non similar triangles for any real value of $k$
Using  $P+Q+R=\pi$, I got $\frac{\tan P+\tan Q}{\tan P \tan Q -1}=\tan R$
Then for isosceles triangle $\tan P=\tan Q=t$, so finally I obtained cubic equation which is 
$\frac{2t^3}{t^2-1}=k$ but don't know how to proceed further. Could someone guide me through this?
 A: Here is a simple answer to your problems.
For isosceles $\triangle PQR$ lets suppose that 
$$\angle P=\angle Q$$
$$\Rightarrow 2P+R=180 ^\circ$$
$$\tan 2P=-\tan R$$We also know from your given condition that 
$$2\tan P+\tan R=k$$
$$\Rightarrow 2\tan P-\tan 2P=k$$
$$2tanP-\frac{2 \tan P}{1-\tan ^2P}=k$$on solving we get
$$2\tan^3P-k\tan^2P+k=0$$
Lets take $\tan P=x, x>0$$(\because A<90 ^\circ)$ 
$$\Rightarrow f(x)=2x^3 -kx^2+k\tag1$$
$$\Rightarrow f'(x)=6x^2-2kx=0$$
$$\Rightarrow x=0,k/3$$
Now the following cases arise up from the deduced results.
$(1) k<0$, three graphs of cubic equation (equation $(1)$) are possible

Here in all these cases only one triangle is possible and the condition for the triangle is $f(0)<0$ as $k<0$ so for $k<0$ there is only one isosceles triangle possible.
For $k>0$, also we have three graphs of the cubic equation

Here two such triangles are possible in first figure with the condition
$f(k/3)<0$
$$\Rightarrow k\left(1-\frac{k^2}{27}\right)<0$$
$$\Rightarrow k>3\sqrt{3}$$
For the second figure one such triangle is possible with condition
$$f(k/3)=0$$
$$\Rightarrow k=3\sqrt{3}$$
In third figure no such triangle is possible with condition 
$$f(k/3)>0$$
$$\Rightarrow k<3\sqrt{3}$$

Checking for third condition of $k$ i.e $k=0$
No such triangle is possible (verify that by graph simply)
Hence the solution for mentioned conditions is
$A$$\rightarrow$ either $k<0$ or $k=3\sqrt{3}$
$B$ $\rightarrow$$k>3\sqrt{3}$
$C$ $\rightarrow$ Clearly there can never exist three or more than three non-similar triangles for any real value of $k$ (from above graphs).
