prove that the triangle is isosceles

Question

In a $\triangle ABC$, If

$\begin{vmatrix} 1 & \;\;1\;\;&\;\; 1\;\;\\\\ \displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\ \displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \;\;\displaystyle\tan \frac{A}{2}+\tan \frac{B}{2}\end{vmatrix}=0$

Then prove that the triangle is isosceles.

Try: Let $\displaystyle \tan \frac{A}{2}=p\;\;,\tan \frac{B}{2}=q\;\;,\tan \frac{C}{2}=r$

Using $$\tan\bigg(\frac{A}{2}+\frac{B}{2}\bigg)=\tan\bigg(\frac{\pi}{2}-C\bigg)$$

So $$\sum \tan\frac{A}{2}\tan\frac{B}{2}=1\Rightarrow pq+qr+rp=1$$

So $$\begin{vmatrix}1& 1& 1\\ \displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\ q+r & r+p& p+q\end{vmatrix}=0$$

So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\ \displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\ p(q+r) & q(r+p)& r(p+q)\end{vmatrix}=0$$

So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\ \displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\ 1-qr & 1-rp& 1-pq\end{vmatrix}=0$$

So $$-\begin{vmatrix}p^2& q^2& r^2\\ \displaystyle p & \displaystyle q & \displaystyle r \\ 1 & 1& 1\end{vmatrix}=0$$

So we have $(p-q)(q-r)(r-p)=0.$

So either $p=q$ or $q=r$ and $r=p.$

Could some help me some short way to solve it? Thanks.

Answer 1

With transformation $R_3 \longrightarrow R_3- \left(\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{B}{2}\right)+\tan\left(\dfrac{C}{2}\right)\right)R_1$

we get $\begin{vmatrix} 1 & \;\;1\;\;&\;\; 1\;\;\\\\ \displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\ \displaystyle -\tan\frac{A}{2} &\;\;\displaystyle -\tan \frac{B}{2} & \;\;\displaystyle -\tan \frac{C}{2}\end{vmatrix}$

Now factor out $(-1)$ from Row $3$ and multiply the columns by $\displaystyle \tan\frac{A}{2}, \displaystyle \tan\frac{B}{2}, \displaystyle \tan\frac{C}{2}$ respectively to obtain

$-\begin{vmatrix} \displaystyle \tan\frac{A}{2} &\;\;\displaystyle \tan \frac{B}{2} & \;\;\displaystyle\tan \frac{C}{2}\;\;\\\\ 1 & \;\;1\;\;&\;\; 1 \\\\ \displaystyle \tan^2\frac{A}{2} &\;\;\displaystyle \tan^2 \frac{B}{2} & \;\;\displaystyle\tan^2 \frac{C}{2}\end{vmatrix}$ which can be simplified to the Vandermonde Determinant and hence it evaluates to $\displaystyle \left(\tan\frac{A}{2} - \tan\frac{B}{2}\right)\left(\tan\frac{B}{2} - \tan\frac{C}{2}\right)\left(\tan\frac{C}{2} - \tan\frac{A}{2}\right)$

Answer 2

I have got this here for your determinant: $$\tan \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right)-\cot \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)-\tan \left(\frac{A}{2}\right) \cot \left(\frac{C}{2}\right)+\cot \left(\frac{A}{2}\right) \tan \left(\frac{C}{2}\right)+\tan \left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)-\cot \left(\frac{B}{2}\right) \tan \left(\frac{C}{2}\right)=0$$ With $$C=\pi-A-B$$ we get $$-\frac{8 (\sin (A-B)+\sin (2 A+B)-\sin (A+2 B))}{-\sin (2 A+2 B)+\sin (2 A)+\sin (2 B)}=0$$ You have to solve $$\sin(A-B)+\sin(2A+B)=\sin(A+2B)$$ One solution is given by $$A=B$$

Answer 3

Hint:

Applying $C_2'=C_2-C_1,C_3=C_3-C_1$

$$\begin{vmatrix}1& 1& 1\\ \displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\ q+r & r+p& p+q\end{vmatrix} =\begin{vmatrix}1& 1-1& 1-1\\ \displaystyle \dfrac1p & \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\ q+r & r+p-(q+r)& p+q-(q+r)\end{vmatrix}$$

$$=1\cdot\begin{vmatrix} \displaystyle \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\r+p-(q+r)& p+q-(q+r)\end{vmatrix} $$

$$=\dfrac{(p-q)(p-r)}{pq}-\dfrac{(p-r)(p-q)}{pr}=?$$