Brocard Angles-Image taken from Wikipedia.org

The angles denoted by $\omega$ are the Brocard angles. Recently i came to know about the Brocard Angles and also their property i.e $\cot{\omega}=\cot{A}+\cot{B}+\cot{C}$. In my previous question I got the answer on proving the identity. But I want to prove this identity only through the sines and cosine formula(excluding any excessive use of geometry parts)

Now I tried the question this way:

In $\triangle APC$, $\angle CAP=A-\omega; \angle CPA=\pi-A$.

Thus using the sine rule we can write $\frac{\sin{(A-\omega)}}{CP} = \frac{\sin{A}}{b}$.

Similarly using the same rule in other triangle we can write:

For $\triangle CPB$ $\frac{\sin{(C-\omega)}}{PB} = \frac{\sin{C}}{a}$

For $\triangle APB$ $\frac{\sin{(B-\omega)}}{AP} = \frac{\sin{B}}{c}$

Now in the respective sine formulae I expanded the expressions of $\sin{(A-\omega)}$, $\sin{(B-\omega)}$ and $\sin{(C-\omega)}$

This gave me- $$\frac{CP}{b}=\cos{\omega}-\sin{\omega}\cot{A}\tag{1}$$ $$\frac{PB}{a}=\cos{\omega}-\sin{\omega}\cot{C}\tag{2}$$ $$\frac{AP}{c}=\cos{\omega}-\sin{\omega}\cot{B}\tag{3}$$

Adding the equations $(1),(2)$ and $(3)$ $$\frac{CP}{b}+\frac{PB}{a}+\frac{AP}{c}=3\cos{\omega}-\sin{\omega}(\cot{A}+\cot{B}+\cot{C})$$

I got stuck after this. Please tell me whether I can proceed further or is my method completely inconclusive.

  • $\begingroup$ The image here has been taken up from www.wikipedia.org $\endgroup$ – Harsh Sharma Jun 21 '16 at 10:51
  • $\begingroup$ @Roby5 I got it in one more way also. $\endgroup$ – Harsh Sharma Jul 7 '16 at 6:26


$$\cot{\omega}=\cot{A}+\cot{B}+\cot{C} \Longleftrightarrow \sin{(A-\omega)}\sin{(B-\omega)}\sin{(C-\omega)}=\sin^3{\omega}\tag{1}$$

Now using sine rule, we get




Multiplying $(2),(3)$ and $(4)$ out, we get $(1)$

Proof of Hint:

$$\quad \quad \cot{\omega}=\cot{A}+\cot{B}+\cot{C}$$
$$\Longleftrightarrow \cot{\omega}-\cot{A}=\cot{B}+\cot{C}$$
$$\Longleftrightarrow \dfrac{\sin{(A-\omega})}{\sin{A}\sin{\omega}}=\dfrac{\sin{(B+C)}}{\sin{B}\sin{C}}=\dfrac{\sin{(\pi-A)}}{\sin{B}\sin{C}}=\dfrac{\sin{A}}{\sin{B}\sin{C}}$$
So, $$\sin{(A-\omega)}=\dfrac{\sin^2{A}\sin{\omega}}{\sin{B}\sin{C}}\tag{5}$$
Similarly, $$\sin{(B-\omega)}=\dfrac{\sin^2{B}\sin{\omega}}{\sin{A}\sin{C}}\tag{6}$$
and $$\sin{(C-\omega)}=\dfrac{\sin^2{C}\sin{\omega}}{\sin{A}\sin{B}}\tag{7}$$ Multiplying $(5),(6)$ and $(7)$ out, we get $(1)$.

  • $\begingroup$ Though your proof is sufficient ,but is there any where through which i can take up my own attempt forward $\endgroup$ – Harsh Sharma Jun 21 '16 at 13:35
  • 1
    $\begingroup$ @HarshSharma $\frac{CP}{b}=\cos{\omega}-\sin{\omega} \cot{A}$ and similar results make your approach too far-fetched and difficult. Moreover, unnessarily involving sides of a triangle makes your method far too complicated. $\endgroup$ – Dragonemperor42 Jun 21 '16 at 13:45

Here I also got one more thing, this can be proved in one more way.

In the figure enter image description here

We can see that $$\frac{\sin(A-\omega)}{PC}=\frac{\sin \angle CPA}{b}$$ Here through basic geometry we can easily conclude that $\angle CPA=\pi - A$, hence the above equation can be written as: $$\frac{\sin(A-\omega)}{PC}=\frac{\sin A}{b}$$ Also in $\triangle PCB$ we can use sine rule $$\frac{\sin(\omega)}{PC}=\frac{\sin( \angle CPB)}{a}\tag1$$ Here also we can conclude that $\angle CPB=\pi -C$ Hence $$\frac{\sin(\omega)}{PC}=\frac{\sin(C)}{a}\tag2$$

Now dividing equation $(1)$ by equation $(2)$ $$\frac{\sin (A-\omega)}{sin(\omega)}=\frac{a \sin A}{b \sin C}$$ Using sine rule we can also write: $$\frac{\sin (A-\omega)}{sin(\omega)}=\frac{\lambda \sin A \cdot \sin A}{\lambda \sin B \cdot \sin C}$$ $$\frac{\sin (A-\omega)}{sin(\omega)}=\frac{ \sin A \cdot \sin (\pi -(B+C)}{ \sin B \cdot \sin C}$$ $$\frac{\sin A \cos \omega- \cos A \cdot \sin \omega}{sin(\omega)}=\frac{ \sin A \cdot (\sin B \cdot \cos C + \sin C \cdot \cos B)}{ \sin B \cdot \sin C}$$ $${\sin A \cot \omega- \cos A }={ \sin A \cdot ( \cot C + \cot B)}$$ Dividing both sides $sinA$ $$cot\omega = \cot A + \cot B + \cot C $$


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