Triangle Solving – Problem 8: Compute

A point $P$ inside $\triangle ABC$ is called a Brocard point if $\angle PAB=\angle PBC=\angle PCA=\theta$; $\theta$ is the Brocard angle. It satisfies $$\cot \theta=\cot A+\cot B+\cot C\quad(\text{note that }\tan x\cdot\cot x=1).$$ Compute $\frac{PA}{c}+\frac{PB}{a}+\frac{PC}{b}$.

A. $2\sin \theta$ B. $2\cos \theta$ C. $2\tan \theta$ D. $2\cot \theta$

Step 1. In $\triangle ABP$, $\angle PAB=\theta$ and $\angle ABP=B-\theta$, so $\angle APB=\pi-B$.

Step 2. By the Law of Sines in $\triangle ABP$, $$\frac{PA}{\sin(B-\theta)}=\frac{AB}{\sin(\pi-B)}=\frac{c}{\sin B},$$ so $$\frac{PA}{c}=\frac{\sin(B-\theta)}{\sin B}=\cos\theta-\sin\theta\cot B.$$

Step 3. In $\triangle BCP$, $\angle PBC=\theta$ and $\angle PCB=C-\theta$, so $\angle BPC=\pi-C$. Thus $$\frac{PB}{a}=\frac{\sin(C-\theta)}{\sin C}=\cos\theta-\sin\theta\cot C.$$

Step 4. In $\triangle CAP$, $\angle PCA=\theta$ and $\angle PAC=A-\theta$, so $\angle APC=\pi-A$. Hence $$\frac{PC}{b}=\frac{\sin(A-\theta)}{\sin A}=\cos\theta-\sin\theta\cot A.$$

Step 5. Therefore $$\frac{PA}{c}+\frac{PB}{a}+\frac{PC}{b}=3\cos\theta-\sin\theta(\cot A+\cot B+\cot C).$$ Using $\cot\theta=\cot A+\cot B+\cot C$, we get $$3\cos\theta-\sin\theta\cot\theta=3\cos\theta-\cos\theta=2\cos\theta.$$

Step 6. So the correct choice is B.

B