You are given rays $OA$ and $OB$. segment $OP$ is located between the 2 rays and has a length of $1$. Angle $POA$ is $45^\circ$ and angle $POB$ is $30^\circ$. A line is drawn through the point P that intersects rays $OA$ and $OB$ respectively at the points $Q$ and $R$. Find the maximum value of $\frac{1}{PQ} + \frac{1}{PR}$.
Okay, so I called angle $OQP$ to be $\alpha$ then by the law of sines we get that $\frac{\sin(45^\circ)}{PQ} = \frac{\sin(\alpha)}{1} $ and $\frac{\sin(30^\circ)}{PR} = \frac{\sin(105^\circ-\alpha)}{1}$. Then we get that $\frac{1}{PQ} + \frac{1}{PR} = \sqrt2\sin(\alpha) + 2\sin(105^\circ -\alpha)$. Now we just take the derivative and equate it to $0$. $\sqrt2\cos(\alpha) – 2\cos(105^\circ -\alpha) = 0$, from this we get that $\alpha $ is equal to $45^\circ$ and plugging back into to the original equation we get that the maximum value is $\sqrt3 + 1$. What I cannot find is a purely geometric way to solve this.
Best Answer
You can avoid the derivative by noticing that $$ \sin105°={\sqrt2\over2}{1+\sqrt3\over2}, \quad \cos105°={\sqrt2\over2}{1-\sqrt3\over2} $$ and inserting that into your formula gives: $$ {1\over PQ}+{1\over PR}=(1+\sqrt3)\sin(45°+\alpha). $$ Hence maximum is reached for $\alpha=45°$.