My 25 followers problem

Relevant Figure

Let $P$ and $Q$ be the points representing the two objects, and let $PQ$ meet the road ( $AB$ ) in $B$ .

If $A$ be the point at which the greatest angle is subtended, then $A$ must be the point where a circle drawn through $P$ and $Q$ touches the road $^{\dagger \textit{ see below for proof}}$ .

Let $\angle QAB = \theta$ . Then, as the circle is tangent to the road at $A$ , $\angle APQ = \angle QAB = \theta^{\ddagger \textit{ see below for proof}}$ .

Now, $\begin{aligned} \pi &= \text{ Sum of angles of }\Delta PAB\\&= \theta+ \left(\alpha+ \theta \right)+ \beta\\\implies \theta &= \frac{\pi}{2}- \frac{\alpha+\beta}{2}\end{aligned}$

Applying sine rule to $\Delta PAQ$ and $\Delta QAB$ , $\begin{aligned} \frac{PQ}{AQ} &= \frac{\sin \alpha}{\sin \theta}\\\frac{AQ}{c} &= \frac{\sin \beta}{\sin \angle AQB} = \frac{\sin \beta}{\sin \left( \theta + \alpha \right)}\end{aligned}$ In the last equality, exterior angle property has been used.

Multiplying these equations, we get $\begin{aligned} \frac{PQ}{c} &= \frac{\sin \alpha \sin \beta}{\sin \theta \sin \left(\theta +\alpha \right)}\\&= \frac{\sin \alpha \sin \beta}{\cos \frac{\alpha +\beta}{2} \cos \frac{\alpha -\beta}{2}} \quad \left( \because \theta = \frac{\pi}{2}- \frac{\alpha+\beta}{2} \right) \\\implies PQ &= c\sin \alpha \sin \beta \sec \left( \frac{\alpha +\beta}{2} \right) \sec \left (\frac{\alpha -\beta}{2} \right) \end{aligned}$

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$^{\dagger}$ To see this, take any point $A' \neq A$ on $AB$ and join it to $P$ cutting the circle at $B'$ , and join $A'Q$ and $B'Q$ .

Then, $\angle PA'Q < \angle PB'Q = \angle PAQ\\\implies \angle PA'Q < \angle PAQ$

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$^{\ddagger}$ Construct $P'A \perp AB$ where $P'$ lies on the circle.

Now, $\angle QAB +\angle P'AQ =\fracπ2$ .

Also,

$\angle P'QA =\fracπ2 \quad (\because \angle P'QA \text{ is an angle inscribed in a semicircle })\\\implies \angle AP'Q+ \angle P'AQ =\fracπ2 \\(\because \text{ sum of angles in }\Delta P'AQ \text{ is } π)\\\implies \angle QAB=\angle AP'Q=\angle APQ$

The last equality follows by the inscribed angle theorem .

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Note: This problem, its solution, and the presented image have been taken from a book ( Plane Trigonometry ) by S. L. Loney .