# Find the distance between two points, given maximal angles subtended by them

Problem:

$A$ and $B$ are two points on same side of straight line $L$, such that $AB$ is not parallel to $L$. Thus line $AB$, when produced, cuts $L$ at point $P$. $AB$ subtends two maximal angles on $L$, one on either side of $P$. Let these angles be $\alpha$ and $\beta$, and the two points be $M$ and $N$. Distance between $M, N$ is $c$.

To prove,

$d=c\sec{(\frac{\alpha+\beta}{2})}\sqrt{\sin{\alpha}\sin{\beta}}$

where d is the distance between A and B.

My attempt:

I realised that to subtend maximal angles, $M$ must be the point of tangency of the circle passing through $A,B$ and tangent to $L$. Reasons are somewhat obvious, and I'm not stating them. Same goes for $N$.

Using simple geometry, I found $MBA$, $BMP$ are equal. Similarly $NAB$, $BNP$ are also equal.

Also, I got $MP=PN=c/2$. (I proved this by coordinate geometry)

Using these informations, I tried solving the problem by trigonometry, assuming angles $MAB=x, NAB=y$. But the process gets too complicated, and I cannot eliminate x, y from the final expressions. Any suggestions/hints or solutions would be very helpful. Thank you!

First,as you said in your post: $\angle BNM = \angle BAN = \gamma$ and $\angle BMN = \angle BAM = \theta$

Using law of sines in $\triangle ABN$ and $\triangle ABM$ we get $$\frac {AB}{BN} = \frac{\sin \beta}{\sin \gamma}$$ $$\frac{AB}{BM}=\frac{\sin \alpha}{\sin \theta}$$

So length of $AB$ will be $$AB = \sqrt{\dfrac{BN.BM}{\sin \theta .\sin \gamma} \sin\alpha\sin\beta}=\sqrt{\dfrac{BN.BM}{\sin \theta .\sin \gamma}}\times\sqrt{\sin\alpha\sin\beta}$$

Now use law of sines, this time on $\triangle BMN$ $$\frac{BM}{\sin \gamma}=\frac{BN}{\sin\theta}=\frac{MN}{\sin \angle MBN}$$

Then $$AB=\sqrt{\dfrac{BN.BM}{\sin \theta .\sin \gamma}}\times\sqrt{\sin\alpha\sin\beta} = \frac{MN}{\sin\angle MBN}\sqrt{\sin\alpha\sin\beta}$$

So we have to calculate $\angle MBN$ \begin{align} \angle MBN & = \pi - (\gamma + \theta) \\ & = \pi - \left(\frac{\pi - (\alpha + \beta)}{2}\right) \\ & = \frac{\pi +\alpha+\beta}{2} \end{align}

So $$\sin \angle MBN = \sin \frac{\pi +\alpha+\beta}{2} = \cos \frac{\alpha+\beta}{2}$$

Substituting this \begin{align} AB & = MN\sec\left(\frac{\alpha+\beta}{2}\right)\sqrt{\sin\alpha\sin\beta}\\ \Longrightarrow d & = c\sec{\left(\frac{\alpha+\beta}{2}\right)}\sqrt{\sin{\alpha}\sin{\beta}} \end{align}