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In other words, I want to prove that angle is at its maximum when it lands at A. Each of these angles goes through a third of the diameter.

That is, if we set O as the center of the circle, Angle A reaches its maximum when AO is perpendicular to the diameter in the figure.

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3 Answers 3

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By the sine rule (see figure below) we have two equations: $$ {\sin\theta_1\over a}={\sin(\phi-\theta_1)\over r}, \quad {\sin\theta_2\over a}={\sin(\phi+\theta_2)\over r}, $$ which can be solved to $$ \tan\theta_1={a\sin\phi\over r+a\cos\phi}, \quad \tan\theta_2={a\sin\phi\over r-a\cos\phi}. $$ Hence: $$ \tan(\theta_1+\theta_2)= {\tan\theta_1+\tan\theta_2\over1-\tan\theta_1\cdot\tan\theta_2} ={2ra\over r^2-a^2}\sin\phi. $$ From this it follows that $\theta_1+\theta_2$ is maximum when $\sin\phi$ is maximum, that is for $\phi=90°$.

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  • $\begingroup$ Wow! that's cool\ $\endgroup$
    – S.Y.Li
    Commented May 14, 2023 at 15:30
  • $\begingroup$ Nice thing about this proof is that it can be easily adapted to work also for spherical and hyperbolic geometries ( where the sine rule is almost the same) $\endgroup$
    – orangeskid
    Commented May 15, 2023 at 16:48
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On the boundary of the union of the green circles the angle is constant. Inside is large, outside is smaller. That shows how moving on the blue circle the angle increases up to the top, then decreases again.

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  • $\begingroup$ And if you construct the circle through $A$ and the two red points, it will be internally tangent to the blue circle at $A$ so that all other points on the blue circle are outside this new circle -- and therefore the angles at those points are smaller. $\endgroup$
    – David K
    Commented May 15, 2023 at 4:39
  • $\begingroup$ @David K: Yes, you are very right! The more general circles get us the monotony of the angle. So the level curves ( circles in this case) tell us a lot. $\endgroup$
    – orangeskid
    Commented May 15, 2023 at 4:48
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Here's an alternative path to the key relation in @intelligenti's answer. The setup is more involved, but the derivation is quick.


Consider $\bigcirc O$ of radius $r$, and $\triangle PQS$ with $S$ on the circle and $O$ the midpoint of $PQ$ diameter. Define $|OP|=|OQ|=s$ (which need not be one-third of the diameter), $p := |SQ|$, $q:=|SP|$.

Extend $\overline{SO}$, $\overline{SP}$, $\overline{SQ}$ to meet the circle again at $S'$, $P'$, $Q'$. Note that $\angle P'$ and $\angle Q'$ are right angles; also, $\square SPS'Q$ is a parallelogram, so that $\angle S'QQ'\cong\angle S\;(\cong\angle S'PP')$.

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Considering the area of $\triangle PQS$ and an appropriate ratio in right $\triangle S'QQ'$, we can write $$\tan S = \frac{\sin S}{\cos S} = \frac{2\,|PQS|/(pq)}{p'/q} = \frac{2\,|PQS|}{pp'} = \frac{2\,|PQS|}{r^2-s^2} \tag{$\star$}$$ where the last equality is provided by the Power of a Point Theorem. (We'd get the same result if we'd used $\cos S = q'/p$ from $\triangle S'PP'$.)

For fixed $P$ and $Q$, the denominator of $(\star)$ is constant, so we see that $\tan S$ (and therefore also $\angle S$) is maximized when $|PQS|$ itself is maximized. As base $\overline{PQ}$ is fixed, the maximum area is attained when $S$ reaches its maximum height; ie, when $\overline{SO}\perp\overline{PQ}$. $\square$

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