TITLE: Largest angle between three circle points
QUESTION [0 upvotes]: Given a circle $k$ with centre $M$ and another point $B$ inside the circle. Find those points $V$ on $k$ for which the angle $BVM$ has maximum size.

I think that the angle $MBV$ has to be $90°$ so that the other can have maximum size. Has anyone an idea for an approach?

REPLY [0 votes]: Let's take any $V$, then construct points $V_1$ and $V_2$ on $k$ such that $V_1$ lies on continuation of $VB$ and $V_2$ lies on continuation of $VM$. Then $\angle BVM=\angle V_1VV_2=\frac{\angle V_1MV_2}{2}\leq 90°$.
But If $\angle BVM=90°$, then $\angle V_1MV_2=180°$ then $V_1V_2$  is diameter. $VV_2$ is also diameter. Then $V$ is the same as $V_1$ which is contradictory. So angle $90°$ cannot be obtained.
Let's consider point $U$ which is center of circumscribed circle of triangle $BVM$. Then $\angle BVM=\frac{\angle BUM}{2}$. Let $D$ is middle of $BM$, then $\angle BUM=2\angle BUD$, $\angle BVM=\angle BUD$. Point $U$ is such that $BU=MU=VU$. Using triangle rule, we can say that $2 BU = MU+VU \geq MV = R$, then $BU \geq \frac{R}{2}$. Then $\sin BVM=\sin BUD=\frac{BD}{BU}=\frac{BM/2}{BU} \leq \frac{BM}{R}$. As $BVM \leq 90°$, maximum sine corresponds to maximum value of angle. Then $\angle BVM \leq \arcsin \frac{BM}{R}$. Note that maximum angle value corresponds to $MU=BU=\frac{R}{2}$ and $MU+VU=MV$.
To find point $V$, one need to construct $U$ which is intersection of circle with radius $MU=\frac{R}{2}$ and perpendicular bisector of $BM$ where $BU=MU$. Then point of intersection of $MU$ continuation with $k$ will be $V$. There are two possible location of this point symmetric about $BM$ line.
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