# A problem on circle

Consider a circle $C$, such that $\overline{AB}$ is a chord. $P$ be a moving point on the circumference of the circle.

(i) How to find the point $P$ such that $\overline{PA}\cdot \overline{PB}$ is maximum?

(ii) How to find the point $P$ such that $\overline{PA}+\overline{PB}$ is maximum?

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（i) is $PA*PB$ max? what is your try? – chenbai Jun 8 '14 at 10:43
No idea. Can provide any hints for trying atleast? – Rudstar Jun 8 '14 at 10:44

Notation: since the circle is fixed, take the radius of the circle to be $R$ and the angle opposite $AB$ as $\alpha$. Take angles $\angle PAB$ and $\angle PBA$ to be $\beta$ and $\gamma$ respectively.

For $(i)$, $PA\times PB=2R\sin \beta\times 2R\sin\gamma$. Since the circle is fixed, we need to maximize $$\sin \beta\times \sin\gamma$$ or maximise$$\cos(\beta-\gamma)-\cos (\beta+\gamma)$$ or $$\cos (\beta-\gamma)+\cos\alpha$$ Now, $\alpha$ is fixed. So we need to take the max value of $\cos (\beta-\gamma)$, which is $1$, and occurs when $\beta=\gamma$, or when $P$ is the point of intersection of the circle and the perpendicular bisector of the chord AB in the major segment.

For $(ii)$,

Using the sine rule, we need to maximize $$2R(\sin\beta+\sin\gamma)$$ or $$(\sin\beta+\sin \gamma)$$ since $R$ is fixed. Now $$(\sin\beta+\sin \gamma)=2\sin\left( \dfrac{\beta+\gamma}{2}\right)\times \cos \left(\dfrac{\beta-\gamma}{2}\right)=2\cos \dfrac {\alpha}{2}\times\cos\dfrac {\beta-\gamma}{2}$$ Since $\alpha$ is fixed, in order to maximize the required expression, we take the max value of $\cos \left(\dfrac{\beta-\gamma}{2}\right )$, which is $1$,and occurs when $\beta=\gamma$, or when $P$ is the point of intersection of the circle and the perpendicular bisector of the chord $AB$ in the major segment.

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hint:

$S_{ABP}=\dfrac{1}{2}PA*PB \sin{\angle APB}= \dfrac{1}{2}h*AB$

$PA=2R*\sin{B},PB=2R*\sin{A}$

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(i) Hint:

Let $\alpha$ be the fixed angle APB. Let x be the variable angle PBA.

$$\frac{\sin x}{|PA|}=\frac{\sin \alpha}{|AB|}$$

$$\frac{\sin(\alpha+x)}{|PB|}=\frac{\sin \alpha}{|AB|}$$

The problem is then to maximize:

$$sinx\cdot sin(\alpha+x)$$

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It's a good idea to linearize at $O$. Then one sees that this function is the restriction to the sphere of the map $\vec{v}\mapsto R^2-\vec{v}\cdot\overrightarrow{OA+OB}+\vec{OA}\cdot\vec{OB}$ where $R$ is the radius of the circle.

This function admits critical points when the plane tangent to the circle is orthogonal to $\overrightarrow{OA+OB}$. If $A$ and $B$ are not opposite points then there are only two such points, otherwise the function is constant...

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I have no idea what it means to "linearize at $O$". It looks like you're maximizing the dot product $\overrightarrow{PA} \cdot \overrightarrow{PB}$. That wasn't the question; part (i) asks to maximize the product $\overline{PA} \cdot \overline{PB} = \lVert \overrightarrow{PA}\rVert \cdot \lVert \overrightarrow{PB} \rVert$ of the lengths. – epimorphic Dec 26 '15 at 20:55