04/05/2026

$AB $ is the diameter of a circle and $AB =8$. $P$ is a point on the circle. Find the maximum value of $PA+PB$.

Solve:

\(\begin{multline} \shoveleft \text{Extend }AP \text{ to } Q \text{ such that }PB=PQ\\ \shoveleft \implies \dfrac{AB}{sin(45^{\circ})}=\dfrac{AQ}{sin(45^{\circ}+\angle{ABP})}\\ \shoveleft \implies AQ=\dfrac{AB}{sin(45^{\circ})}sin(45^{\circ}+\angle{ABP})\\ \shoveleft \implies AQ \le \dfrac{AB}{sin(45^{\circ})}=\bbox[5px, border: 1px solid black]{8\sqrt{2}} \end{multline}\)

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