Problem - 2995
There are four points on a plane as shown. Points $A$ and $B$ are fixed points satisfying $AB=\sqrt{3}$. Points $P$ and $Q$ can move, as long as $AP=PQ=QB=1$. Let $S$ and $T$ be the area of $\triangle{APB}$ and $\triangle{PQB}$, respectively. Find the maximum value of $S^2+T^2$.
Firstly, we have $$S=\frac{\sqrt{3}}{2}\cdot PA \cdot AB=\sin{A}$$ and $$T=\frac{1}{2}\cdot PQ\cdot QB=\frac{1}{2}\sin{Q}$$
Therefore, we find: $$S^2 + T^2 = \frac{3}{4}\sin^2{A} + \frac{1}{4}\sin^2{Q}$$
Now, applying the law of cosine on $PB$ in $\triangle{PAB}$ and $\triangle{QPB}$:
$$PA^2=PA^2 + AB^2 - 2\cdot PA \cdot AB\cdot\cos{A}=4 -2\sqrt{3}\cos{A}$$
and
$$PA^2 = PQ^2 + QB^2 - 2PQ\cdot QB\cos{Q}=2-2\cos{Q}$$
Therefore:
$$4-2\sqrt{3}\cos{A} = 2-2\cos{Q}\implies \cos{Q} = \sqrt{3}\cos{A} - 1$$
It follows that
\begin{align*}
S^2+T^2 &= \frac{3}{4}\sin^2{A} + \frac{1}{4}\sin^2{Q} \\
&= \frac{3}{4}(1-\cos^2{A}) + \frac{1}{4}(1-\cos^2{Q})\\
&= -\frac{3}{2}\cos^2{A} + \frac{\sqrt{3}}{2}\cos{A} + \frac{3}{4}\\
&= -\frac{3}{2}(\cos{A}-\frac{\sqrt{3}}{6})^2 + \frac{7}{8}\\
&\le \boxed{\frac{7}{8}}
\end{align*}