Maximum area!

(I hope that there exist a more simple solution.) In this question, $\large P$ is the orthocenter of $\large \triangle ABC$ .

Proof: First, fix the line $BC$ . Now, the locus of $A$ is a circle with center $P$ and radius $3$ . As the line $BC$ is fixed, the area of $\triangle ABC$ achieve its maximum value when its height is maximum( which means length of $A$ to line $BC$ achieve maximum value). Draw the line $BC$ and the circle with center $P$ , it is obvious that $AP$ must perpendicular to line $BC$ so that the distance from $A$ to line $BC$ achieve maximum value.

By the diagram above, $\triangle AZP \sim \triangle CXP, \triangle AYP \sim \triangle BXP$ . Which means $\frac{PZ}{PX}=\frac{3}{5}, \frac{PY}{PX}= \frac{3}{4}$ , this imply that $PX:PY:PZ=1: \frac{3}{5}: \frac{3}{4}=20:12:15$ . Let $PX=20k,PY=12k,PZ=15k$ . From the diagram above, we can get $BC^2-BY^2=PC^2-PY^2 \space\space-(eqn1)$ $BC=\sqrt{PC^2-PX^2}+ \sqrt{PB^2-PX^2}\space\space-(eqn2)$ Let $BC=a$ , from $eqn1$ , we get $\begin{aligned} a^2-(4+15k)^2&=5^2-(15k)^2 \\ a^2-25&=(15k+4)^2-(15k)^2 \\ &=(15k+4+15k)(15k+4-15k) \\ a^2&=120k+41 \end{aligned}$ From $eqn2$ , we get $\begin{aligned} a^2&=(\sqrt{25-(20k)^2}+\sqrt{16-(20k)^2})^2 \\ 20k^2+3k&=\sqrt{1-16k^2}\sqrt{1-25k^2} \\ (20k^2+3k)^2&=(\sqrt{1-16k^2}\sqrt{1-25k^2})^2 \\ 120k^3+50k^2-1&=0 \\ k&=0.12413 \end{aligned}$ ( $k=0.12413$ is the only positive solution.)

Maximum of area of $\triangle ABC$ is $\begin{aligned} &0.5×BC×AX \\ =&0.5×\sqrt{120k+41}×(20k+3) \\ =&20.49 \\ \approx& 20.5 \end{aligned}$