Find the minimum of $\max($area of any of $3$ parts of a right triangle$)$ 
Let $\triangle ABC$ be a right triangle with $BC=AC=1$. Let $P$ be any point on $AB$. Draw perpendiculars $PQ$ and $PR$ on $AC$ and $BC$ respectively from $P$. Define $M$ to be the maximum of the areas of $\triangle BPR$, $\triangle APQ$, and rectangle $PQCR$. Find $\min(M)$.

I tried to find $M$ but couldn't do it.

edit: Original Question
 A: 
Let $x= \angle PCB$. Then, $PR = \frac1{1+\cot x}$, $PQ = \frac1{1+\tan x}$ and the areas are respectively,
$$[APQ]= \frac1{2(1+\tan x)^2}, \>\>\>
[BPR]= \frac1{2(1+\cot x)^2} \>\>\>
[PQCR]= \frac1{(1+\tan x)(1+\cot x)}$$
Due to the symmetry, the minimum of the maximum among the three areas above is when
$[BPR] = [PQCR]$ or $[APQ] = [PQCR]$. Use the former to establish
$$\text{min}(M)= \frac1{2(1+\cot x)^2}= \frac1{(1+\tan x)(1+\cot x)}$$
Solve to get $\cot x =\frac12$ and $\text{min}(M)= \frac29$.
A: 
Let $\triangle ABC$ be a right triangle with $BC=AC=1$. Let $P$ be any point on $AB$. Draw perpendiculars $PQ$ and $PR$ on $AC$ and $BC$ respectively from $P$. Define $M$ to be the maximum of the areas of $\triangle BPR, \triangle APQ, \sqsubset \! \sqsupset PQCR$. Find $\min(M)$.


Let $BP=x$, then all you have to find is the minimum of the vaguely angular curve $XYZ$ sitting on the top of all in $x\in[0,\sqrt2]$. That is, find which one of $Y$ or $Z$ is lower by finding the intersection of the curves.

Check answer: $\frac29$
A: I am getting the answer as $\frac29$ based on boundary conditions.

Hope this helps!
