A geometry problem by Yahia El Haw

The correct answer is $2 : (3 - 2\sqrt { 2 } )$ It is given that $RS$ passes through $L$ and $RQ$ passes through $M$ .
On drawing the line $RSL$ and $RQM$ , the diagram shown above results. From the diagram, it is clear that $MNLR$ has to be a square and $P$ has to be the mid-point of the arc.

As $P$ is the mid-point of the arc, $RN$ passes through $P$ ,

$RP$ = $a (\sqrt { 2 } - 1)$

where $MN$ = $PN$ = $a$

$\Rightarrow$ $PQ$ = $\frac { RP }{ \sqrt { 2 } }$

= $a(\sqrt { 2 } -1)/\sqrt { 2 }$

${ MN }^{ 2 } : { PQ }^{ 2 }=2 : 3-2\sqrt { 2 }$

Let the side length of square $MLNR$ be $X$ and the side length of square $QPSR$ be $Y$ , then $RL=X\sqrt{2}$ and $RP=Y\sqrt{2}$ .

$RL=RP+X$ $\implies$ $X\sqrt{2}=Y\sqrt{2}+X$ $\implies$ $Y\sqrt{2}=X\sqrt{2}-X$ $\implies$ $Y=\dfrac{X\sqrt{2}-X}{\sqrt{2}}$

Squaring both sides, we have $Y^2=\dfrac{X^2(3-2\sqrt{2})}{2}$ .

Finally, the ratio of the areas is $\dfrac{X^2}{Y^2}=\dfrac{X^2}{\dfrac{X^2(3-2\sqrt{2})}{2}}=\dfrac{2}{3-2\sqrt{2}}$ and the desired answer is $\boxed{2:3-2\sqrt{2}}$

If LP= r is the radius of the circle, LR the diagonal of LMRN= $\sqrt2$ r and PR=RL-PL, the diagonal of PQRS = ( $\sqrt2$ -1)r.
So the ratio= $\left\{\dfrac{\sqrt2}{\sqrt2-1}\right \}^2\ \implies\ 2 :\ 3-\sqrt2.$