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If the length of the sides of the small square is $x$, the sides of this triangle are:
$x/2$, $x + r\sqrt{2}/2$ and $r$.
By pythagoras, \[{({1\over 2}x)^2} + {( {r\sqrt2\over2} + {x})^2} = {r^2} \] \[{x^2\over4} + {2r^2\over4} + {xr\sqrt2} + {x^2} = {r^2} \] \[{x^2} + {2r^2} + {4xr\sqrt2} + {4x^2} = {4r^2} \] \[{5x^2} + {4xr\sqrt2} = {2r^2} \] Substitute $y = r\sqrt{2}$ (length of the side of the large square)
\[{5x^2} + {4xy} = {y^2} \] \[{5x^2} + {4xy}  {y^2} ={ 0} \] \[{x} = {{4y \pm\sqrt{{(4y)^2} + {20y^2}}}\over 10} \] \[ {x} = {{4y \pm\sqrt{36y^2}} \over 10} \] \[{x} = {{4y \pm 6y}\over 10} \] \[ {10x} = {2y or 10y} \]
Ratio must be positive, therefore $10x = 2y$, therefore $y =
5x$
QED
Solution to part B.
Submitted by Samantha Gooneratne, Colombo International School, Sri Lanka. Well done Samantha! Her teacher also solved the problem using similar triangles, which I have include below Samantha's solution.
Let $X$ be the mid point of $PQ$, $C$ the center of the circle and $r$ the radius.
$RX + XC = r$ Hence $PR \sin 60^{\circ}+ CM \cos 60^{\circ}=r$ \[{{\sqrt{3}\over 2}+ {{1 \over2}r} = {r}} \] Therefore \[{ {\sqrt{3}.PQ = {r}}} \] Now \[{{LM} ={2(r \sin 60)}} \] so
\[{{LM} ={\sqrt{3}.r}} \] \[{{LM} = {\sqrt{3}.(\sqrt{3}.PQ)}}
\] \[{{LM} = {3PQ}} \]

Solution using similar triangles:
$NR$ is a diameter so $NLR = 90^{\circ}$
but $NLM = 60^{\circ}$ so $RLP = 30^{\circ}$
Now $LPR = 120^{\circ}$ so $LRP = 30^{\circ}$
Hence $LP = PR$ but $PR = PQ$
so $LP = PQ = QM$
$LM = 3 PQ$