$x/2$, $x+r\sqrt{2}/2$ and $r$

By pythagoras,

${\displaystyle (\frac{1}{2}x{)}^2+(\frac{r\sqrt{2}}{2}+x{)}^2=r^2}$

${\displaystyle \frac{x^2}{4}+\frac{2r^2}{4}+\mathrm{xr}\sqrt{2}+x^2=r^2}$

${\displaystyle x^2+2r^2+4\mathrm{xr}\sqrt{2}+4x^2=4r^2}$

${\displaystyle 5x^2+4\mathrm{xr}\sqrt{2}=2r^2}$

Substitute $y=r\sqrt{2}$ (length of the side of the large square)

${\displaystyle 5x^2+4\mathrm{xy}=y^2}$

${\displaystyle 5x^2+4\mathrm{xy}-y^2=0}$

${\displaystyle x=\frac{-4y\pm \sqrt{(4y{)}^2+20y^2}}{10}}$

${\displaystyle x=\frac{-4y\pm \sqrt{36y^2}}{10}}$

${\displaystyle x=\frac{-4y\pm 6y}{10}}$

${\displaystyle 10x=2y\mathrm{or}-10y}$

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 o + CM cos 60 o = r ${\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}r=r}$ Therefore ${\displaystyle \sqrt{3}.\mathrm{PQ}=r}$ Now ${\displaystyle \mathrm{LM}=2(r\sin 60)}$ so ${\displaystyle \mathrm{LM}=\sqrt{3}.r}$ ${\displaystyle \mathrm{LM}=\sqrt{3}.(\sqrt{3}.\mathrm{PQ})}$ ${\displaystyle \mathrm{LM}=3\mathrm{PQ}}$

Solution using similar triangles:

NR is a diamter so NLR = 90 ^o
but NLM = 60 ^o so RLP = 30 ^o
Now LPR = 120 ^o so LRP = 30 ^o
Hence LP = pr but PR = PQ
so LP = PQ = QM
LM = 3 PQ