Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.

Question

Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.

What I Tried: Here is a picture :-

I made an attempt solving this and found rather a weird conclusion.

We have that $A,P,C,R$ are collinear, and also $AP = CR$. Let $DL = LS = BK = KQ = x$ . We have $AP = x\sqrt{2}$ . Since $PC$ = $(4 - 2\sqrt{2})$ , $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ .

Now, $PF = FC$ . So by Pythagoras Theorem on $\Delta PFC$ , $PF = FC = \sqrt{12 - 8 \sqrt{2}}$.

$AL = LS = \sqrt{12 - 8 \sqrt{2}} + 2x$ and $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ . So now by Pythagoras Theorem :-

$$2\bigg(\sqrt{12 - 8 \sqrt{2}} + 2x\bigg)^2 = (4 - 2\sqrt{2} + 2x\sqrt{2})$$ $$\rightarrow 2\bigg(12 - 8\sqrt{2} + 4\sqrt{12 - 8\sqrt{2}}x + 4x^2\bigg) = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$ $$\rightarrow 24 - 16\sqrt{2} + 8\sqrt{12 - 8\sqrt{2}}x + 8x^2 = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$ $$\rightarrow \sqrt{12 - 8\sqrt{2}}x = 2\sqrt{2}x - 16x$$ $$\rightarrow \sqrt{12 - 8\sqrt{2}} = 2\sqrt{2} - 16$$ But this is obviously false.

So I literally ended up in a completely false equation, might be making a mistake in this complicated calculations, and I cannot find it. So can anyone tell me where I made the mistake, or I am solving this in the correct way or not?

Answer 1

Using the fact that the side of the outer square and diagonal of the inner square are equal to the diameter of the circle,

$AD = \frac{AL + PE}{2} = \frac{2r + r\sqrt2}{2} = \frac{(2 + \sqrt2)(2 - \sqrt2)}{2} = 1$

Answer 2

Let radius of circle be $r$. So side of square $a=2r$.

Let the circle touch $KR$ at $X$. So $KX=a/2=r$

Let $KQ=x$. So $PF=a-2x=2(r-x)$

Use power of point of $Q$, $$PQ\cdot QF=QX^2$$