Square inside a square

Let $x$ the redius of the circle

$x\sqrt{2}=x+r$ so $r=x\sqrt{2}-x$

Then

$x+x+r=4\sqrt{2}$

$2x+x\sqrt{2}-x=4\sqrt{2}$

$x(\sqrt{2}+1)=4\sqrt{2}$

$\boxed{x=\frac{4\sqrt{2}}{\sqrt{2}+1} \approx 2.343}$

From the figure we see that $rcos45^\circ + r = 4$

$\implies$ $\dfrac{r}{\sqrt{2}} +r = 4$

Solving this equation we would get $r = \dfrac{4\sqrt{2}}{\sqrt{2}+1} \approx 2.343 units$ .

Let the top and right sides of the square be the x and y axis respectively (in the third quadrant). Let the radius of the circle be "r". We can write an equation of the circle. The center is (-r,-r) because it is touching both axis in the third quadrant. The radius is r as we give it. Therefore, the equation is (x+r)^2 + (y+r)^2 = r^2.

Now we need to substitute in a value. As it touches the tip of that square, we can substitute in that value. On the co-ordinate plane, that is equivalent to (-4,-4). We then get the equation r^2 - 16r + 32 = 0. Solving, we get 8 - 4*sqrt(2) which is equivalent to 2.343.

$r +$ $\frac{1}{\sqrt{2}}$ $r + 2$ $= 6$

$r +$ $\frac{1}{\sqrt{2}}$ $r$ $= 4$

$\frac{1 + \sqrt{2}}{\sqrt{2}}$ $r$ $= 4$

$r =$ $\frac{4\sqrt{2}}{\sqrt{2}+1}$

We get that $x$ is $\frac1{\sqrt{2}}$ . So $2 = (6-2r)+(r-\sqrt{2}r)$ , giving us $r=\frac{4}{1+\sqrt{2}}$ .

Am I the only one who has used analitic geometry?

Placing the corner $B$ at the origin of the $xy-$ plane, the circle can be modeled as $[x-(6-r)]^2 + [y-(6-r)]^2 = r^2$ . The circle intersects the smaller square at $(x,y) = (2,2)$ , which gives;

$[2-(6-r)]^2 + [2-(6-r)]^2 = r^2$ ;

or $2(r-4)^2 = r^2;$

or $\sqrt{2}(r-4) = \pm r$

or $r = \frac{4\sqrt{2}}{\sqrt{2} \mp 1}$ ;

or $r = 4\sqrt{2} \cdot (\sqrt{2} \pm 1) = 8 \pm 4\sqrt{2}$ .

Since the minus root makes physical sense for this nested configuration, the radius is just $\boxed{r = 8-4\sqrt{2} \approx 2.343}.$

$h+r=4 .$

$\frac{2r}{2\sqrt{2}}=\frac{2h}{2}$

$h=\frac{r}{\sqrt{2}}$

$\Rightarrow r+\frac{r}{\sqrt{2}}=4$

$r=\frac{8}{2+\sqrt{2}} \approx 2.343$

I defined the distance from point O to line BC as x and the radius of the circle as r. Then using the length of one side: x = 6 - r. I then drew an isosceles right triangle in the bottom left of the large square. From the pythagorean theorem (x)^2 + (x)^2 = (r + 2*sqrt(2))^2. You can substitute 6-r into x in the equation above and then solve the resulting equation for r. Sorry I don't know how to use math formatting.

By the pythagorean theorem, we have

$R^2=(R-4)^2+(R-4)^2$

$R^2=R^2-8R+16+R^2-8R+16$

$0=R^2-16R+32$

$R=4(2-\sqrt{2})$ or $R=4(2+\sqrt{2})$

The desired answer is $R=4(2-\sqrt{2})\approx 2.343145751$