Finding lengths when circles and squares tangents. [closed]

Question

Should one approach by coordinates or by euclidean geometry?

By pure geometry, I am not able to solve.

Answer 1

Assume the larger radius is 1 first (the square's side is 2). Then the smaller radius $r=FB$ satisfies $$\sqrt{(1+r)^2-1}+r=2$$ from which we solve and obtain $r=\frac23$.

Setting up coordinates such that $A$ is the origin, we find $G=(3/2,1/2),GB=\sqrt2/2$ and $x=\frac{2\sqrt2}3$. Since $GB=3$ in the picture, scaling yields the desired $x$ of $$\frac{3\cdot2\sqrt2/3}{\sqrt2/2}=4$$

Answer 2

As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $\triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.

Again, denote the side of the sqaure = $x$ and the radius of small semi circle = $r'$. So, the radius of larger semi circle = $r = \frac{x}{2}$.

From $\triangle QEF$, we get

$EQ^2 + QF^2 = EF^2$

$(x-r')^2 + (\frac{x}{2})^2 = (\frac{x}{2} + r')^2$

$x^2 -2xr' +r'^2 + \frac{x^2}{4} = \frac{x^2}{4} + xr' + r'^2 \implies x^2 = 3xr' \implies x = 3r'$

Hence, $r' = \frac{x}{3} \implies r' = \frac{2r}{3}$

$BD$ is the diagonal of the square $ABCD$ and $\angle CBD = \angle LBG = 45^\circ$. So, here we get that $\triangle GLB$ is an isosceles triangle.

Now, by the pythagorian theorem from $\triangle GLB$,

$GL^2 + LB^2 = GB^2 \implies GL^2 + GL^2 = 3^2 \implies 2GL^2 = 9 \implies GL = \frac{3}{\sqrt2}$

Next, $\triangle CFB \sim \triangle CGL$ and from both tne triangle it can be written that

$\frac{BC}{BF} = \frac{x}{y} = \frac{3y}{y} = 3$

and similarly,

$\frac{CL}{GL} = 3 \implies CL = 3 × \frac{3}{\sqrt2} \implies CL = \frac{9}{\sqrt2}$

From that, $CB = CL + LB = CL + GL = \frac{9}{\sqrt2} + \frac{3}{\sqrt2} = \frac{12}{\sqrt2} = 6\sqrt2$. So, the side of the square $ABCD$ = $x$ = $6\sqrt2$

After that,$\triangle CDE \sim \triangle HME$ and \triangle BAD \sim \triangle HMD$. From the similarity of first two triangles, we get

$\frac{CD}{HM} = \frac{DE}{ME}$

Likewise from the next similarity,

$\frac{AB}{HM} = \frac{AD}{MD} \implies \frac{CD}{HM} = \frac{CD}{MD}$.

So, we can write that

$\frac{DE}{ME} = \frac{CD}{MD}$

$\frac{r}{a} = \frac{2r}{r-a}$......(By denoting $ME = a$)

$2ra = r^2-ra \implies 6\sqrt2x = (3\sqrt2)^2 -3\sqrt2x \implies 6\sqrt2x = 18 - 3\sqrt2x = 9\sqrt2x = 18 \implies x = \frac{18}{9\sqrt2} \implies x = \sqrt2$

Then, $DM= 3\sqrt2 - a = 3\sqrt2 - \sqrt2 = 2\sqrt2$

Notice that, $\triangle DMH$ is an isosceles triangle and so,

$DH^2 = 2DM^2 = 2(2\sqrt2)^2 = 2×8 = 16$

And finally, we get $DH = \sqrt16 = 4$.

It could have been solved by any easier effort. But I made it very difficult. I hope that you will understand or if you have any problem, please let me know.