# Finding lengths when circles and squares tangents. [closed]

Should one approach by coordinates or by euclidean geometry?

By pure geometry, I am not able to solve.

• You can find the ratio between the radius of the large semicircle and the radius of the small circle by applying the Pythagorean theorem to triangle $AEF$. After that, similar triangles will finish the job. Feb 28, 2019 at 7:23
• at corner $B$ construct a square with diagonal $BG=3$. Use similar triangles to show the side of this square $= a/2$ where $AB=2a$. This shows $a=3\sqrt{2}$. Similarly at corner $D$ to get the value of $x=2a\left.\sqrt{2}\right/3$ Feb 28, 2019 at 9:50

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 $$\frac{3\cdot2\sqrt2/3}{\sqrt2/2}=4$$

• I solved it by another way, but got the same answer. Feb 28, 2019 at 8:06
• @Anirban Niloy Yes, of course. Post it. I don't want to post my solution because it still is very ugly. I used a trigonometry. I see some nice fact, but I still don't see how to use it. Feb 28, 2019 at 8:22
• @MichaelRozenberg Thank you for your opinion. Same case of mine. I used a lot of trigonometry but at last reached to conclusion. But my solution is too broad and I think it won't be acceptable in some extent. Feb 28, 2019 at 9:39
• @Anirban Niloy We can prove that $\measuredangle ECF=45^{\circ}$ and $HG^2=DH^2+BG^2,$ but I don't see how to use it for a simple solution. Feb 28, 2019 at 12:25

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.