In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.

In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $$2$$. Find the radius of the circle.

I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $$\frac{1}{4}$$, but the correct answer was $$\frac{2}{9}$$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.

• $2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius? May 6, 2019 at 11:41
• @HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222\ldots$ is in that ballpark.
– Blue
May 6, 2019 at 12:00
• @Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$. May 6, 2019 at 13:05
• Sounds like the start of a joke... May 7, 2019 at 12:43

3 Answers

Look at the picture:

From $$\triangle ABE$$ we have $$(2+r)^2= 2^2+(2-r)^2$$ so $$r=1/2$$. From $$\square ECGF$$ we have $$CG^2=(1/2+s)^2-(1/2-s)^2= 2s$$. From $$\square ADGF$$ we have $$GD^2= (2+s)^2-(2-s)^2= 8s$$. So $$2=CG+GD=3\sqrt 2\sqrt s$$, hence $$s=2/9$$.

• It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius. May 6, 2019 at 16:10
• Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me. May 6, 2019 at 16:26
• @Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $\triangle FHE$ to find $FH$. And the same trick for $GD$.
– SMM
May 6, 2019 at 16:37
• @BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
– SMM
May 6, 2019 at 16:46
• @SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though. May 6, 2019 at 21:17

@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.

Let the quarter-, semi-, and full-circles have radius $$a$$, $$b$$, $$c$$, respectively.

From the right triangle, we have $$a^2+(a-b)^2=(a+b)^2 \quad\to\quad a=4b \tag{1}$$

Considering the side of the square a circle of curvature $$0$$, that special case of the Kissing Circles theorem implies $$\frac1{c} = \frac{1}{a}+\frac{1}{b}\pm 2\sqrt{\frac{1}{a}\cdot\frac{1}{b}} = \frac{5}{4b}\pm 2\sqrt{\frac{1}{4b^2}} = \frac{5\pm 4}{4b}\quad\to\quad c = \frac49 b \;\text{or}\; 4b\;\text{(extraneous}) \tag{2}$$

Then, with $$a=2$$, we have $$b=1/2$$, so that $$c=2/9$$. $$\square$$

let the side of the square be $$a$$.

Let's find the radius x of the semicircle

We have $$(a+x)^2 = a^2 + (a-x)^2$$ $$x=\frac{a}{4}$$

Now, a lemma.

If circles of radiuses R and r are touching externally, then the length of their common tangent is $$2\sqrt{Rr}$$

Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $$(R+r)^2 = h^2 + (R-r)^2$$, from where $$h = 2 \sqrt{Rr}$$.

Now, let's use the lemma. Let $$y$$ be the radius of the small circle. We have $$a = 2\sqrt{ay} + 2 \sqrt{\frac{a}{4}y}$$ $$\sqrt{a} = 3 \sqrt{y}$$ $$y = \frac{a}{9}$$

• I like this one because it taught me something useful. May 7, 2019 at 19:03