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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.

enter image description here

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.

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  • $\begingroup$ $2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius? $\endgroup$ – hmakholm left over Monica May 6 '19 at 11:41
  • $\begingroup$ @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. $\endgroup$ – Blue May 6 '19 at 12:00
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    $\begingroup$ @Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$. $\endgroup$ – hmakholm left over Monica May 6 '19 at 13:05
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    $\begingroup$ Sounds like the start of a joke... $\endgroup$ – Asaf Karagila May 7 '19 at 12:43
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Look at the picture:

enter image description here

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 $s=2/9$.

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    $\begingroup$ It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius. $\endgroup$ – BlueRaja - Danny Pflughoeft May 6 '19 at 16:10
  • $\begingroup$ 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. $\endgroup$ – Hammerite May 6 '19 at 16:26
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    $\begingroup$ @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 $GD$. $\endgroup$ – SMM May 6 '19 at 16:37
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    $\begingroup$ @BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square. $\endgroup$ – SMM May 6 '19 at 16:46
  • $\begingroup$ @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. $\endgroup$ – BlueRaja - Danny Pflughoeft May 6 '19 at 21:17
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@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.

enter image description here

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 $c=2/9$. $\square$

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let the side of the square be $a$.

Let's find the radius x of the semicircle

enter image description here

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 $2\sqrt{Rr}$

enter image description here

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 $$a = 2\sqrt{ay} + 2 \sqrt{\frac{a}{4}y}$$ $$\sqrt{a} = 3 \sqrt{y} $$ $$y = \frac{a}{9}$$

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  • $\begingroup$ I like this one because it taught me something useful. $\endgroup$ – richard1941 May 7 '19 at 19:03

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