# Find the radius of the circle?

Two Circle of an equal of an radii are drawn , without any overlap , in a semicircle of radius 2 cm.
If these are the largest possible circles that the semicircle can accomodate , then

what is the radius of each of the circles?

Thanks in advance.

-
How would you fit two radius-$0.5$ circles inside a semicircle of radius 1? –  Henning Makholm Apr 3 '12 at 15:37
@Henning I solved it considering a circle.I am sorry. –  vikiiii Apr 3 '12 at 15:40
If you're allowed to use coordinates, then it becomes slightly easy... –  Ｊ. Ｍ. Apr 3 '12 at 15:46
@J.M. ya.we can solve using any of the methods. –  vikiiii Apr 3 '12 at 15:50

## 3 Answers

This is a difficult problem if you are expected to give a full formal argument. However, it is intuitively quite reasonable to assume that the "best" circles will be tangent to the diameter of the semicircle, tangent to each other, and tangent to the circle of radius $2$. Under the assumption that these are indeed the best circles, we will find the answer. (Alternately, we could look at the problem with a full circle instead of a semicircle, and four small circles, and assume that the symmetrical configuration is best.)

Draw a picture, and concentrate on the small circle $\mathcal{K}$ on the right. Let $O$ be the centre of the semicircle, and let $C$ be the centre of $\mathcal{K}$. Let $T$ be the point where $\mathcal{K}$ is tangent to the semicircle.

Note that the line $OT$ is perpendicular to the common tangent line at $T$. Note also that the line $CT$ is perpendicular to the common tangent line at $T$. It follows that the line through $O$ and $T$ is the same line as the line through $C$ and $T$. In other words, the radius $OT$ of the semicircle passes through $C$.

Let $r$ be the radius of $\mathcal{K}$. Drop a perpendicular from $C$ to the point $P$ on the diameter of the semicircle. Note that $OP=r$. By the Pythagorean Theorem, $(OC)^2=(PO)^2+(PC)^2=2r^2$, so $OC=\sqrt{2}r$. It follows that $$2=OT=OC+CT=\sqrt{2}r +r.$$ Now we are finished. We have $r=\frac{2}{\sqrt{2}+1}$. It may be that in the answer given, the denominator was rationalized by multiplying top and bottom by $\sqrt{2}-1$. That gives the answer $r=2(\sqrt{2}-1)$.

Added: Thanks to David Mitra for the diagram!

-
$\frac{2}{\sqrt{2}-1}\approx 4.828$. It should probably be $1+\sqrt{2}$ in the denominator. –  example Apr 3 '12 at 17:04
@example: Thanks for pointing out the typo. Fixed! My fingers were looking forward to the rationalization of the denominator. –  André Nicolas Apr 3 '12 at 17:08
Would you mind if I added a diagram to your answer? –  David Mitra Apr 3 '12 at 17:29
@DavidMitra: Not at all, and thanks in advance. But if one does handwaving, as I do, then probably one should go all the way and handwave to the $4$ circle version. So the argument I give is not really worth illustrating. –  André Nicolas Apr 3 '12 at 17:35
I had already drawn it, but you posted first; so I'll just add it to your answer. –  David Mitra Apr 3 '12 at 17:40

Due to symmetry two circles in a semicircle is the same problem as one in a quatercircle or four in a full circle. If we look at a quatercircle originating at the origin, with radius $r$ and completely contained in the first quadrant, then the circle has to be centered at a point $(c,c)$, touching the x axis, the y axis and the quatercircle at $(r/\sqrt{2},r/\sqrt{2})$.

To find out how large the circle may be, without crossing the x and y axes, consider the distance from the center of this circle to one axis being equal to the distance to the point $(r/\sqrt{2},r/\sqrt{2})$ $$c=\sqrt{2\left(\frac{r}{\sqrt{2}}-c\right)^2}=\sqrt{2} \left(\frac{r}{\sqrt{2}}-c\right)$$ $$\Rightarrow c = \frac{r}{1+\sqrt{2}}$$

For $r=2\,\text{cm}$, $c\approx 0.83\,\text{cm}$.

-

By construction, the maximum radius of two circles of equal size in a circle of radius r is 1/2r. The maximum radius of 4 circles of equal size in a circle of radius x is 1/4r.

-
Have you got the right diagram - I think I can get $\frac r {1+\sqrt2}$ for four circles –  Mark Bennet Apr 3 '12 at 16:06