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Three identical circles are placed such that they touch each other. A larger circle is drawn around the smaller circles such that it touches them, as shown in the diagram. Can you find the ratio between the radii of the smaller circles and the larger circle?

enter image description here

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The ratio is

$\frac{2+\sqrt{3}}{\sqrt{3}}$

Proof

Let the radius of the small circles be $r$. For any two circles which are tangent to each other, the centres of the circles and the tangent point are collinear. This is easily seen by drawing the tangent line and realising that it is perpendicular to both radii. With that in mind, if we join each of the centres of the small circles we obtain an equilateral triangle with side length $2r$.
Then the radius of the large circle is given by the circumradius of the equilateral triangle plus the radius of the small circle (because, as before, the centres and the tangent point are collinear), see diagram below
enter image description here
The circumradius of an equilateral triangle is just the side length divided by $\sqrt{3}$ which in this case is $\frac{2r}{\sqrt{3}}$ so that the radius of the large circle is $\frac{2r}{\sqrt{3}} + r = \frac{2+\sqrt{3}}{\sqrt{3}} r$, whence the resuly

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I'll answer this in the form of a poem:

Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.

Just follow the instructions, and you'll have the answer.

All credit to:

Frederick Soddy who wrote up this proof in the form of a poem, published in 1936 in Nature magazine.

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    $\begingroup$ Wow I was not aware of this. Nice! $\endgroup$ – Dmitry Kamenetsky Jan 19 at 10:45
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    $\begingroup$ It's funny that I used 0:19:01 and hexomino used 0:19:02 to provide an answer to this puzzle. :) Pretty tight race... $\endgroup$ – CG. Jan 20 at 10:21

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