5
$\begingroup$

enter image description here

Is there any easy way to get the radius of the central radious?

$\endgroup$
3
$\begingroup$

This is a more general approach : We consider several layers of circles with $n$ circles per layer. The layers are numbered $ 1 , 2, 3, ... k, ...$. The small inner circle is not included ( there are $n$ circles in the layer number 1 ).

The next figure shows the notations and how the recurrence formula is derived, leading to the relationship between the radius of the circles of the k-ieme layer and the radius of the circles of the first layer.

enter image description here

Comming back to the case considered in the question raised by "Complex Guy", there are 2 layers. The radius of the circles in layer 2 is given. Applying the above formula leads to the formula for the radius of the small inner circle.

enter image description here

$\endgroup$
1
$\begingroup$

Let $(0,0)$ be the centre of the figure and $(a,0)$ the center of one of the radius $r$ circles and let the tiny circle be a unit circle. Let $(b,c)$ be the center of one of the meium sized circles touching the $x$-axis. Then by Pythagoras $$\tag1b^2+c^2=(c+1)^2$$ and $$\tag2(a-b)^2+c^2=(r+c)^2.$$ Moreover, by the symmetries of the figure $$\tag3r:a = c:(c+1) = \sin 36^\circ=\frac{\sqrt{2(5-\sqrt 5)}}{4}.$$ From $(3)$ we find $$c=\frac{\sqrt{2(5-\sqrt 5)}}{4-\sqrt{2(5-\sqrt 5)}},$$ then with $(1)$ $$ b=\sqrt{\frac{4+\sqrt{2(5-\sqrt 5)}}{4-\sqrt{2(5-\sqrt 5)}}}.$$ (I was too lazy to simplify these expressions). Using these values and $(3)$ to eliminate $a$, $(2)$ becomes a quadratic equation in $r$ (which looks a bit involved, so one should really spend a few thoughts on simplifying all the surd expressins first). Ultimately, we have computed $r$ as the ratio of the radii of the big circles and the small circle. Invert to compute the radius of the small circle if the big radius is given.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.