Problem of the Week

Problem 21 Solution

Congratulations to Keo Osman from Simon Balle School in Hertford, Bruno Cerda-Mendoza from Sawston Village College, Rian Sarkar from Highgate School for successfully solving this problem!
(i)
Consider the triangle $OC_1C_2$ , where $C_1$ and $C_2$ are the centres of two adjacent smaller circles.

Let angle $C_2 O C_1 = \theta$ . Since similar triangles to $OC_1C_2$ can be made for each pair of adjacent circles, there are $n$ of these triangles, all with an angle $\theta$ .

Thus, $\theta = \frac{360}{n}$ . Taking one of these triangles we can split it into two right-angled triangles (as shown below).

Now considering triangle $C_1OM$ (where $M$ is the midpoint of $C_1$ and $C_2$ ), we see that $C_1O = (r + 1)$ , as the radius of $A$ plus the radius of one of the smaller circles.

$C_1M = 1$ , as the radius of one of the smaller circles. Let angle $C_1OM$ (and by symmetry $MOC_2$ ) = $\alpha$ .

From the previous conditions, we get that $\sin(\alpha) = \frac{1}{r+1} \implies \alpha = \arcsin(\frac{1}{r+1})$ . From our previous condition of $\theta = \frac{360}{n}$ , we also get $\frac{360}{n} = 2\alpha \implies n = \frac{360}{2\alpha} = \frac{180}{\alpha}$ .
$\implies n = \frac{180}{\arcsin(\frac{1}{r+1})}$ .
$\arcsin(\frac{1}{r+1}) = \frac{180}{n} \implies \sin(\frac{180}{n}) = \frac{1}{r+1}$
$\implies r = \frac{1}{\sin(\frac{180}{n})} - 1$ .

Note that, there were other valid formulas that we accepted, but this is the official solution.

(ii)
Circle B has radius $r + 2$ , and thus area $\pi(r+2)^2$ .
The area of each individual smaller circle is $\pi$ , thus their combined area is $n\pi$ .

It follows that $F = \frac{n}{(r+2)^2} \implies r = \sqrt{\frac{n}{F}} - 2$

Plugging in our equation for $n$ found in part (i) $n = \frac{180}{\arcsin(\frac{1}{r+1})}$

$r_{i+1} = \sqrt{\frac{\frac{180}{\arcsin\left(\frac{1}{r_i + 1}\right)}}{F}} - 2$

Starting with $r=15$ (the value of r is irrelevant here), after iterating many times you get $r \approx 2.863603305$

Plugging this back into our previous equation of $n = F (r+2)^2$ yields $n=12$ (if you iterate enough to find a good approximation of r).

Problem of the Week

Problem 21 Solution

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