Five spherical balls of diameter $$10\text{ cm}$$ each fit inside a closed cylindrical tin with internal diameter $$16\text{ cm}$$. What is the smallest possible height of the tin can be?
Solution
Each ball has radius $$5$$, while its centre can be at most $$8-5=3$$ units from the cylinder axis. Two consecutive centres can therefore be separated horizontally by at most $$6$$, so their vertical separation is at least $$\sqrt{10^2-6^2}=8$$. Five centres require four such gaps, and adding the top and bottom radii gives $$5+4\mathbin{\times}8+5=42$$.
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