2011/12 AHSMC Part 1 Problem 11

Problem 11 of 16

$ABCD$ is a quadrilateral with $AD = BC$ and $AB$ parallel to $DC$ . It is only given that the lengths of $AB$ and $DC$ are $20$ and $15$ centimetres respectively. Adrian puts $n$ copies of this tile together so that the edge $BC$ of each copy coincides with the edge $AD$ of the next, and the edges $DC$ of all copies together form a regular $n$ -sided polygon. Then the value of $n$ is

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Solution

We can draw a regular polygon of any number of sides such that the side length is $20$ centimetres. We can then draw a regular polygon of the same number of sides but with side length $15$ centimetres, placed centrally inside the larger polygon. Then a tile can be chosen which can pave the ring-shaped region inside the larger polygon but outside the smaller one. Hence the answer is (E).

University of Alberta

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