$1$ and vertices numbered clockwise from $1$ to $6.$ She draws segments connecting each pair of vertices that differ by $2$ or $4,$ then cuts the paper along these segments. She does the same with the newly found hexagon and keeps going. If the smallest area she can cut is $\frac{1}{1000000},$ how many hexagons will she cut in total?
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Have you tried drawing a diagram?
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What is the area of the initial hexagon?
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Try finding the side lengths of the new hexagon created.
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… by deducing the length of the cuts Alice makes.
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How does the ratio of side length between successive hexagons help you find the ratio of successive areas?
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How could you express “the smallest area she can cut is
$\frac{1}{1000000}$” as a mathematical inequality? -
… in terms of
$n,$where$n$is the number of cuts Alice makes.
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We first find the area of the initial hexagon to be
$\frac{3 \sqrt 3}{2}$(a diagram may be useful for visualisation).We then find the length of a cut that joins two vertices that differ by
$2$or$4$by any suitable method (e.g. using cosine rule). This gives the length of the cuts to be$\sqrt 3,$from which we can deduce that the side length of the next hexagon is$\frac{\sqrt 3}{3}=\frac{1}{\sqrt 3}.$Using this side length ratio, we deduce that the ratio of areas between successive hexagons is
$\frac{1}{3}.$Hence Alice will make$n$cuts, where$n$is the largest number of cuts such that$\frac{3\sqrt3}{2}\frac{1}{3^n} \geq 10^{-6}.$We can rearrange this to get$3^{\frac{3}{2}-n} \geq 2 \times 10^{-6}$or$n = \lfloor \frac{3}{2} + 6 \log_3 10 - \log_3 2\rfloor.$We can approximate$6 \log_3 10 \approx 12.1$and$\log_3 2 \approx 0.6$to get that$n=13.$