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Solutions: 5

V = π rBy convention, the volume and the circular base area are integer multiples of π, so r^{2}hS = 2π rh + 2π r

^{2}

By equability,

rA right cylinder is a right prism with a circular base, so h > 2 as is established elsewhere. It follows that r > 2 in any equable solution. We then have^{2}h = 2rh + 2r^{2}

(r - 2)(h - 2) = 4 [1]

and

is an integer. Then r - 2 divides 16. Therefore, r ∈ {3,4,6,10,18} and we have the following five equable solutions:

It follows from [1] that rh = 2(r + h), so every equable cylinder is created from a solid of rotation of an equable

r h r ^{2}2rh V/π=S/π 3 6 9 36 54 4 4 16 32 64 6 3 36 36 108 10 5/2 100 50 250 18 9/4 324 81 729

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