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Solutions: 1 (rational height)
Proof. For an equable bicone with radius r and overall height h, we have
which reduces to
for an equable solution. By convention, W = V/π = r^{2}h/3 is a positive integer. Then r^{2} = 3W/h > 9, so h < W/3 and [1] can be written
(r^{2} - 9)h^{2} = 36r^{2}, r > 3 [1]
(3W/h - 9)h^{2} = 36(3W/h)The equable solutions for h are the zeroes of the cubic3h^{3} - Wh^{2} + 36W = 0, 0 < h < W/3
f_{W}(h) = 3h^{3} - Wh^{2} + 36WSince
d/dh f_{W}(h) = 9h^{2} - 2Whand
d^{2}/dh^{2} f_{W}(h) = 18h - 2W,f_{W} has critical points at h = 0 and h = 2W/9 > 0. The first is a relative maximum and the second a relative minimum, and f_{W}(0) = 36W > 0. Therefore, there are two positive real zeroes when
In conclusion, there are infinitely-many equable bicones with real dimensions, and volumes that are integers of at least 47, times π.
and any equable solution requires that h > 6. It follows that
[2]
We will show that h must be an integer. Let the height h = N/D for positive integers N and D. Since h > 6,
[3]
and we can write [3] as
D < N/6, [4]
Each factor in the denominator divides the numerator. Since D < N by [4], then if D divides N^{3}, it also divides N and h is an integer. Otherwise, D divides 3. If D = 1, then h is again an integer, so assume D = 3. Then [5] becomes
[5]
Since h = N/3 > 6, we have N > 18 and the fraction on the right is a positive integer, so Each factor of the denominator in [6] divides the numerator (this is a necessary but not sufficient condition for W to be an integer). By long division, so both N - 18 and N + 18 are factors of 5832, subject to 19 ≤ N ≤ 324:
[6]
N ∈ {19,20,21,22,24,26,27,30,36,42,45,54,72,90,99,126,180,234,261} ⋂ {36,54,63,90,144,198,225,306} = {36,54,90}Evaluating in [3], the only solution for which W is an integer is h = 12. From [2], the corresponding value for r^{2} is so r is irrational. In this case, the volume and surface area are 48π.h = N/3 ∈ {12,18,30}
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