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Consider the "cylindrical cone" consisting of a cone of radius R and height h surmounting a cylinder of the same radius and height. There is one such polytope that is equable. The total volume of the cone + cylinder is V, where
V/π = (1/3)R2h + R2h = (4/3)R2hThe lateral surface area of the cone is Scone, where
The lateral surface area of the cylinder is Scyl, where![]()
The surface area of the cylinder base is Sbase, where
Scyl/π = 2Rh [1]
The above four quantities are positive integers by convention. We have
Sbase/π = R2 [2]
so R is rational. By [2], R2 is an integer, so it follows that R is an integer.![]()
The equability assumption is
Any equable solution will have a positive left side. Squaring both sides, we have![]()
If R = 1, then h = 12/5 and Scyl/π = 24/5 is not an integer. If R = 2, then h < 0. Therefore, R ≥ 3 and
(16R2 - 48R + 27)h = 24R2 - 36R [3]
so both the right side and the coefficient of h in [3] are strictly positive, and a rational solution for h is![]()
From [1],
![]()
[4]
so![]()
for some integer k. We previously established that the denominator is positive, and the numerator is also positive since R ≥ 3, so k is positive. The preceding equation can be written as a quadratic in R:![]()
Since R is an integer, the discriminant of this quadratic must be a square:
(16k - 72)R2 - (48k - 81)R + 27k = 0 [5]
576k2 + 6561 = m2 for some integer mThe possibilities are summarized below:(m - 24k)(m + 24k) = 6561
The only solution is k = 15, and thus [5] can be written
m - 24k < m + 24k Sum = 2m m k = [(m + 24k) - m]/24 1 < 6561 6562 3281 (6561 - 3281)/24 is not an integer 3 < 2187 2190 1095 (2187 - 1095)/24 is not an integer 9 < 729 738 369 (729 - 369)/24 = 15 27 < 243 270 135 (243 - 135)/24 is not an integer
168R2 - 639R + 405 = 0,which has the sole integer solution R = 3. From [4], we have h = 4 and
V = 48 πThe equable cylindrical cone is a solid of revolution of a non-equable quadrilateral consisting of aScone = 15 π
Scyl = 24 π
Sbase = 9 π
Scone2 = Sbase2 + (Scyl/2)2Therefore, any Pythagorean triple a2 + b2 = c2, for which
generates the non-equable solution
![]()
[6]
For any Pythagorean triple, an alternate solution with a and b exchanged is possible as long as [6] is satisfied. Applying the limits on a and b, we have![]()
so![]()
h2 > 9/16 + 81/624 ≈ 0.6923Any element of a Pythagorean triple is at least 3, so R2 = a ≥ 3 and
which reduces to![]()
V > SconeTherefore, all solutions for which Scone ≤ M for some maximum M will include as a subset all solutions with V ≤ M. The 23 solutions having volume less than 2000 π are listed below:
R h Scone/π Sbase/π Scyl/π V/π 2 3/2 5 4 6 8 3 4 15 9 24 48 * 4 3 20 16 24 64 6 5/2 39 36 30 120 3 40/3 41 9 80 160 4 15/2 34 16 60 160 6 9/2 45 36 54 216 12 17/12 145 144 34 272 4 63/4 65 16 126 336 6 8 60 36 96 384 5 12 65 25 120 400 8 6 80 64 96 512 6 77/6 85 36 154 616 12 7/2 150 144 84 672 6 35/2 111 36 210 840 12 5 156 144 120 960 10 15/2 125 100 150 1000 6 80/3 164 36 320 1280 8 15 136 64 240 1280 9 12 135 81 216 1296 10 21/2 145 100 210 1400 7 24 175 49 336 1568 12 9 180 144 216 1728
*: Equable
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