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Solutions: 1

V/π = (1/3)RThe lateral surface area of the cone is S^{2}h + R^{2}h = (4/3)R^{2}h

The lateral surface area of the cylinder is S

The surface area of the cylinder base is S

S _{cyl}/π = 2Rh[1]

The above four quantities are positive integers by convention. We have

S _{base}/π = R^{2}[2]

so R is rational. By [2], R

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 S

(16R ^{2}- 48R + 27)h = 24R^{2}- 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)R ^{2}- (48k - 81)R + 27k = 0[5]

576kThe possibilities are summarized below:^{2}+ 6561 = m^{2}for some integer m(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]/241 < 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

168Rwhich has the sole integer solution R = 3. From [4], we have h = 4 and^{2}- 639R + 405 = 0,

V = 48 πThe equable cylindrical cone is a solid of revolution of a right trapezoid with sides 3, 4, 5, 8 and diagonal 5, rotated around its longest side and having perimeter 20 and area 18.S

_{cone}= 15 πS

_{cyl}= 24 πS

_{base}= 9 π

STherefore, any Pythagorean triple a_{cone}^{2}= S_{base}^{2}+ (S_{cyl}/2)^{2}

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

hAny element of a Pythagorean triple is at least 3, so R^{2}> 9/16 + 81/624 ≈ 0.6923

which reduces to

V > STherefore, all solutions for which S_{cone}

R h S _{cone}/πS _{base}/πS _{cyl}/πV/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

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