This problem is a generalization of
Hypotenuse from Cone.
An acute triangle is rotated about each of its three sides to produce three double-cones of differing volumes.
The three double-cones have volumes of 55440*pi cm^3, 47520*pi cm^3, and 36960*pi cm^3.
Determine the perimeter of the triangle.
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Submitted by Brian Smith
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Solution:
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Let a, b, and c be the sides the triangle. Let Va, Vb, and Vc be the volumes produced by rotating the triangle about sides a, b, and c respectively. Let A be the area of the triangle. Then the altitudes of the triangle are 2A/a, 2A/b and 2A/c
Because the two halves of each double cone are coaxial, the volume of the double-cone uses the same formula as a regular cone: V=(pi/3)*r^2*h, where h is the combined height of the double cone and r is the radius where the two halves meet.
For each of the double-cones h is equal to the side rotated about {a, b, c} and r is the altitude perpendictular to that side {2A/a, 2A/b, 2A/c}.
Then Va, Vb, and Vc can be expressed as:
Va = 55440*pi = (pi/3)*(2A/a)^2*a = (4*pi*A^2)/a
Vb = 47520*pi = (pi/3)*(2A/b)^2*b = (4*pi*A^2)/b
Vc = 36960*pi = (pi/3)*(2A/c)^2*c = (4*pi*A^2)/c
Take ratios Va/Vb and Vb/Vc:
Va/Vb = (55440*pi)/(47520*pi) = [(4*pi*A^2)/a]/[(4*pi*A^2)/b]
Va/Vb = 7/6 = b/a
Vb/Vc = (47520*pi)/(36960*pi) = [(4*pi*A^2)/b]/[(4*pi*A^2)/c]
Vb/Vc = 9/7 = c/b
From the simplified ratios there must be a value k so that the sides equal a=6k, b=7k, and c=9k; the perimeter is then 22k. By Heron's formula A = sqrt[11k*5k*4k*2k] = k^2*sqrt(440).
Plugging this back into Va:
55440*pi = (4*pi*A^2)/a
55440*pi = (4*pi*[k^2*sqrt(440)]^2)/(6k)
55440*pi = (880*pi/3)*k^3
k = 3*cbrt(7)
From this the answer is determined: the perimeter is 66*cbrt(7) cm. The triangle's three sides have lengths of 18*cbrt(7) cm, 21*cbrt(7) cm, and 27*cbrt(7) cm. |
