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.
It took me a few days but I finally found the right approach.
Call the sides a,b,c and corresponding altitudes x,y,z.
Since V=(1/3)*pi*x^2*a etc we get the following equations:
ax^2=116320
by^2=142560
cz^2=110880
Using Heron's formula and A=bh/2 for the area of the triangle
(let n=(a+b+c)(-a+b+c)(a-b+c)(a+b-c))
A= √(n)/4 = ax/2
x= √(n)/(2a)
thus
ax^2 = n/(4a) = 166320
by^2 = n/(4b) = 142560
cz^2 = n/(4c) = 110880
So b = 166320a/142560 = (5/6)a
and c = 166320a/110880 = (3/2)a
The perimeter sought is then a+b+c = (11/3)a
To find a, turn to the area formulas again this time using the sides in terms of a:
A= √((11a/6)(7a/6)(2a/3)(a/3)) = a^2*√(77/162) = ax/2
thus
x=2a*√(77/162)
and
ax^2 = 4a^3*77/152 = a^3*154/81 = 232848
a^3 = 87480
a = ∛87480 = 18∛15 = 44.39
(Unfortunately not an integer)
perimeter = a+b+c = 11a/3 = 66∛15 = 162.78
for good measure A=1358.61
I'm very curious how Brian Smith picked his numbers.
Edit: fixed the error Brian pointed out, but something is still not right.
Edited on February 25, 2016, 11:13 am
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Posted by Jer
on 2016-02-23 11:06:29 |
Solution
