[Monte314]

For a right circular cone to completely enclose a sphere, its volume must be at least X times the volume of the sphere:

....../\............................/\
For /O\ to happen, volume /..\ >= X times volume O

What is X?

(This problem can be done without the use of calculus)

[HIDE="Artistic rendering that puts Rembrandt to shame"]

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[/HIDE]

[themuzicman]

( Show Spoiler )

OK, the key is calculating the radius of the sphere that touches both the point and the base of the cone. This problem can be reduced to the 2d equivalent of a triangle in a circle.

If we assume a circle with radius (R) of 2, a triangle whose base intersects the center of the circle will have a base (B) of 4, and height (H) of 2. A theoretical triangle with a B = 0 will have H = 4.

Given these two points (and a rash assumption), the radius, then, should be:

(1/2B + H)/2 (Average length of the base and height)

And the area of the circle:

A = ((1/2B + H)i/2)^2 * pi.

Blowing this out to the cone and sphere, then, given the diameter of the base of the cone as D, height of cone as H, we have:

V = 4/3 * ((1/2D + H)/2)^3 * pi.

If my rash assumption is incorrect, then we'll have to dive into trigonometry...

Which I have neither the time nor patience to do ATM, although my instincts are telling me that I am to use the Cosine of one and the Sine of the other...

[Malle]

I have a long solution which does make use of calculus

( Show Spoiler )

Place the sphere of radius r such that it tangents the xz-plane
x² + (y-r)² = r²
Since the problem is rotationally symmetric, study the xy-plane. Place the cone (a triangle in the xy-plane) around the sphere, such that the cone has its tip at x=0, y=h and its width at it's base, where y=0, is 2b.

For this we can write the volume of the two bodies as
V_s = (4/3)(πr³)
V_c = (1/3)(πb²h)
V_r = V_c / V_s = (1/4)(b²h/r³)

Since the font isn't very good, it should be noted that π = pi

We then study all possible such triangles. We do this by studying the tangent lines to the circle for 0 < x < r.

It's straightforward to find the slope of the tangent line that tangents in the point x0:
(dy/dx)|(x=x0) = -x0 / √(r^2-x0^2)
where (f)|(x=x0) is f evaluated for x = x0

Using the equation
x² + (y-r)² = r²
and the equation for a straight line
y = (dy/dx)|(x=x0) * x + h
it's possible to solve for h which gives
h = r + r²/√(r²-x0²)

Finding where the tangent line intersects the x-axis (or xz-plane) we find our b as
b = (r² + r√(r²-x0²))/x0

Proceed to solve the equation
(d/dx0) ((1/4)(b²h/r³)) = 0
which I had to use Wolfram Alpha for, or I'd make plenty of errors and take a long time to solve it, we can find the simple solution that
x0 = r(2√2)/3

Substituting this into the equation for V_r all dependencies on r are eliminated and we end up with
V_r = 2

Thus, a cone that circumscribes a sphere must have a volume that is at least twice as large as the sphere's volume.

I'll keep thinking on a solution that does not require calculus, because this solution is not pretty.

[themuzicman]

( Show Spoiler )

D'oh...

From the 2d version (Flatten out the cone and sphere into triangle and circle), we can form a triangle using the difference between the radius (r) and the height (cone=h) of the triangle as height (ht) and half the base (cone=b) of the triangle (bt).

(a picture would be great here, but alas... if someone can send me an email, I can send it to you so it can be uploaded to an image host and i could refer to that address... can't upload images here at work)

Thus:

ht = |h-r| (doesn't matter at this point whether the height is greater or less than, I think)
bt = 1/2 b

The hypotenuse is the radius of the circle.

Thus (using an old friend):

r^2 = bt^2 + ht^2

or

r^2 = (1/2 b)^2 + (h - r)^2

r^2 = 1/4 *b^2 + h^2 -2h * r + r^2 (uh oh!)

0 = 1/4 b^2 + h^2 - 2hr

2hr = 1/4 b^2 + h^2

r = b^2/8h + h^2/2h (combining 1/4 with 1/2h while dividing by 2h)

r = b^2/8h + h/2

Thus, the area of the sphere is:

V= 4/3 (b^2/8h + h/2) * pi^3

(maybe?)

[BingeArtist]

No calculus.

( Show Spoiler )

Answer: X = 2.

Solution: We assume the sphere is radius 1. Let h and r be the height and radius of the cone. What we want to do is minimize r^2*h.

In the plain (first quadrant..I think..with both x & y nonnegative), we can represent the sphere by the equation x^2 + (y-1)^2 = 1, and we can represent the cone by the line y = -(h/r)x + h.

Let P be the point where the line (tangently) intersects the circle. From similar triangles, we have D(P, (0,h)) = h/r. Applying the Pythagorean Theorem to the right triangle defined by P, (0,h), and (0,1), we have (h/r)^2 + 1 = (h-1)^2. Hence, r^2 = h/(h-2). So, since we're trying to minimize r^2*h, we're trying to minimize h^2/(h-2).

Here's where we have to use a few tricks to avoid the dreaded calculus. To minimize h^2/(h-2), we want to maximize (h-2)/h^2 = 1/h - 2/(h^2). Set m = 1/h. Hence, we're trying to maximize m - 2m^2. And this is just a parabola with a vertex at (1/4, 1/8). So our minimum value of r^2*h is 8, and occurs when h = 4 (and hence, when r^2 = 2).

So the volume of the cone is 8 pi over 3, and the volume of the sphere is 4 pi over 3. So the ratio is 2:1. ###

[K27]

[HIDE="spoiler"]Let the radius of the base of the cone be b, height of the cone be h and radius of the sphere be b.
Looking at the intersection of vertical half of the cone and the sphere, and drawing a line from the centre of the sphere to the touching point of the sphere to the cone, we obtain two similar triangle.

By similar triangle, we get:
r / (h - r) = b / sqr(h^2 - b^2)

Expend and do some arithmetic, we get:
b^2 = (r^2)h / (h - 2r) ..... (1)

Substitute into the volume of cone, we get:
(1/3)(pi)(r^2)[h^2/(h-2r)]

We have to minimise h^2/(h-2r). I initially used calculus, but after peeking into BingeArtist's solution I did the same thing... Maximise (h - 2r) / h^2 and take x = 1/h... Then got h = 4r in my solution.

Then sub h = 4r into (1) we get b^2 = 2r^2.

So the volume of the cone is:
(1/3)(pi)(b^2)h
= (1/3)(pi)(2r^2)(4r)
= (8/3)(pi)(r^3)
= 2(4/3)(pi)(r^3)

Therefore, X=2![/HIDE]

I think BingeArtist's way of avoiding calculus is brilliant!

[themuzicman]

( Show Spoiler )

to finish:

V(s) = 4/3 (b^2/8h + h/2) * pi^3 volume of Sphere
V(c) = 1/3 pi * b^2 * h

So, the ratio (X) is:

X=V(s)/V(c)

or

X= [4/3 (b^2/8h + h/2)^3 * pi] / [1/3 pi * b^2 * h]

(reducing a bit)

X=[4(b^2/8h + h/2)^3 ] / [b^2 * h]

[envirodude]

I got the answer, 2, by drawing a section through the centre of the cone&sphere, and defining alpha as the base angle of the isoceles triangle formed. Simple trig and similar triangles leads to the equation for the volume ratio as:

X = (cos alpha +1)³ / ( sin² alpha · cos alpha)

a bit of thinking leads to the constraint: 0 < alpha < 90°

To minimise X, I used Solver in Excel, and confirmed with Wolfram Alpha [minimize (cos a +1)^3/(sin a)^2/cos a]. That's not calculus, is it?

alpha = 70.5287°; X = 2

[Monte314]

Nice use of automated tools!

[rufsketch1]

Oh that's easy. The ratio of the cone's volume to that of the sphere is V/(4r³/pi) : 1

Where V is the lowest point between 0 and 1.57 on the function

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---------- Post added 03-02-2011 at 11:32 PM ----------

Or as you earthlings would say: "2".

[Monte314]

Wow; it's obvious when you put it that way!

*Dog begins hyperventilating*