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This one's a little tough. Try it out.

A cube with sides of length L rests on a table. A hollow right cone without a bottom, of height 3 and radius 1, is placed over the cube so that each of the top four corners of the cube just touch the interior rim of the cone when the bottom circumference of the cone rests level upon the table (so that the cube is completely concealed). Find L.

By Seeker (Seeker) on Saturday, March 22, 2003 - 06:51 am: Edit

Is the answer L = 6/(3rt2 + 2)? Seemed pretty simple to me...

By Quarky (Quarky) on Saturday, March 22, 2003 - 10:55 am: Edit

OK... this is a good prob. First of all, many people will simply draw a 2D diagram where there is a triangle (cone) and a square (cube) inscribed in it, so that its upper corners touch the sides of the triangle. That's srong because then the cube would "stick" out of the cone (at least, its corners would). What you have to do to obtain a similar picture, but to follow the condition that the cube is completely cocealed, is to cut the objects by a diagonal plane. The way I do this is look at the top view, draw a diagonal of the cube, and bisect the figures with a 2D plane lying along this diagonal. So the picture you get in a 2D side view is an isosceles triangle and a RECTANGLE (because the length of it is greater than the hight due to the section we just did). The length of the rectangle = the length of the cube's side's diagonal. It is equal to the sqrt(2*L^2). = sqrt(2)L. (pythag thm).

Now, let's work with the triangle. Its right-hand side has a slope of -3 because we know that h=3 and r=1. Furthermore, we know that point (0,3) belong to the side, so the equation of the line representing this side is y=3-3x. Somewhere on this line, lies a point at which the rectanlge's corner meets the side. At that point, we know that the height is simply L, or y (TWO x because the y=axis bisects the rectangle into two parts vertically). The horizontal length of the rectangle is 2*x and is equal to sqrt(2)L. So x=sqrt(2)L/2


y=3-3x
L=3-sqrt(2)/2*L

L = 6-3sqrt(2) = 1.76

Is this right?

By Seeker (Seeker) on Saturday, March 22, 2003 - 12:30 pm: Edit

Augh, Quarky's right. Sigh I'm so stupid.

By Pat57575 (Pat57575) on Saturday, March 22, 2003 - 12:32 pm: Edit

Hmm, well I saw a problem similar to this somewhere, but I just made this one up using my own numbers. When I worked the question and rationalized the answer, I got (18sqrt2-12)/14, which I think is seeker's answer rearranged (it comes out to .9611 I think).

Quarky, I'm not sure where your method went wrong... it looked good. (though I worked it completely differently)

Correction: Unless your name is seeker, you might find this problem difficult.

By Pat57575 (Pat57575) on Saturday, March 22, 2003 - 12:37 pm: Edit

No, I don't think that's possible. Because if 1.76 was L, then the diagonal of one of the cube faces would be 1.76*sqrt2= 2.49. Since the diameter of the base of the cone is only 2, a cube of these dimensions wouldn't fit.

By Brd (Brd) on Saturday, March 22, 2003 - 01:30 pm: Edit

Imagine x-axis running through two of the four corners of the base of cube, and the y-axis running vertically through the center of the cube and the vertex of the cone. Now then, we have three equations in 3 unknowns:

* First, the possible acceptable values of x and y are constrained by the equation for the line that bounds the cone:

y == 3 - 3x

* Further, we are concerned with what happens when a top corner of the cube just touches the side of the cone. This constraint is expresed by:

y == L

* Finally, we can deduce what the x value is when the top corner just touches the cone. Since the x axis is a diagonal of the top face, the x value is just half the length of the diagonal (which depends on L), i.e.:

x == (Sqrt[2]*L)/2

Solving for L we find that

L == 6 / (2 + 3*Sqrt[2]) ~= 0.961132


The thing to realize is that the calculation has to be invariant with respect to any particular coordinate system. So in that case, you get to pick the coordinate axes to whatever you want, so best to pick a system that makes the problem easier. For instance, you could also pick the x-axis to run perpendicular to the center of two of the faces of the base of the cube -- but then the problem is a little harder to solve.

By Pat57575 (Pat57575) on Saturday, March 22, 2003 - 02:17 pm: Edit

Though the x/y coordinate system method looks much more elegant than my solution, I guess I will share anyways.

Let Z be the angle formed between two opposite sides of the circular base at the vertex. With that, you know:

tan (Z/2) = r/h (radius divided by height)
tan (Z/2) = 1/3

You also know:

tan (Z/2) = ([{L*sqrt2}/2]/[3-L]) (half of the diaganol of the cube divided by the vertical distance between the top face of the cube and the vertex of the cone)

Setting the equations equal, you get:

1/3 = ([{L*sqrt2}/2]/[3-L])

L = 6/(2+3sqrt2)

By Quarky (Quarky) on Saturday, March 22, 2003 - 10:52 pm: Edit

I feel stupid!

By Yeah10 (Yeah10) on Sunday, March 23, 2003 - 12:36 am: Edit

you guys are fags

By Quarky (Quarky) on Sunday, March 23, 2003 - 12:09 pm: Edit

Actually, I do not feel stupid anymore. Two reasons:

1) Yeah10 is even more stupid and incorrect
2) The only reason my final answer came out to be incorrect is:


"y=3-3x
L=3-sqrt(2)/2*L

L = 6-3sqrt(2) = 1.76

Is this right?"

As you can see, L=3-sqrt(2)/2*L should have been L=3-3*sqrt(2)/2*L, as the equation y=3-3x tells me to do. I just forgot the three, that's all.

By Pat57575 (Pat57575) on Sunday, March 23, 2003 - 03:06 pm: Edit

Yeah10, some people find satisfaction in putting their brains to the test. Though I am confident you are completely unfamiliar with a ritual as outlandish to your kind as such, I suggest you start with some multipication flash cards. Remember, persistence always pays off.

By Incognito (Incognito) on Sunday, March 23, 2003 - 03:28 pm: Edit

To Yeah10 :
Why are you calling them fags? Why are you even getting upset? They're just working on a math problem, and then you come along and make an obnoxious comment. What's your problem?

By Brd (Brd) on Sunday, March 23, 2003 - 04:14 pm: Edit

Well, I can give credit where credit is due: he (or she) is not *completely* wrong -- I am gay, after all -- but I think I'd chalk that up to dumb luck rather than any prescient observational prowess. And certainly zero points for originality, at any rate.

BTW Pat, I think your calculation based on a similar traingles argument is quite clever. It's also good to show that there is not just one right way of doing things.

By Yeah10 (Yeah10) on Friday, April 18, 2003 - 02:59 am: Edit

aaahahaha what you guys are trying to put me down by telling me how stupid i am and you dont even know me? im just telling it like it is. you assume im some kind of idiot because i dont enjoy putting my brains to the test, please..