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Icarus
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Three Grapefruit and an Orange
« on: Oct 29th, 2003, 7:16pm » |
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Three grapefruit, each spherical with a 3 inch radius, are sitting on a counter so that they touch. Also sitting on the counter in the middle of the grapefruit is a spherical orange. If the orange is in contact with all three grapefruit, what is its radius?
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Speaker
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Re: Three Grapefruit and an Orange
« Reply #1 on: Oct 29th, 2003, 9:44pm » |
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Well, just to give it a shot, I drew three grapefruit size circles and an orange size circle in a word document which gave me an answer. Then, because this wasn't the mathamatical way (geometric way) which I don't know how to do and probably what Icarus is looking for, I wasn't going to put in my answer. But, then I realized that the question has more dimensions than I first thought, so now I have no answer, except the first one.
Now I really don't know how to do it. 
But, I am guessing that the two dimensional size I found is too small.
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towr
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Re: Three Grapefruit and an Orange
« Reply #2 on: Oct 30th, 2003, 1:00am » |
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I'm surprised it gives such a nice result..
Then again it is in the easy section..
hint: You can make it a lot easier on yourself by cleverly choosing coordinates for the centers of the objects, where the ones for orange contains some unknowns
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aero_guy
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Re: Three Grapefruit and an Orange
« Reply #3 on: Oct 30th, 2003, 7:54am » |
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Wow, that does come out remarkably easy. If only I could figure out a way to save all those prisoners so quickly.
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Speaker
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Re: Three Grapefruit and an Orange
« Reply #4 on: Oct 30th, 2003, 4:37pm » |
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Well, I am glad that it is easy for someone. Staying with the pencil and paper route (in the MS Word document). I drew a straight line under two of the grapefruit and called it the table. Then, I drew an orange that would be just large enought to touch the table and each of the two grapefruit. This is still in only two dimensions, but It seems to me that it would work in three.
I measured the fruit by formatting them, and eyeballing them.
I got .75 inch for the orange's radius.
Is that an answer?
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| « Last Edit: Oct 30th, 2003, 4:38pm by Speaker » |
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Icarus
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Re: Three Grapefruit and an Orange
« Reply #5 on: Oct 30th, 2003, 6:34pm » |
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While our orange is small, it is not so small as that. I don't think this can be solved by projecting it into 2 dimensions.
For towr & aero_guy, I offer the related puzzle:
Balls of radius 1, 2, 3, and 4 cm are sitting in a hemispherical bowl. If all of the balls are in contact with each other and with the bowl, how big is the bowl? (This one has an irrational answer, so it's just not as nice as the first.)
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Sir Col
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Re: Three Grapefruit and an Orange
« Reply #6 on: Oct 31st, 2003, 5:04pm » |
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What a lovely problem your initial challenge was, Icaurs. I don't know if those who solved it, did it for the general case, but a remarkable result is obtained. If you haven't already done so, and not to detract from Icarus' new challenge...
In the original problem, what must be true about the radius of the grapefruit for the orange to have an integral radius?
Then...
Four identical spherical grapefruit are placed on the counter and a spherical orange is in contact with the counter and the four grapefruit. If the orange has an integral radius, what can you say about radius of the grapefruit?
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towr
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Re: Three Grapefruit and an Orange
« Reply #7 on: Nov 1st, 2003, 9:44am » |
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on Oct 31st, 2003, 5:04pm, Sir Col wrote:| In the original problem, what must be true about the radius of the grapefruit for the orange to have an integral radius? |
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Well, that's easy since radius scales linearly So any integral multiple will work..
Quote:| Four identical spherical grapefruit are placed on the counter and a spherical orange is in contact with the counter and the four grapefruit. If the orange has an integral radius, what can you say about radius of the grapefruit? |
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This is easier than the original problem, since now 3 oranges can be totally ignored, rather than just one.. (Though if I had know where the center of a equilateral triangle was I could have ignored another one there..)
on Oct 30th, 2003, 6:34pm, Icarus wrote:For towr & aero_guy, I offer the related puzzle:
Balls of radius 1, 2, 3, and 4 cm are sitting in a hemispherical bowl. If all of the balls are in contact with each other and with the bowl, how big is the bowl? (This one has an irrational answer, so it's just not as nice as the first.) |
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I'm working on it, but the numbers are a bit too ugly to do it comfortably without my own computer.. (I'm making too many mistakes on paper..)
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| « Last Edit: Nov 1st, 2003, 9:54am by towr » |
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Icarus
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Re: Three Grapefruit and an Orange
« Reply #8 on: Nov 1st, 2003, 6:41pm » |
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I suppose that at least the follow-up really does not belong in the "Easy" section. There is a very nice formula involving the radii of spheres that touch like this. If you find the formula, both of these problems are easy applications (with a bit of interpretation for the original). As a small hint: the corresponding formula for touching circles is easier to find, and should suggest the appropriate form for the spheres formula.
I was moved to post this puzzle because of a poem that goes with the formula. Once the problem is solved, I will give it.
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TenaliRaman
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Re: Three Grapefruit and an Orange
« Reply #9 on: Nov 2nd, 2003, 3:08am » |
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Those trying to solve on their own need not look at the hidden part.(This is not an answer anyways)
::
Some names to mention here,
Descartes' Circle Theorem
Kissing Circle's Theorem
Soddy's Circle Theorem and its extension to higher dimensions
::
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| « Last Edit: Nov 2nd, 2003, 3:09am by TenaliRaman » |
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towr
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Re: Three Grapefruit and an Orange
« Reply #10 on: Nov 2nd, 2003, 10:28am » |
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on Oct 30th, 2003, 6:34pm, Icarus wrote:| Balls of radius 1, 2, 3, and 4 cm are sitting in a hemispherical bowl. If all of the balls are in contact with each other and with the bowl, how big is the bowl? |
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I get a radius of ::6 sqrt(645)/5 + 30::
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Icarus
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Re: Three Grapefruit and an Orange
« Reply #11 on: Nov 2nd, 2003, 7:47pm » |
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That sounds like what I got, but when I tried to do the calculation again, I got a horrible monstrosity. I must of messed up somewhere. I'll try again. In the meantime, here is the poem I mentioned. All but the last stanza are by Frederick Soddy, the Nobel prize-winning British chemist who discovered Isotopes. The fourth stanza is by Thorold Gosset, an English Barrister:
For pairs of lips to kiss maybe
Involves no trigonometry.
'Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There is now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
is half the square of their sum.
To spy out spherical affairs
An oscular surveyor
Might find the task laborious,
The sphere is much the gayer,
And now besides the pair of pairs
A fifth sphere in the kissing shares.
Yet, signs and zero as before,
For each to kiss the other four
The square of the sum of all five bends
Is thrice the sum of their squares.
And let us not confine our cares
To simples circles, planes, and spheres,
But rise to hyper flats and bends
Where kissing multiple appears
In n-ic space the kissing pairs
Are hyperspheres, and Truth declares -
As n + 2 such osculate
Each with an n + 1-fold mate.
The square of the sum of all the bends
Is n times the sum of their squares.
(What is called here the "bend" is actually the circle/sphere/hypersphere's curvature.)
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towr
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Re: Three Grapefruit and an Orange
« Reply #12 on: Nov 3rd, 2003, 1:02am » |
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Actually, mathworld defines a bend as a special kind of curvature, a signed one (depending on internal or external contact)..
It keeps surprising me how much terminology I know nothing about..
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