20070909, 16:25  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Hole and sphere.
A cylindrica hole six inches long has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere? Mally 
20070909, 17:44  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3×19×109 Posts 
I remember seeing this puzzle years ago and was surprised to discover that (IIRC) the radius of the sphere is completely cancelled out and the volume is the same no matter how big we make the radius.
But I think the question should be posed a little differently to avoid confusion. Perhaps something like "What is the volume of the remainder in the sphere"? Last fiddled with by retina on 20070909 at 17:44 
20070909, 18:05  #3 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Presentation!
Well Retina, suit yourself as to the presentation of the problem. It was first posed by the puzzilist Sam Loyd and adapted in several other publications. Well , at least you did not comment that it has insufficient data to be solved. So what is the the volume of the rest of the sphere after the hole has been bored? Hint: Dont go into calculus and the rest. There is also a purely logical solution too! Without that, one must at least solve it by pure geometry though you must arrive at the volume of the caps which requires a formula not too well known. Mally P.S. I'm sorry I did not take note of your spoiler (though I read it} and concentrated on the rewording of the problem. You are quite right, though. Last fiddled with by mfgoode on 20070909 at 18:09 
20070909, 20:30  #4  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
47·229 Posts 
Quote:
As the question doesn't mention the diameter of a sphere, the answer must be independent of the size of the sphere, provided that the hole can be six inches long. Therefore, chose a sphere of six inches diameter. The hole must then be of zero diameter and so of zero volume. Consequently, the answer is the same as the volume of a six inch diameter sphere. The volume of a sphere of radius r is 4\pi r^3/3. Therefore the answer to the question is 4\pi 3^3/3 cubic inches, or 36\pi cubic inches. Paul 

20070909, 20:39  #5 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 

20070909, 22:09  #6  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:


20070910, 08:27  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Same problem  different twist!
Quote:
You don't have too OT. Here is a more rigorous solution by an analogy in two dimensions. Take a cross section of the 'completed' residue of the sphere. You will get two concentric circles. Let us call it a circular track. Pl. draw a figure for a better view. Now given the longest straight line that can be drawn on the circular track of ANY dimensions. This is a tangent to the smaller circle. The area of the track will be equal to the area of a circle having the straight line as a diameter. No assumptions here as it can be proved! Convince your self if you like by proving it. It should be called a theorem! Now reduce the dia. of the central smaller circle. The straight line (Tangent) will become longer. In the limit the straight line becomes the dia. and the area of the smaller circle zero. We are given the depth of the hole is 6 inches and so convert the problem from area of the cross section to volume of the sphere of dia. 6 inches. And this is 36 pi. a constant. In other words the residue is constant regardless of the hole's diameter or the size of the sphere. The diameter of the sphere will be the same as the depth of the hole. Vary these if you like to make it general. Mally ATTN: Akruppa: I doubt if this view was given in the last time this problem was presented in '06 Last fiddled with by mfgoode on 20070910 at 08:35 

20070910, 11:01  #8 
Jun 2003
The Texas Hill Country
2101_{8} Posts 
Actually, this last portion of your statement is incorrect. The diameter of the sphere is greater than (or in the limiting case equal to) the depth of the hole. The hole removes a right cylindrical plug whose height is the depth of the hole PLUS two "end caps". But, as noted, the residue is of a constant volume even though the diameter of the sphere varies.

20070910, 12:27  #9  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
End caps.
Quote:
Thank you Wacky! I stand corrected and an astute observation on your part. I welcome this type of correction instead of posters taking digs at presentation or mis spelling or such trivia which breeds unpleasantness. Mally 

20070910, 14:09  #10 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
By additionally considering the intersection of
a plane tangential to the cylinder with the sphere (a circle) I can deduce from Mally's theorem that the largest area of Mally's cross section is that of this circle (9PI) whatever the diameter of the sphere. However it is far from obvious that the area of other rings stacked in the 6 inch height are similarly independent of the sphere's diameter. Am I missing something? David Last fiddled with by davieddy on 20070910 at 14:56 Reason: Sometimes handwaving aruments are more trouble than they are worth 
20070911, 10:28  #11 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Let the radius of the sphere be R inches.
Take its intersection with a plane h inches from the centre of the sphere, and let the radius of this circle be r inches. r^2=R^2  h^2 The square of the radius of the cylinder is (R^2  3^2) inches^2 To find the area of the ring ("track" as Mally called it) we subtract these values (*PI). Lo and behold, R leaves the problem. I think Mally half showed this for h=0. David Last fiddled with by davieddy on 20070911 at 11:04 
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