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| Chappy... |
 Posted: Thu Oct 29, 2009 11:54 pm |
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Enigma 1563 - Drawing a line
New Scientist magazine, 16 September 2009.
By Susan Denham.
"Enigma" is a piece of contemporary art. It
consists of a solid white cuboid of sides
18 centimetres, 24 centimetres and some
other length (which is not a whole number of
centimetres).
On each face of the cuboid there is a
straight red line a whole number of
centimetres long going from the mid-point of
one edge to the mid-point of another. The
six lines together form a continuous route
over the surface of the cuboid. The shortest
two of the lines are of different lengths.
What are the lengths of the six lines?
Ciao,
Chappy. |
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| James Dow Allen... |
| Posted: Fri Oct 30, 2009 5:36 am |
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On Oct 30, 4:54 pm, Chappy <petergregorychap... at (no spam) hotmail.com> wrote:
Possible Spoiler.
OP as spoiler warning space.
Quote: Enigma 1563 - Drawing a line
New Scientist magazine, 16 September 2009.
By Susan Denham.
"Enigma" is a piece of contemporary art. It
consists of a solid white cuboid of sides
18 centimetres, 24 centimetres and some
other length (which is not a whole number of
centimetres).
On each face of the cuboid there is a
straight red line a whole number of
centimetres long going from the mid-point of
one edge to the mid-point of another. The
six lines together form a continuous route
over the surface of the cuboid. The shortest
two of the lines are of different lengths.
What are the lengths of the six lines?
Ciao,
Chappy.
Solution:
The lines are 18, 24, 31, 31, 32, 32.
Clearly the ambiguous term "cuboid" here denotes rectangular
parallelepiped.
The cuboid's faces comprise two each of three sizes. Each
is traversed once diagonally or laterally. Inspecting
possible routes we see that either all traverses are diagonal,
or that four are, with two unequal lateral traverses of
same-sized faces. In the first case, line lengths form
equal pairs and the condition that the shortest two are
different cannot be satisfied, so there are lateral sides
and they must be from the 18x24 faces, or one would be
non-integer.
Letting the sides of the cuboid be 18, 24 and 2Y, the line
lengths are therefore 18, 24 and two each of sqrt(81+Y^2)
and sqrt(144+Y^2). We need therefore non-integer Y with
81+Y^2 and 144+Y^2 perfect squares; that is to say, perfect
squares which differ by 144-81 = 63. Inspection finds
that only (32^2,31^2) are satisfactory, i.e. Y = sqrt(880).
James Dow Allen |
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| Philippe 92... |
| Posted: Fri Oct 30, 2009 10:42 am |
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Chappy wrote :
Quote: Enigma 1563 - Drawing a line
New Scientist magazine, 16 September 2009.
By Susan Denham.
"Enigma" is a piece of contemporary art. It
consists of a solid white cuboid of sides
18 centimetres, 24 centimetres and some
other length (which is not a whole number of
centimetres).
On each face of the cuboid there is a
straight red line a whole number of
centimetres long going from the mid-point of
one edge to the mid-point of another. The
six lines together form a continuous route
over the surface of the cuboid. The shortest
two of the lines are of different lengths.
What are the lengths of the six lines?
The shortest two of the lines are 18 and 24 cm !
Without that last condition that they are of different lengths, there
would be exactly two solutions.
BTW the largest two of the segments *are* of equal lengths...
And also are the two intermediate segment lengths.
Regards.
--
Philippe Ch., mail : chephip+news at (no spam) free.fr
site : http://mathafou.free.fr/ (recreational mathematics) |
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| Philippe 92... |
| Posted: Fri Oct 30, 2009 10:59 am |
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James Dow Allen wrote :
Quote: On Oct 30, 4:54 pm, Chappy wrote:
...
...that is to say, perfect
squares which differ by 144-81 = 63. Inspection finds ...
Hi,
while I was composing my reply, you posted the same.
I deleted then my answer.
Just react to "by inspection".
To find m^2 - n^2 = (m + n)(m - n) = 63 = 3^2 * 7
That is to choose two "complementary" dividers of 63
1*63 = 3*21 = 7*9
Hence three solutions (m,n) = (8, 1), (12, 9), (32, 31)
(just one solution fits because we require also m > 12 and n > 9)
An interesting question is to discard some conditions of the problem :
* the third dimension may be an integer
* The *total length* of the route is an integer (not the individual
segments)
* discard "the two shortest lengths are the same"
This gives many solutions. To find all solutions, or describe the
family if there are infinitely many.
Regards.
--
Philippe Ch., mail : chephip+news at (no spam) free.fr
site : http://mathafou.free.fr/ (recreational mathematics) |
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