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Greek geometry, Euclidean geometry
Egyptian geometry
Type of Math Object: 
Major Section: 

Mathematics Subject Classification

51-00 no label found51-01 no label found


The 2,000 BCE Moscow Mathematical Papyrus (MMP) lists
an arithmetic method that found the area of a truncated
pyramid as discussed by many over the years. The scribal
method took small slices of the pyramid, and added all the
pieces to find the exact area.

Jumping forward to Archimedes, following the 300 BCE
method of Eudoxus, many additional geometric shapes'
areas and volumes could be found, also taking small
slices and adding them for the correct areas and volumes.
Eudoxus is said to have used a 1/4th geometric series
to achieve his accurate results. This method has often
been called the first form of calculus, as later improved
by Archimedes, Newton, Gauss and Leibnitz.

The only Hellene example was left of this method was offered
by Archimedes within his famous Lemma, finding the volume
of a parabola, first, by:

4A/3 = A + A/4 + A/16 + A/64 + ... + A/4n + ...

the second proof offered by Archimedes, was first read by
Heiberg in 1904, as reported by Dijksterhuis, within his
biography of Archimedes. This was an Egyptian (unit) fraction
proof, using only arithmetic. It stated:

4A/3 = A + A/4 + A/12

as easily shown to have been derived from Egyptian fraction
methods that converted any rational number, vulgar and prime,
into a concise and exact unit fraction series, as discussed on

and elswhere, by several others.

What is the size of the largest square that will fit in the unit cube?

i think that given by the diagonals of 4 sides

I dont think so, becuase 4 diagonals of a cube wont render you a square

First of all, I know that the answer is 9/8

what I know in advance is that the area of the largest rectangle that will fit in the box is sqrt 2.

So somehow we have to think about a square whose sides are somewhere between 1 and sqrt 2 long.

I thought to take the geometric mean which rendered me the correct answer i.e. sides of length 3/4 of the sqrt of 2 which is 1.0606 and the area would be 9/8 which is 1.125.

But I dont know how to prove that it is the biggest absolute possible.

Can you please do that?

I will ping back if I am able to come up with something useful - however, in the meantime, if you would like to 'see' the square you are looking for, you should head over here:

Here is a very simple approach to the largest square in a cube.

The center of the square must coincide with the center of the cube. Let us take this point O as the origen of coordinates x, y, z.
A vertex P of the square must lay on one of the edges of the cube, if it is to be the largest square. So its coordinates are (x, 1/2, 1/2), for example. The coordinates of another vertex Q are (1/2, y, 1/2). There are two conditions for these vertices to generate a cube:
1 - they must be at the same distance from O, so |x| = |y|.
2 - the dot product of the vectors OP and OQ must vanish (right angle).
These two conditions determinate completely x and y, and give 9/8 for the area of the square.

> The center of the square must coincide with the center of
> the cube.

Maybe there is some simple argument using symmetry, but I am not seeing it so far. Can you explain this?

> > The center of the square must coincide with the center of
> > the cube.
> Maybe there is some simple argument using symmetry, but I am
> not seeing it so far. Can you explain this?

Let ABCD be a square in a cube. AB is a vector from A to B (BA = -AB). The diagonals are AC and BD, AC.BD = 0.
Define 4 points A', B', C', D' such that

OA' = CA/2 OB' = DB/2 OC' = -OA' = AC/2 OD' = -OB' = BD/2

Now prove that A'B'C'D' is a square of center O, size ABCD and lies entirely into the cube.

Thanks dh2718 for your reply. I wanted to confirm with you just one more thing, what value of x did you get?

@dh2718, I think your argument just simulates a square whose centre coincides with the centre of the cube.

But we need to prove that; for getting maximum area of square, centre of square should coincide with the centre of cube.

Please correct me if I am wrong

1 - The vector construction I described proves thar for any square in the cube, there is another square of the same size located at the center of the cube (center of square coincident with the center of the cube), and that is enough: if there is a maximal square somewhere, not at the center, there is also a centered square of the same size.

2 - You asked the value of the x coordinate in my first post on this subject: the orthogonality condition gives (x + y)/2 + 1/4 = 0, and with |x| = |y|, we get x = y = -1/4. The distance R of the vertex to the center is R^2 = x^2 + 1/4 + 1/4, and the area 2R^2 is 9/8.

3 - If we try to apply the same method for higher dimensions (hypercube), it works up to 4 dimensions where we get x = -1/2 and an area of 2. But for higher dimensions, we get x greater than 1/2, that is, out of the cube.


Awesome; Hats off to you, sir

Great... I'd recommend that you add this as a PM entry. This would be a good example showcasing the use of geometry, algebra, and some analysis to solve a problem.

> Great... I'd recommend that you add this as a PM entry. This would be a good example showcasing the use of geometry, algebra, and some analysis to solve a problem.

I agree. I have found this thread very interesting to read, and it would be nice to have all the information in one PM entry rather than scattered through many posts. Moreover, I am glad that a solution was obtained. :-)

I apologize, I have just posted a note on this subject at the wrong thread (linking scope).

No problem. I'm sure it's a mistake that I would make. Just so that the relevant post is added to this string (to make it easier to find), I am putting a link to it here:

BTW, are you a LaTeX-analphabet by choice, or is this an issue that you would like to remedy?

In any case, when I get a chance (which ought to be soon, but you never know), I will sift through all these posts and add an entry.

Given a square piece of property of unit side you wish to build fences so that it is impossible to see through the property, ie there is no sightline connecting two points outside the property and passing through the property that does not intersect a fence. The fences do not have to be connected and several fences can come together at a point. What is the minimum total length of fencing required and how is it arranged. For example you could place fencing along all four sides. This would have total length 4 but is not the best possible.

I got a length of sqrt(3) + 1 by minimizing the fence in the figure below, hoping it is still clear despite Notepad's limitation. It is certainly less than 4, but I am not sure that this is the best which can be done.
|\ /|
| \__/ |
| / \ |

Thanks dh2718 for your answer, can you please give a detail answer of how did you get sqrt(3) + 1 and also if possible can you please give a more clear diagram if possible

Thanks & Regards

The figure I drew with Notepad is really a beautiful piece of modern art, but not very related to the problem; so I am going to try to describe it:

Let ABCD be the square; AB = 1. Let M and N be the middle points of AB and CD respectively. On the line MN, let P be a point such that MP = x < 1/2, and Q such that NQ = x. Our fence is:
APQD + BP + QC. Its length is 2sqrt(1 + 4x^2) + 1 - 2x
By differentiation, the minimum is found at x = 1/2sqrt(3) and the minimum length is sqrt(3) + 1.

What you are solving is known as the Steiner minimal tree problem for
4 points.

I didn't know that nice problem, Thomas. (I had really not understood PARASHAR's statement)
and dh2718's answer is correct!

Thanks a lot dh2718 for such a descent explaination.
Thanks Thomas for that great link

An thanks to PARASHAR for post so interesting problem!

Hi dh2718, Thomas, Perucho

sqrt(3) + 1 is not the minimal possible solution. There are more smaller answers possible.
1. 2 + sqrt(2)/2 i.e. 2 edges and one fence to center
2. [2*sqrt(2) + sqrt(6)]/2

so I think, we need to Start from the cross(2*sqrt(2)) and play around with the center to get the minimal possible solution. Can you please help?

Sincerely, I don't finish to understand the statement of the problem (because my short English, I guess). I assume that in all the 4 corners of the square, at least one fence it must concur, because if were not so, then it will always be possible connect two external points traspassing the property. So I think that dh2718-Mathprof (5 fences) answer is correct. Nevertheless, I ask myself, as it is possible to cross (side-by-side) the property and connect *two external points* if I place a fence throughout a square's diagonal?(I consider intern the two fence's ends) I don't understand that.

Hi Parashar,

You are right. Fence 1 is disconnected, a possibility which I didn't
take into consideration. Can you please describe the geometrical
shape of fence 2?

I am also not sure about the best possible minimal solution, but one thing is for sure that the same "playing around with the center" that changes the crossed diagonals into the back-of-envelope can also be used to further optimize the solution.
One more optimised solution is sqrt(2) + sqrt(3/2)

I would guess that Parashar's fence 2 is like this:

Let c = 1/2 - sqrt(3)/6.

Four fences:

(c,c) to (0,0)
(c,c) to (1,0)
(c,c) to (0,1)
(1,1) to (.5,.5)

yark, you are right. But how to prove that this is the most minimal possible?

I already could understand your problem, finally! Your statement was very clear but my English is terrible!!! Sorry my friend.
The solution given by yark is very ingenious. He places at any corner a fence to \pi/4 (L_1=1/sqrt{2}), thus avoiding to watch from the opposite diagonal. Next he places, from the remaining 3 corners, connected fences and proceeds to minimize its subtotal length L_c, say. (c=1/2(1-1/\sqrt{3}), L_c=1/\sqrt{2}(1-1/\sqrt{3})+2\sqrt{2}/3). So the total length is L=L_1+L_c=5\sqrt{2}/3-1/\sqrt{6}=1.948..., clearly lesser than L=1+\sqrt{3}=2.732... (5 fences).
Intuitively, it seems not easy to find a better solution. (if there were it)

perucho writes:

> total length is L=L_1+L_c=5\sqrt{2}/3-1/\sqrt{6}=1.948...

It's actually sqrt(2) + sqrt(3/2) = 2.63895843..., as PARASHAR said in an earlier post.

I am sorry but I didnt understand your comment becuase the mathematical denotions are totally messed up. What do you mean by "/" and "\" sign. "/" is used to denote division. What is "\" for?
Yark, if you have understood his denotion, can you please make it more clear?

PARASHAR writes:

> What is "\" for?
> Yark, if you have understood his denotion,
> can you please make it more clear?

\sqrt{x} is the TeX code for sqrt(x).

What do you say, is L=L_1+L_c=5\sqrt{2}/3-1/\sqrt{6}=1.948 the minimal possible solution? can you please comment on the authenticity of this answer? because although we have sqrt(2) + sqrt(3/2) as one of the answer, we have not proved that it is the minimum possible solution

Many thanks yark for correct my mistake. Here: \sqrt(2+2/3), and mentally, I put: 2\sqrt{2}/"3" ???.(Next time I'll use a paper)
So I must give all the worth to PARASHAR and apologize with him, again!
I really sorry my friend. Thanks yark; as you posted those coordinates I could understand this nice problem proposed by PARASHAR.
A greeting for both.
PS. As far as the question that PARASHAR did you, I think that the minimal distance to block the sight from the opposite diagonal is L_1, and since L_c is also a minimum, then the minimal possible length's fences is L.

PARASHAR writes:

> yark,
> What do you say, is L=L_1+L_c=5\sqrt{2}/3-1/\sqrt{6}=1.948
> the minimal possible solution?

No, that was just perucho's error.

As a comparative example, the symmetric quadratic function
is involved with the 3 connected fences' length L_c. Its minimization leads to the value a=b=c=1/3. (CF. yark's reply on 2007-07-13 09:57:18)
This value outcomes the solution
which is not better than L=2.6389... ones given by yark-PARASHAR.

That is indeed a good relevance. Thanks for that.

There are 5 platonic solids, the tetrahedron (4 vertices, 4 triangular faces, 6 edges), the cube (8 vertices, 6 square faces, 12 edges), the octahedron (6 vertices, 8 triangular faces, 12 edges), the dodecahedron (20 vertices, 12 pentagonal faces, 30 edges) and the icosahedron (12 vertices, 20 triangular faces, 30 edges). Consider open models of these solids with wire edges connecting the vertices. Suppose each wire has unit resistance. For each case find the total resistance between a pair of adjacent vertices. Express each answer as a rational number.

Hint: The answers can be found by brute force but there is a way to use symmetry.

PARASHAR! What's up my friend!
You and your ponder (IBM) challenge problems! That's good to exercise the mind! But my tired mind this time is going to do mutis (you can say me "coward", no problem, he-he) as I have an alibi because my tired mind and myself will go to USA the next Saturday on vacations by one month.
(please don't take my modest entries as the next past year [of course, not you!]; wait until I'll be back).
You can also post this problem in PlanetPhysics, perhaps bloftin et al be interested in solving it. BTW, looking at the web I have found the following related article:
I don't know if this one helps, but anyway.
So you are talking about ``regular polyhedra'', believe it or not! I didn't know the ``platonic solids'' nomenclature. ( I don't know if there exist in Spanish: "s\'olidos plat\'onicos"; we usually name it: "poliedros regulares")

Hi Perucho

As usual, you are the speed champ. Actualy I have reached the correct path, but something is confusing me which I am not able to figure out.

Lets for example take octahderon.
In this case, V = 6, F = 8, N = 12. Now as per the question, resistance between each vertice is 1 unit considering the open model.
As there are 12 edges, total resistance is 12 units.

According to the formula, if total resistance of octahedron is 12 units, resistance between any adjacent vertice is 5*5/12 i.e. 5 units.

My first question is, is this answer right?

And second question is for reaching the final answer, the final answer is to expressed in rational number.

But in my case, the final answer would be???? I am confused. Can you please help

Hi Parashar, you are indeed confused: the resistance between two adjacent vertices must be less than 1: 1 ohm in parallel with something. That seems to confirm Perucho's observation that this problem is more appropriate to Planet Physics than here.
For the tetrahedron the answer is 1/2 and for the cube it is 7/12. I give up for the over polyhedra, my spatial vision is too weak. Besides, each polyhedron demands a special treatment; the symmetries of the tetrahedron, for example, don't help to solve the cube's problem.
"Brute force" here means, I guess, using a circuit simulation program like Pspice, which gives the answers at a glance, with decimal numbers.
A few years ago, I submitted to IBM a similar challenge: to find the resistance between two opposite vertices of an n-dimensional hypercube (5/6 for the 3-dimension cube). They didn't accept it; I guess that they don't like n-dimensional spaces where spatial intuition cannot work.

K light-emitting points are placed on a three-dimensional integer lattice (N*N*N cube). Three spectators observe these lights from three different perspectives. Each of them watches a projection on two of three coordinates: XY; YZ; and ZX. For example, the first observer sees the point (2,7,3) as (2,7); the second sees it as (7,3); and the third as (3,2).

A set of light points is defined as revealing if given any point (u,v) in the N*N square, there is an observer who sees a light at that point in his projection and one can uniquely reveal both the exact source of light and the identity of the observer. Each point of light (u,v) is seen by exactly one spectator and he or she sees it exactly once (no light hides behind another light).

For which N is there a revealing set? Show how to construct such a set for all feasible N and prove the nonexistence of them for all other N.

Anyone interested??:-)

Two circles with centres P and Q intersect at 2 points A and B such that none of P and Q fall in the area of intersection between the two circles. Also both the circles have same radiues. Find out minimum range which describes all possible values of angle APQ?

1) 0 to 30
2) 0 to 45
3) 0 to 70
4) 0 to 90

Any thoughts??

I believe it is 0 to 60. The extrema are, first, the case where the circles are tangent, when angle APQ is 0, and second, the case where P lies on the circle around Q, and Q on the circle around P. Then PA and PQ are radii of the P-circle, so are equal, while AQ and QP are radii of the Q-circle, so are equal. Thus they are all three equal and APQ is an equilateral triangle.

I'm having serious troubles with all the your three problems.
1.- Circles
As Roger, I readily got 0-60 but this doesn't appear in your set of solutions!
2.- Denominations
I don't see another solution (by using all the three denominations) that:
3.- Minimum value
With the restriction x,y,z>0, this problem reduces to positive the octant.
But i) the plane x+y+z=0 and the plane bisectors ii) x=y, y=z and z=x, all of them are passing for the origin (0,0,0), and we have the family of spheres f=x^2+y^2+z^2=c^2, so I don't see a minimum value other than (0,0,0) which is absolute. See this pahio's entry:
So I understand nothing!
Do you want place me in a corner of the ring, PARASHAR?
Your friend,


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