Non-congruent Polyhedrons with same Volume and Surface Area?

Question

This is inspired by Manjyomesando1's very interesting question:
http://answers.yahoo.com/question/index;_ylt=Aqs8_vrKQ9.drMgE6zse.wnty6IX;_ylv=3?qid=20090216140728AAIDqbR&show=7#profile-info-VS9BoXrHaa

As shown there, infinitely many pairs of non-congruent polyhedrons with the same volume (V) and surface area(S) exist, I repeat here some of them for convenience:
Octahedrons:
http://farm4.static.flickr.com/3004/2802617542_794c3a85db_o.gif
Hexahedrons:
http://farm4.static.flickr.com/3582/3294907454_839e2907fa_o.gif
/both #2 and #3 can be obtained truncating Kepler's Stella Octangula:
http://en.wikipedia.org/wiki/Stella_octangula
removing 6 out of its 8 thorns/
Dragan K's cuboids: 1 x 1 x (p(p+1)/2) and p x p x ((p+1)/(2p)) /p≠1/

Now the question: what is the combined minimal number of faces of such pair?

I began to play with a right triangular prism with right isosceles bases and lateral faces in planes x=0, y=0, x+y=1 as shown here:
http://farm4.static.flickr.com/3570/3382194753_04922ac58f_o.gif
Taking red triangles instead of black ones as bases, we obtain a pair of pentahedrons (#1 and #2) with the desired property - we can even join the rightmost vertexes in #2 to obtain a pyramid for the following result:
Prism: vertexes: (1, 0, 0), (1, 0, -2), (0, 1, 0), (0, 1, 2), (0, 0, ±1);
faces x = 0, y = 0, x + y = 1, x - y + z - 1 = 0, x - y + z + 1 = 0;
Pyramid: vertexes: (1, 0, 0), (0, 1, ±2), (0, 0, ±1);
faces x = 0, y = 0, x + y = 1, x - y + z - 1 = 0, x - y - z - 1 = 0
V = 1, S = 4 + 2√2 + √3 for both of them.

Finally, is 10 faces the combined minimum?

JB · Accepted Answer

I'm betting the minimal answer is 8.  Now, back to work, to see if I have to eat my words.
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EDIT:  OK, barring arithmetic errors here is an example of two non-congruent tetrahedra with the same volume and same surface area.  Let the vertices be (2a,0,0), (0,2a,0), (0,0,3b), and (0,0,0).  Then (if I calculate correctly), 

V = 2a^2 b
SA = 6ab + a√(4a^2 + 18b^2)

First let a=11 and b=12, and second let a = 21.97444168338800490094 and b = 3.006982615620947182871 . In both cases

V = 2904
SA = 1402.078683449930087786

The exact value for SA is 792+11√3076.  Although I have given the results as floating point approximations, one can make a continuity argument that shows there is a real number, a, near 22 for which the solution is exact.  In fact, I believe a is the root of a polynomial of degree 10 from which it follows that b is then the root of a certain polynomial also.

I am sure there are easier examples.  This is just the first thing I came up with.
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EDIT 2:  Slightly simpler, in the same family of tetrahedra, let b = 36/a^2.  Then the volume is 72 and the surface area is (2/a)(108+ √(a^6+5832)).  This surface area is equal to 126 exactly for both a = 3 and for a = 3x, where x is the root of x^5 + x^4 + x^3 + x^2 - 48x + 8 nearest to 2. This second value of a is approximately 6.7552093628.
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EDIT 3:  Duke has kindly pointed out that I have made errors under EDIT 2.  However, the idea was right. Please ignore EDIT 2.  I will fix it here:

Slightly simpler, in the same family of tetrahedra, let b = 36/a^2.  Then the volume is 72, independent of a, and the surface area is (2/a)(a^3 + 108 + √(a^6+5832)).  This surface area is equal to 144 exactly for both a = 3 and for a equal to the root of the polynomial 2a^3 + 3a^2 - 63a + 27 which is nearest to 5, and whose approximate value is a ≈ 4.64992867166.  This polynomial arises by equating (2/a)(a^3 + 108 + √(a^6+5832)) to 144 and solving for a, which involves squaring and removing the root a = 3.
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EDIT 4:  Here is a much simpler example.  The pyramid (tetrahedron) with an equilateral triangle of edge e for base, and isosceles triangles with equal edges s for sides, has (if I calculate correctly), 

Volume = (e^2/12)√(3s^2 - e^2) and
Surface Area = (e/4)(e√3 + 3√(4s^2 - e^2))

Now consider two (non-congruent) such pyramids, the first with e=√2, s=√(146/3) and the second with e=4, s=√(73/12), and compute.  You will find they both have Volume=2, and Surface Area = 9√3.

Anonymous · Answer

Total surface area of cylinder = 2*pi*r*(r+h), where r is radius and h is height. Since r = h Total surface area of cylinder = 4*pi*r^2 ---------------(1) Volume of cylinder = pi*r^2*h = pi*r^3 (since r = h) -------------- (2) Equate (1) and (2) 4*pi*r^2 = pi*r^3 So r = 4 units And since r = h, the height is also 4 units. So the cylinder has the raduis as well as height of 4 units.

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Non-congruent Polyhedrons with same Volume and Surface Area?

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