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The pyramid of cubes

(A math problem to distract thinking)

Statement: We

built a pyramid of wooden blocks, each of them has a volume of 1 dm ³. On the upper floor there is a cube, located in the center of the second floor, consisting of four cubes. These four blocks are located in the center of another apartment, consisting of nine cubes.

We paint the visible part of the pyramid, ie, not paint or faces that are under or parts of faces that are covered by another floor.

a) What is the area we'll have to paint?

b) And if we add a new floor, consisting of 16 cubes, What would be the surface we would have to paint?

c) Would you know if generalize to any number would No Flats?

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Solution:

a) En primer lugar, a partir del volumen de los cubos, obtenemos el valor de la arista ( a ):

En relación al cubo del primer piso ( n =1), la superficie lateral ( S l ) would be to paint all 4 sides and top:

In relation to the 4 cubes of the second floor ( n = 2), the lateral surface (S l) to paint would be the 8 sides more than the free surface, which is that of the four cubes, less the area of \u200b\u200ba cube , which is the surface that is covered by the floor n = 1. So we have:

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Regarding the 9 cubes on the third floor ( n = 3), the lateral surface (S l) would be to paint the 12 faces side more than the free surface, which is that of the nine cubes, less the area of \u200b\u200bthe four buckets, which is the surface that is covered by the floor n = 2. So we have:

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understand, from the problem statement, the BOP will not be painted. So do not add here the surface of the base (9 to ² )

Then, the total area to paint is:

Substituting the value of the edge ( to = 1DM), we finally obtain that:

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b) to the solution of the previous section we add the surface of the fourth floor. Then, the lateral surface (S l) to paint, compared to the fourth floor ( n = 4), would be to the 16 sides plus free surface superior, which is that of the sixteen cubes, less the area of \u200b\u200bnine cubes, which is the surface that is covered by the floor n = 3. So we have:

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Thus, we obtain the total surface to be painted in the case of a pyramid of blocks of four floors:

Substituting the value of the edge ( to = 1DM), we obtain:

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c) To generalize to the case of n floors, we will analyze the evolution of the series obtained in the two preceding paragraphs.

For each floor, we obtained the surface of the sides, you sumábamos the upper surface and Reset will contact the surface of the floor above. And construct the following table with the coefficient of the surface side face of a cube ( to ²)

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The surface of the floor above we have to subtract is always the same as calculated upper surface on the floor above, so we only have the surface n ², as top surface on the top floor (equivalent to what would be the pyramid from the top, right from the vertical) Thus, we see the surface of the sides is provided for each floor, 4 times the number of floors: 4 n .

That is, the side surface to paint should be:

Fixing the expression obtained, it appears that the surface will have to paint in a general case of a pyramid of n story cube of edge formed by to is :

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Finally, we will then analyze the evolution of this surface to be painted in the particular case of to = 1.

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's it:

"With this and a biscuit! Until tomorrow morning at eight!"

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