Surface Area Problem Math Contest

Question

What is the number of square units in the least possible surface area of a model made with 15 unit cubes?

I have absolutely no idea how to solve this problem.

Answer 1

Assume there is a model with $15$ cubes and surface area $<40$. We may assume of course that the model is connected. The model fits into a bounding box of dimensions $a\times b\times c$ where wlog $a\le b\le c$ and of course $abc\ge 15$. Then necessarily $c\ge 3$, and if $c=3$, then $b=3$, $a=2$.

The projection of the model to one of the three primary planes fits into a $a\times b$, resp. $a\times c$, resp. $b\times c$ rectangle, touches all four sides of that rectangle, and is a connected "polyomino". Hence the three projections have area at least $a+b-1$ etc., which means that the surface of the model is at least $$\tag 1A\ge 2(a+b-1)+2(b+c-1)+2(a+c-1)=4(a+b+c)-6.$$ As we are interested in surface $<40$ only, we need only investigate $a+b+c\le 11$. Hence we are left only with bounding boxes of the following dimensions:

Case 1: $a=1$. As $bc=15$, we have $b+c\ge 8$. The projection to the $b\times c$ plane covers exactly $15$ squares, the other projections must cover $b$ and $c$ squares, respectively. Then the model surface is at least $2\cdot(15+b+c)\ge 2\cdot (15+8)=46$.

Case 2: $a=b=2$, $4\le c\le 7$: Consider the $c$ "sheets" of dimension $2\times 2$. Within each sheet, each cube contributes at least $2$ to the surface by its two faces that touching the boundary box. This already gives us $30$ surface units. Each not completely filled sheet contributes at least two additional units. Additionally, we have at least $6$ units corresponding to the projection to the $2\times 2$ plane, and in fact $8$ units if at least one sheet is filled completely. As $15$ is not a multiple of $4$, we cannot fill all sheets and obtain as lower bound for the model surface

• $\ge 30+8+2=40$ if some, but not all sheets are filled
• $\ge 30+6+2c\ge 44$ if no sheets are completely filled

Case 3: $2\times 3\times 3$. As the box volume is $18$, there are only three gaps to complete the box. A square-gap in one of the three projections "costs" at least two cube-gaps, hence each projection can have at most one such gap. But as two of the three projection directions require all three gaps to be aligned in a row for this to happen, actually at most one of the three projections can have a gap. Thus the model surface is at least $2(2\cdot 3+2\cdot 3+3\cdot 3-1)=40$.

Case 4: $2\times 3\times c$ with $4\le c\le 6$. TODO

Case 5: $2\times 4\times c$ with $4\le c\le 5$. TODO

Case 6: $3\times 3\times c$ with $3\le c\le 5$. TODO

Case 7: $3\times 4\times 4$. TODO