# Surface Area Problem Math Contest

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.

• There are not so many interesting configurations to consider, so a good start might be to consider them. For instance, if we remove a corner from a $2\times 2\times 4$ box, the surface area of such object is the same as the surface area of the original box, i.e. $40$. Can you do better than $40$? – Jack D'Aurizio Nov 4 '18 at 0:06
• my advice is to find, say, 16 identical cubes, and see what happens with 3 cubes, then 4, and keep going. Better if there is some way to attach them along the faces. I am a big fan of doing things with one's own hands. – Will Jagy Nov 4 '18 at 1:18
• "I have absolutely no idea how to solve this problem." Again? How comes you never have any idea about how to solve the problems you post here? It might be time to switch to problems more adapted to what you can solve... – Did Jan 12 at 10:01

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