Pick's Blocks

Since we are solving for $4$ variables $a$ , $b$ , $c$ , and $d$ , we can use $4$ equations to solve the system. With the given example, since $V = 11$ , $B_e = 72$ , $I_e = 6$ , and $I_p = 1$ , we have $1$ equation $11 = 72a + 6b + c + d$ . An additional $3$ equations are needed, and differently sized cubes are sufficient examples.

A unit cube has $V = 1$ , $B_e = 12$ , $I_e = 0$ , and $I_p = 0$ , and so $1 = 12a + d$ ; a $2$ x $2$ x $2$ cube has $V = 8$ , $B_e = 48$ , $I_e = 6$ , and $I_p = 1$ , and so $8 = 48a + 6b + c + d$ ; and a $3$ x $3$ x $3$ cube has $V = 27$ , $B_e = 108$ , $I_e = 36$ , and $I_p = 8$ , and so $27 = 108a + 36b + 8c + d$ . Solving this system of $4$ equations gives $a = \frac{1}{8}$ , $b = \frac{1}{2}$ , $c = -\frac{1}{2}$ , and $d = -\frac{1}{2}$ , and $a + b + c + d = \frac{1}{8} + \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -\frac{3}{8} = \boxed{-0.375}$ .

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This relationship $V = \frac{1}{8}B_e + \frac{1}{2}I_e - \frac{1}{2}I_p - \frac{1}{2}$ can be proved inductively for all unit block polyhedra. If we assume the relationship is true for one unit block polyhedra, then all the possible ways of adding $1$ unit block will result in changes to the number of border edges $B_e$ , the number of interior edges $I_e$ , and the number of interior points $I_p$ in such a way that according to the formula, the volume $V$ will always increase by $1$ .

There are $6$ cases to examine:

(1) Adding $1$ unit block so that $1$ face is shared. Then $B_e$ would increase by $8$ , so $V$ would increase by $\frac{1}{8} \cdot 8 + \frac{1}{2} \cdot 0 - \frac{1}{2} \cdot 0 = 1$ .

(2) Adding $1$ unit block so that $2$ adjacent faces are shared. Then $B_e$ would increase by $4$ and $I_e$ would increase by $1$ , so $V$ would increase by $\frac{1}{8} \cdot 4 + \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 0 = 1$ .

(3) Adding $1$ unit block so that $3$ adjacent faces around a corner are shared. Then $I_e$ would increase by $3$ and $I_p$ would increase by $1$ , so $V$ would increase by $\frac{1}{8} \cdot 0 + \frac{1}{2} \cdot 3 - \frac{1}{2} \cdot 1 = 1$ .

(4) Adding $1$ unit block so that $3$ adjacent faces in a row are shared. Then $I_e$ would increase by $2$ , so $V$ would increase by $\frac{1}{8} \cdot 0 + \frac{1}{2} \cdot 2 - \frac{1}{2} \cdot 0 = 1$ .

(5) Adding $1$ unit block so that $4$ adjacent faces around $2$ corners are shared. Then $B_e$ would decrease by $4$ , $I_e$ would increase by $5$ , and $I_p$ would increase by $2$ , so $V$ would increase by $\frac{1}{8} \cdot (-4) + \frac{1}{2} \cdot 5 - \frac{1}{2} \cdot 2 = 1$ .

(6) Adding $1$ unit block so that $5$ faces are shared. Then $B_e$ would decrease by $8$ , $I_e$ would increase by $8$ , and $I_p$ would increase by $4$ , so $V$ would increase by $\frac{1}{8} \cdot (-8) + \frac{1}{2} \cdot 8 - \frac{1}{2} \cdot 4 = 1$ .

(Note that adding $1$ unit block so that $2$ opposite faces are shared is not allowed since it would form a polyhedron with a hole in it, adding $1$ unit block so that $4$ adjacent faces in a row are shared is not allowed since it either assumes a starting polyhedron with a hole in it or it would form a polyhedron that is hollow, and adding $1$ unit block so that $6$ faces are shared is not allowed because it assumes a starting polyhedron that is hollow.)