Cut Cube Volume

Label the diagram as follows:

The plane has an equation of $x + by + cz = d$ , and since it goes through $A(0, \frac{1}{2}, 1)$ , $B(\frac{1}{2}, 0, 1)$ , $C(1, 1, \frac{1}{2})$ , we know that $0 + \frac{1}{2}b + c = d$ , $\frac{1}{2} + 0 + c = d$ , and $1 + b + \frac{1}{2}c = d$ , which solves to $b = 1$ , $c = 3$ , and $d = \frac{7}{2}$ , and gives an equation of $x + y + 3z = \frac{7}{2}$ or $2x + 2y + 6z = 7$ .

$G$ is also on the plane $2x + 2y + 6z = 7$ with $x = 1$ and $y = 0$ , so its coordinates solve to $(1, 0, \frac{5}{6})$ , which means $EG = 1 - \frac{5}{6} = \frac{1}{6}$ .

Now extend $AB$ , $DH$ , and $FC$ to meet at $I$ and $AB$ , $EH$ , and $GC$ to meet at $J$ to form right-angled tetrahedron $IJHC$ :

$J$ is on the plane $2x + 2y + 6z = 7$ with $y = 1$ and $z = 1$ , so its coordinates solve to $(-\frac{1}{2}, 1, 1)$ , which means $JE = \frac{1}{2}$ . By symmetry, $ID = JE = \frac{1}{2}$ .

The volume of the cube that lies above the cutting plane is $V = V_{IJHC} - V_{BJEG} - V_{IADF} = \frac{1}{6} \cdot HI \cdot HJ \cdot HC - \frac{1}{6} \cdot EB \cdot EJ \cdot EG - \frac{1}{6} \cdot DI \cdot DA \cdot DF = \frac{1}{6} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} - \frac{1}{6} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} - \frac{1}{6} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} = \frac{25}{144} = (\frac{5}{12})^2$ . Therefore, $a = 5$ , $b = 12$ , and $a + b = \boxed{17}$ .

Label the cube as shown. If we look along $E'C'$ , we note that $\triangle AEE'$ and $\triangle ADD'$ are similar. Then $\frac {EE'}{DD'} = \frac {AE'}{AD'} \implies EE' = \frac {AE'}{AD'} \times DD' = \frac 13 \times \frac 12 = \frac 16$ . Then we have $C \left(0,1,\frac 56\right)$ and $E\left(1,0,\frac 56\right)$ .

Now extend the top of the cube until the cut meet the extended edge of $FF'$ at $F''$ . From similar triangle we have $F'F''=\frac 16$ . Color the block whose volume $V$ we need to find yellow; and the volume of extended blocked of height $\frac 16$ to top red $V_1$ and blue $V_2$ .

We note that $V+V_2$ is half the volume of the cuboid $C'''D''E'''F''CDE''F''$ which has a height of $FF''=\frac 16 + \frac 12 = \frac 23$ . Since the volume of the cuboid is $1 \times 1 \times \frac 23 = \frac 23$ , then $V+V_2 = \frac 12 \times \frac 23 = \frac 13$ .

We also note that $V_1 + V_2 = 1 \times 1 \times \frac 16 = \frac 16$ or a unit cuboid of height $\frac 16$ .

And that $V_1$ is a pyramid with a base area $\frac 12 \times \frac 12 \times \frac 12 = \frac 18$ and height $\frac 16$ . That is $V_1 = \frac 13 \times \frac 18 \times \frac 16 = \frac 1{144}$

Then we have:

$V = (V+V_2) - (V_1+V_2) + V_1 = \frac 13 - \frac 16 + \frac 1{144} = \frac {25}{144} = \left(\frac 5{12} \right)^2$

Therefore $a+b = 5+12 = \boxed{17}$ .