Cut Cube Surface Area

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 $E\left(1,0,\frac 56\right)$ . The the cut out surface is a pentagon $ABCDE$ of the shape of a cut diamond.

We can find the side lengths using the coordinates of the end points by $\ell = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ . And they are $AB=\frac 1{\sqrt 2}$ , $BC=EA = \frac {\sqrt{10}}6$ , $CD=DE=\frac {\sqrt{10}}3$ , and $CE=\sqrt 2$ . We note that the are of the pentagon $[ABCDE]$ is the of areas of trapezoid $ABCE$ and $\triangle CDE$ . From the side view along $EC$ , we note that the height of $ABCE$ , $h_1$ , and the height of $\triangle CDE$ , $h_2$ , are $h_1:h_2 = 1:2$ . And $h_1+h_2 = \sqrt{\left(\frac 34\sqrt 2\right)^2+\left(\frac 12\right)^2} = \frac {\sqrt{11}}{2\sqrt 2}$ . $\implies h_1 = \frac {\sqrt{11}}{6\sqrt 2}$ and $h_2 = \frac {\sqrt{11}}{3\sqrt 2}$ . Then the area of pentagon $ABCDE$ is given by:

$\begin{aligned} [ABCDE] & = [ABCE] + [CDE] \\ & = \frac {AB+CE}2 \times h_1 + \frac {CE}2 \times h_2 \\ & = \frac 12 \left(\frac 1{\sqrt 2} + \sqrt 2\right)\frac {\sqrt{11}}{6\sqrt 2} + \frac 12 \times \sqrt 2\times \frac {\sqrt{11}}{3\sqrt 2} \\ & = \frac {\sqrt{11}}8 + \frac {\sqrt{11}}6 = \frac {7\sqrt{11}}{24} \end{aligned}$

Other surface areas of the top part of the cube are:

$\begin{aligned} [ABC'D'E'] & = 1 \times 1 - \frac 12 \times \frac 12 \times \frac 12 = \frac 78 \\ 2[AEE'] & = 2\times \frac 12 \times \frac 12 \times \frac 16 = \frac 1{12} \\ 2[E'EDD'] & = 2 \times \frac {\frac 16 + \frac 12}2 \times 1 = \frac 23 \end{aligned}$

Therefore the surface area of the top part of the cube is $\dfrac {7\sqrt{11}}{24} + \dfrac 78 + \dfrac 1{12} + \dfrac 23 = \dfrac {39+7\sqrt{11}}{24}$ . $\implies a + b + c + d = 39+7+11+24 = \boxed {81}$ .

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 $EG = 1 - \frac{5}{6} = \frac{1}{6}$ .

The area of $\triangle BEG$ is $A_{\triangle BEG} = \frac{1}{2} \cdot BE \cdot EG = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{24}$ . By symmetry, $A_{\triangle ADF} = A_{\triangle BEG} = \frac{1}{24}$ .

The area of trapezoid $EHCG$ is $A_{EHCG} = \frac{1}{2}(EG + HC)EH = \frac{1}{2}(\frac{1}{6} + \frac{1}{2})1 = \frac{1}{3}$ . By symmetry, $A_{DHCF} = A_{EHCG} = \frac{1}{3}$ .

The area of $\triangle EDH$ is $A_{\triangle EDH} = \frac{1}{2} \cdot EH \cdot DH = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$ .

By the Pythagorean Theorem, the height of trapezoid $ADEB$ is $h_1 = \sqrt{BE^2 - (\frac{1}{2}(DE - AB))^2} = \sqrt{(\frac{1}{2})^2 - (\frac{1}{2}(\sqrt{2} - \frac{\sqrt{2}}{2}))^2} = \frac{\sqrt{2}}{4}$ . The area of trapezoid $ADEB$ is then $A_{ADEB} = \frac{1}{2}(AB + DE)h_1 = \frac{1}{2}(\frac{\sqrt{2}}{2} + \sqrt{2})\frac{\sqrt{2}}{4} = \frac{3}{8}$ .

By the Pythagorean Theorem on $\triangle BEG$ , $BG = \sqrt{BE^2 + EG^2} = \sqrt{(\frac{1}{2})^2 + (\frac{1}{6})^2} = \frac{\sqrt{10}}{6}$ . Also by the Pythagorean Theorem, the height of trapezoid $AFGB$ is $h_2 = \sqrt{BG^2 - (\frac{1}{2}(FG - AB))^2} = \sqrt{(\frac{\sqrt{10}}{6})^2 - (\frac{1}{2}(\sqrt{2} - \frac{\sqrt{2}}{2}))^2} = \frac{\sqrt{22}}{12}$ . The area of trapezoid $AFGB$ is $A_{AFGB} = \frac{1}{2}(AB + FG)h_2 = \frac{1}{2}(\frac{\sqrt{2}}{2} + \sqrt{2})\frac{\sqrt{22}}{12} = \frac{\sqrt{11}}{8}$ .

By the distance formula, $GC = \frac{\sqrt{10}}{3}$ , and by symmetry, $FC = GC = \frac{\sqrt{10}}{3}$ . By Heron's Formula the area of $\triangle GFC$ is $A_{\triangle GFC} = \frac{\sqrt{11}}{6}$ .

The total surface area is $S = A_{\triangle BEG} + A_{\triangle ADF} + A_{EHCG} + A_{DHCF} + A_{\triangle EDH} + A_{ADEB} + A_{AFGB} + A_{\triangle GFC} = \frac{1}{24} + \frac{1}{24} + \frac{1}{3} + \frac{1}{3} + \frac{1}{2} + \frac{3}{8} + \frac{\sqrt{11}}{8} + \frac{\sqrt{11}}{6} = \frac{39 + 7\sqrt{11}}{24}$ . Therefore, $a = 39$ , $b = 7$ , $c = 11$ , $d = 24$ , and $a + b + c + d = \boxed{81}$ .