Revival Of This Problem. Part 1

Let the intersection of the two diagonals $AC$ and $BD$ be $O(0,0)$ the origin, $x$ - and $y$ -axis be parallel and perpendicular to $AB$ respectively and $z$ -axis vertically up, and the side length of the cube be 2. Then $A (-1,-1,1)$ , $B(1,-1,1)$ and $C(1,1,-1)$ and let $\textbf{a}$ , $\textbf{b}$ and $\textbf{c}$ be the respective vectors. By dot product we have:

$\begin{aligned} \textbf{a} \cdot \textbf{b} & = |\textbf{a}||\textbf{b}| \cos x \\ [-1,-1,1] \cdot [1,-1,1] & = \sqrt 3 \times \sqrt 3 \cos x \\ -1+1+1 & = 3 \cos x \\ \implies \cos x & = \frac 13 = \cos z \end{aligned}$

Similarly,

$\begin{aligned} \textbf{b} \cdot \textbf{c} & = |\textbf{b}||\textbf{c}| \cos y \\ [1,-1,1] \cdot [1,1,-1] & = \sqrt 3 \times \sqrt 3 \cos y \\ 1-1-1 & = 3 \cos y \\ \implies \cos y & = - \frac 13 = \cos w \end{aligned}$

Therefore,

$\begin{aligned} v & = \cos^2 x + \cos^2 y + \cos^2 z + \cos^2 w \\ & = \left(\frac 13 \right)^2 + \left(-\frac 13 \right)^2 + \left(\frac 13 \right)^2 + \left(-\frac 13 \right)^2 \\ & = \frac 49 \end{aligned}$

$\implies c+d = 4+9 = \boxed{13}$

Let s = side of the cube and p be the intersection of two diagonals. $\rightarrow s \sqrt{3}$ will be the length of diagonals. Now, since the diagonals bisect each other, therefore, the lengths of Ap = Bp = Cp = Dp = $\cfrac{s \sqrt{3}}{2}$ .

By Law Of Cosines,

$\rightarrow \left( \cfrac{s \sqrt{3}}{2}\right )^2 + \left( \cfrac{s \sqrt{3}}{2}\right )^2 - 2\left( \cfrac{s \sqrt{3}}{2}\right )\left( \cfrac{s \sqrt{3}}{2}\right ) * \cos x = s^2 \\ 2*\left( \cfrac{3}{4}\right )s^2 - 2\left( \cfrac{3}{4}\right )s^2 *\cos x = s^2 \\ \left( \cfrac{1}{2}\right )s^2 = \left( \cfrac{3}{2}\right )s^2 * \cos x \\ \cos x = \cos z = \ \cfrac{1}{3}$

and...

$\rightarrow \left( \cfrac{s \sqrt{3}}{2}\right )^2 + \left( \cfrac{s \sqrt{3}}{2}\right )^2 - 2\left( \cfrac{s \sqrt{3}}{2}\right )\left( \cfrac{s \sqrt{3}}{2}\right ) * \cos a = [\overline{AD}]^2 = [s \sqrt{2}]^2 \\ \left( 2*\cfrac{3}{4}\right )s^2 - \left( 2*\cfrac{3}{4}\right )s^2 *\cos a = 2s^2 \\ \left( -\cfrac{1}{2}\right )s^2 = \left( \cfrac{3}{2}\right )s^2 *\cos a \\ \cos w = \cos y = \ -\cfrac{1}{3}$

$\therefore v = \cos^2 x + \cos^2 y + \cos^2 z + \cos^2 w \\ = \left( \cfrac{1}{3}\right )^2 + \left( -\cfrac{1}{3}\right )^2 + \left( \cfrac{1}{3}\right )^2 + \left( -\cfrac{1}{3}\right )^2 \\ = \cfrac{4}{9} \\ \implies c + d = \ \boxed{13}$

The lines $\overline{AC}$ and $\overline{BD}$ are diagonals of the rectangle $ABCD$ . By geometry of two intersecting lines in a plane, $x=z$ and $w=y$ . Since the scaling is arbitrary, keep it simple with $\overline{AB}=\overline{DC}=1$ , then $\overline{AD}=\overline{BC}=\sqrt{2}$ , applying Pythagoras' Theorem to the diagonal of the unit square (corresponding to a face of the unit cube).

Now we have two sides of $\triangle BDC$ , where it is clear to see that $\angle DCB = 90^{\circ}$ . Define the point E as the intersection point where the four angles of interest are made, (the midpoint of line segment $\overline{BD}$ ), and F as the midpoint of the line segment $\overline{BC}$ . Then as $\frac{BE}{BD} = \frac{BF}{BC} = \frac{1}{2}$ , and $\overline{EF} \parallel \overline{DC}$ so $m\angle EFB = m\angle DCB = 90^{\circ}$ , we show that $\triangle BEF \sim \triangle BDC$ with scale factor $\frac{1}{2}$ .

Using Pythagoras' Theorem in $\triangle BDC$ , $BD = \sqrt{3} \Rightarrow BE = \frac{\sqrt{3}}{2}$ . Since $\triangle BEF$ and $\triangle CEF$ share $\overline{EF}$ , $m\angle EFB = m\angle EFC = 90^{\circ}$ (angles on line $BC$ at point $F$ sum to $180^{\circ}$ ) and $BE = EC = \frac{\sqrt{3}}{2}$ by the symmetry of the rectangle $ABCD$ , we conclude that $\triangle BEF \cong \triangle CEF$ .

Therefore $m\angle BEF = \frac{y}{2}$ . Taking $BF = \frac{\sqrt{2}}{2}$ , $EF = \frac{1}{2}$ and $BE = \frac{\sqrt{3}}{2}$ and applying trigonometry in a right triangle:

$\cos \frac{y}{2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

By the double angle formula:

$\cos y = 2\cos^{2} \frac{y}{2} - 1 = 2(\frac{1}{\sqrt{3}})^{2} - 1 = -\frac{1}{3} \Rightarrow \cos^{2}y = \frac{1}{9}$

The same method can be applied to find $\cos^{2}x$ , by letting the midpoint of $\overline{AB}$ be the point $G$ , proving that $\triangle AEG \cong \triangle BEG$ and using the same trigonometric technique. This yields that $\cos^{2}x = \frac{1}{9}$ also.

Now we simply solve for $v$ . Since we established that $x=z$ and $w=y$ , it is implicit that $\cos^{2}x=\cos^{2}z$ and $\cos^{2}w=\cos^{2}y$ . Therefore:

$v = \cos^{2}w + \cos^{2}x + \cos^{2}y + \cos^{2}z = 2(\cos^{2}x + \cos^{2}y) = 2(\frac{1}{9} + \frac{1}{9}) = \frac{4}{9}$

Finally, $c + d = 4 + 9 = \boxed{13}$