Thinking Inside the Box

Relevant wiki: Pythagorean Theorem

Let the bottom right and top right corner of the face where $Q$ is the intersections of the diagonal be $S$ and $T$ respectively.

The faces of a cube are squares. The diagonals of a square right bisect each other. It follows that $PQ=QT=\frac{1}{2}PT$ . Since the face is a square, $\angle PST=90^°$ and $\triangle PST$ is right-angled. Using the Pythagorean Theorem, $PT^2=PS^2+ST^2=2^2+2^2=8$ . So $PT=\sqrt{8}=2\sqrt{2}$ . Then $PQ=\frac{1}{2}PT=\frac{2\sqrt{2}}{2}=\sqrt{2}$ .

• Because of the 3-dimensional nature of the problem it may not be obvious to all that $\angle RPQ=90^°$ . To help visualize this, note that $\angle RPS=90^°$ because the face of the cube is a square. Rotate $PS$ counterclockwise about point $P$ on the side face of the cube so that the image of $P S$ lies along $P Q$ . The corner angle will not change as a result of the rotation so $\angle RPQ=\angle RPS=90^°$ .

We can now use the Pythagorean Theorem in $\triangle RPQ$ to find the length $RQ$ .

$\begin{aligned} & & RQ^{ 2 } & ={ RP }^{ 2 }{ +PQ }^{ 2 } \\ & & & =2^{ 2 }+\left( \sqrt { 2 } \right) ^{ 2 } \\ & & & =4+2 \\ & & & =6 \\ & \therefore & RQ & =\sqrt { 6 } =1\sqrt { 6 } \\ & \Longrightarrow & x+y & =1+6 \\ & & & =7 \\ & \Longrightarrow & xy & =1(6) \\ & & & =6 \\ & \therefore & x+y-xy & =7-6 \\ & & & =\boxed { 1 } \end{aligned}$