Keeping Things in Perspective

Let $d$ be the distance from the observer to the face of the cube at the origin. Then, if

$\dfrac{2d^2+12d+6}{2d^2+10d+5}=\dfrac{993}{901}$

then

$\dfrac{2d^2+14d+7}{2d^2+12d+6}=\dfrac{1085}{993}$

and so the answer is $1085$

Let the origin $(0,0,0)$ be at the center of cube $O$ and $(0,0,-d-\frac{1}{2})$ be the coordinate of the POV. Then the coordinates of all the vertices of cubes $P, Q, R$ relative to the POV are found by adding $(\pm \frac{1}{2}, \pm \frac{1}{2},\pm \frac{1}{2})$ to $(-3,2,d+\frac{1}{2})$ , $(1,-2,d+\frac{1}{2})$ , $(5,1,d+\frac{1}{2})$ respectively. Thus, if we project the vertices onto the plane $(0,0,-\frac{1}{2})$ , then the adjustment factor between the vertices at $(x, y, -\frac{1}{2})$ and $(x,y,+\frac{1}{2})$ is $\frac{d+1}{d}$ . So, for example, the coordinates of the projected irregular hexagon silhouette of cube $P$ are

$( -\frac{5}{2}, \frac{5}{2} )$
$( -\frac{5}{2}, \frac{3}{2} )$
$( -\frac{7}{2}, \frac{3}{2} )$
$( -\frac{7 (d+1)}{2 d}, \frac{3 (d+1)}{2 d} )$
$( -\frac{7 (d+1)}{2 d}, \frac{5 (d+1)}{2 d} )$
$( -\frac{5 (d+1)}{2 d}, \frac{5 (d+1)}{2 d} )$

which has the area of

$\dfrac{ 2d^2+12d+6}{2d^2}$

We do likewise for cubes $Q$ and $P$ , with areas

$\dfrac{ 2d^2+10d+6}{2d^2}$

$\dfrac{ 2d^2+14d+7}{2d^2}$

respectively

In determining the area the silhouette of cube $R$ , we only need to consider relative ratios of the areas of the silhouette at any plane of projection, so we can drop the term $2d^2$ .

See Shoelace formula for computing areas of irregular polygons given the coorindates of its vertices.

Since all the cubes are on a level surface, the sides of the bottom and top faces of each drawn cube stays constant, but the difference of where they are drawn in relation to each other is proportional to how far away from the origin the cube is. The center of the top square with a fixed side length of $a$ in the first quadrant that is $x$ units to the right and $y$ units up from the origin would be drawn at $(ax, ay)$ and vertices at $(a(x \pm \frac{1}{2}), a(y \pm \frac{1}{2}))$ .

Likewise, the bottom square with a fixed side length of $b$ would have vertices at $(b(x \pm \frac{1}{2}), b(y \pm \frac{1}{2}))$ .

This means that the area of hexagonal silhouette of the perspective cube is the difference between the area of the rectangle with sides $c = a(x + \frac{1}{2}) - b(x - \frac{1}{2})$ and $d = a(y + \frac{1}{2}) - b(y - \frac{1}{2})$ and the areas of the two opposite triangles, one with sides $c - a$ and $d - b$ , and the other with sides $c - b$ and $d - a$ . In other words, the area of the hexagon is $A = cd - \frac{1}{2}(c - a)(d - b) - \frac{1}{2}(c - b)(d - a)$ , and letting $p = \frac{1}{2}(a^2 - b^2)$ and $q = \frac{1}{2}(a^2 + b^2)$ this simplifies to $A = p(x + y) + q$ .

A similar argument can be made for each quadrant to give the general formula of $A = pd + q$ , where $d = |x| + |y|$ . Surprisingly, this means that the hexagonal area is a linear function based on its taxicab distance from the origin.

In this problem, hexagon $P$ has an area of $993$ with a taxicab distance of $d_P = |-3| + |2| = 5$ from the origin, and hexagon $Q$ has an area of $901$ with a taxicab distance of $d_Q = |1| + |-3| = 4$ from the origin. Using the above equation, we get $993 = 5p + q$ and $901 = 4p + q$ , which solves to $p = 92$ and $q = 533$ . Therefore, hexagon $R$ , with a taxicab distance of $d_R = |5| + |1| = 6$ from the origin, has an area of $A = 6p + q = 6 \cdot 92 + 533 = \boxed{1085}$ .