Question

A cube with side length $10$ is suspended above a plane. The vertex closest to the plane is labeled $A.$ The three vertices adjacent to vertex $A$ are at heights $10,$ $11,$ and $12$ above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r - \sqrt{s}}{t},$ where $r,$ $s,$ and $t$ are positive integers. Find $r + s + t.$

Answer

Solution:

Let $h$ be the height of $A,$ and let $e_1,$ $e_2,$ $e_3$ be unit vectors along the three mutually perpendicular edges at $A.$ If $u$ is the upward unit normal of the plane, the height of the vertex along edge $i$ is $h + 10(e_i \cdot u),$ so $10(e_1 \cdot u) = 10 - h,$ $10(e_2 \cdot u) = 11 - h,$ and $10(e_3 \cdot u) = 12 - h.$ Because $e_1, e_2, e_3$ form an orthonormal basis, $(e_1 \cdot u)^2 + (e_2 \cdot u)^2 + (e_3 \cdot u)^2 = 1.$

Therefore $(10 - h)^2 + (11 - h)^2 + (12 - h)^2 = 100,$ which simplifies to $3h^2 - 66h + 265 = 0,$ so $h = \frac{33 \pm \sqrt{33^2 - 3 \cdot 265}}{3} = \frac{33 \pm \sqrt{294}}{3}.$

Since $A$ is the closest vertex to the plane, $h \lt 10,$ forcing $h = \frac{33 - \sqrt{294}}{3},$ and $r + s + t = 33 + 294 + 3 = 330.$

2011 AIME I Problem 13

Solution:

Problem 13 in Other Years