## 3695.11 – A Square in a Cube

The figure shows a cube with unit edge and four points $A, B, C, D$, placed on four edges at a distance $k$ from a vertex, where $0 \le k \le 1$.

(a) The quadrilateral $ABCD$ is planar. Justify this assertion.

(b) Find all values of $k$ that make $ABCD$ a square.

(c) What are the sides of those squares? Solution

(a) For students who are comfortable in 3-dimensional analytic space, it is easily checked that the equation of the plane containing ABCD is $x - y + 2 k z = k$. For others, we are groping for a good old-fashioned Euclidean justification. The old "as any fool can plainly see" is tempting. If any of you out there can help with this, please do so!

(b) $ABCD$ is a square when $AB = BC$. We express both those lengths in terms of $k$, set them equal and solve.

For $AB$ we see from the figure, part I, that it is the hypotenuse of an isosceles right triangle with legs of length $1 - k$. That makes $AB = (1 - k) \sqrt{2}$. See the figure, part II.

As for $BC$, it is the diagonal of a rectangle with sides of length $k \sqrt{2}$ and $1$. This makes $BC = \sqrt{2 k^2 + 1}$.

Setting these two equal, we find that

\begin{aligned} AB^2 &=& BC^2, \\ 2 (1 - k)^2 &=& 2 k^2 + 1, \\ 2 k^2 - 4 k + 2 &=& 2 k^2 + 1,\\ -4 k &=& -1. \end{aligned}

The only value of $k$ that works is 1/4.

(c) For $k = 1/4$, the side is

$AB = (1 - k) \sqrt{2} = \frac{3 \sqrt{2}}{4}.$

Check:

\begin{aligned} BC &=& \sqrt{2 k^2 + 1} \\ &=& \sqrt{2 \frac{1}{16} + 1} \\ &=& \sqrt{\frac{9}{8}} \\ &=& \frac{3}{2 \sqrt{2}} \\ &=& \frac{3 \sqrt{2}}{4}. \end{aligned} 