The point $N$ is the centre of the face $A B C D$ of the cube $A B C D E F G H$, as shown. Also, $M$ is the midpoint of the edge $A E$. If the area of $\triangle M N H$ is $13 \sqrt{14}$, what is the edge length of the cube?

![](https://cdn.mathpix.com/cropped/2024_04_17_d0cbae19e125c7a52478g-5.jpg?height=474&width=507&top_left_y=2018&top_left_x=863)
If we set $M E=x$, then the side length of the cube is $2 x$.

The side $M H$ of $\triangle M N H$ is the hypotenuse of right-angled $\triangle E M H$.

By the Pythagorean theorem, $M H^{2}=M E^{2}+E H^{2}=x^{2}+(2 x)^{2}=5 x^{2}$, and since $M H>0$ and $x>0$, we have $M H=\sqrt{5} x$.

The diagonals $A C$ and $B D$ each have length $\sqrt{(2 x)^{2}+(2 x)^{2}}=2 \sqrt{2} x$ by the Pythagorean theorem.

Since $A B C D$ is a square, the point $N$ is the intersection of $A C$ and $B D$, and is the midpoint of each line.

Therefore, $A N=D N=\sqrt{2} x$.

The side $M N$ of $\triangle M N H$ is the hypotenuse of right-angled $\triangle N A M$, so

$$
M N^{2}=A N^{2}+A M^{2}=(\sqrt{2} x)^{2}+x^{2}=3 x^{2} .
$$

Since $M N>0$, we have $M N=\sqrt{3} x$.

The side $N H$ of $\triangle M N H$ is the hypotenuse of $\triangle D N H$, so

$$
N H^{2}=D N^{2}+D H^{2}=(\sqrt{2} x)^{2}+(2 x)^{2}=6 x^{2} .
$$

Since $N H>0$, we have $N H=\sqrt{6} x$.

We now have that the side lengths of $\triangle M N H$ are $\sqrt{5} x, \sqrt{3} x$, and $\sqrt{6} x$, and that its area is $13 \sqrt{14}$.

There are several ways to compute the area of a triangle in terms of its side lengths.

We will use the cosine law to compute $\cos \angle M N H$, then use the Pythagorean identity to compute $\sin \angle M N H$.

Once $\sin \angle M N H$ is known, the area can be computed as $\triangle M N H$ as $\frac{1}{2}(M N)(N H) \sin \angle M N H$. By the cosine law, we have

$$
M H^{2}=M N^{2}+N H^{2}-2(M N)(N H) \cos \angle M N H .
$$

Substituting the values in terms of $x$ of $M H, M N$, and $N H$, we have

$$
5 x^{2}=3 x^{2}+6 x^{2}-2(\sqrt{3} x)(\sqrt{6} x) \cos \angle M N H
$$

which can be rearranged to get

$$
\cos \angle M N H=\frac{4 x^{2}}{2 \sqrt{18} x^{2}}=\frac{4}{6 \sqrt{2}}=\frac{\sqrt{2}}{3} .
$$

Since $\angle M N H$ is an angle of a triangle, we have $\sin \angle M N H>0$.

By the Pythagorean identity, we have

$$
\sin \angle M N H=\sqrt{1-\cos ^{2} \angle M N H}=\sqrt{1-\frac{2}{9}}=\frac{\sqrt{7}}{3} .
$$

Therefore, the area of $\triangle M N H$ is

$$
\frac{1}{2}(\sqrt{3} x)(\sqrt{6} x) \frac{\sqrt{7}}{3}=\sqrt{\frac{7}{2}} x^{2} .
$$

We are given that the area of $\triangle M N H$ is $13 \sqrt{14}$, which means $13 \sqrt{14}=\sqrt{\frac{7}{2}} x^{2}$.

Rearranging this equation gives $13 \sqrt{2} \sqrt{7}=\frac{\sqrt{7}}{\sqrt{2}} x^{2}$ and then $x^{2}=26$.

Therefore, $x=\sqrt{26}$, so the side length is $2 x=2 \sqrt{26}$.

ANSWER: $2 \sqrt{26}$

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)