The $A B C D E F G H$ cube with unit edge has a point $P$ that trisects the face diagonal $B E$ closer to $E$. How far is $P$ from the plane of the triangle $C F H$? (See the diagram)

![](https://cdn.mathpix.com/cropped/2024_05_02_3dda5ed74f40e5497fecg-1.jpg?height=309&width=331&top_left_y=255&top_left_x=886)
I. solution. Rotate the cube $120^{\circ}$ around the body diagonal $A G$. We know that in this case it maps onto itself, similarly, the equilateral triangles $H F C$ and $E D B$ also map onto themselves.

![](https://cdn.mathpix.com/cropped/2024_05_02_f4872318468eb4c287aag-1.jpg?height=385&width=401&top_left_y=230&top_left_x=865)

Figure 1

Therefore, the plane $E D B$ is parallel to the plane $H F C$, so instead of determining the distance of point $P$ from the plane $H F C$, it is sufficient to determine the distance between the two parallel planes. The centroids of the equilateral triangles $H F C$ and $E D B$ coincide with their centers of gravity and lie on the body diagonal $A G$, which is perpendicular to both planes. Let's calculate the distance between these two centroids. Let the vectors from $A$ to points $B, D$, and $E$ be $\mathbf{x}, \mathbf{y}$, and $\mathbf{z}$, and the centroids be $S_{1}$ and $S_{2}$ (Figure 1).

According to the known relationship,

$$
\begin{gathered}
\overrightarrow{A G}=\mathbf{x}+\mathbf{y}+\mathbf{z}, \quad \overrightarrow{A S_{1}}=\frac{\mathbf{x}+\mathbf{y}+\mathbf{z}}{3} \\
\overrightarrow{A S_{2}}=\frac{\overrightarrow{A C}+\overrightarrow{A F}+\overrightarrow{A H}}{3}=\frac{(\mathbf{x}+\mathbf{y})+(\mathbf{x}+\mathbf{z})+(\mathbf{y}+\mathbf{z})}{3}=\frac{2(\mathbf{x}+\mathbf{y}+\mathbf{z})}{3}
\end{gathered}
$$

Thus,

$$
\overrightarrow{S_{1} S_{2}}=\frac{\mathbf{x}+\mathbf{y}+\mathbf{z}}{3}=\frac{1}{3} \overrightarrow{A G}
$$

This implies that the distance between the centroids is $\frac{1}{3}$ of the body diagonal's length, which is $\frac{\sqrt{3}}{3}$.

Henrietta Somlai (Pápa, Reformed Gymnasium, 11th grade)

II. solution. Place the cube in a three-dimensional coordinate system such that the vertex $D$ is at the origin, and the face $A B C D$ lies in the $x, y$ plane in the first octant (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_f4872318468eb4c287aag-1.jpg?height=395&width=491&top_left_y=1483&top_left_x=814)

Figure 2

Write down the coordinates of points $H, F, C$, and $P$:

$$
H(0 ; 0 ; 1), \quad F(1 ; 1 ; 1), \quad C(0 ; 1 ; 0), \quad P\left(1 ; \frac{1}{3} ; \frac{2}{3}\right)
$$

Next, write down the equation of the plane passing through points $H, F, C$. Its general form is

$$
a x+b y+c z+d=0
$$

Substituting the coordinates of points $H, F$, and $C$:

$$
c+d=0, \quad a+b+c+d=0, \quad b+d=0, \quad C:
$$

From the system of equations, we get $a=d, b=-d, c=-d$, so if $d \neq 0$, the equation of the plane passing through points $H, F, C$ is:

$$
d x-d y-d z+d=0
$$

Dividing by $d \neq 0$ gives

$$
x-y-z+1=0
$$

According to the known relationship, the distance of point $P\left(x_{1} ; y_{1} ; z_{1}\right)$ from the plane $a x+b y+c z+d=0$ is

$$
\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}
$$

Substituting, the distance is $\frac{\left|1-\frac{1}{3}-\frac{2}{3}+1\right|}{\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, as we also saw in the first solution.

András L. László (Veszprém, Industrial and Grammar School, 10th grade)