8,9

The areas of the projections of a certain triangle onto the coordinate planes Oxy and Oyz are $\sqrt{6}$ and $\sqrt{7}$, respectively, and the area of the projection onto the plane $O x z$ is an integer. Find the area of the triangle itself, given that it is also an integer.
Let the vector perpendicular to the plane of the original triangle form angles $\alpha, \beta$, and $\gamma$ with the coordinate axes $O x, O y$, and $O z$ respectively. Then

$$
\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=1
$$

Let the area of the original triangle be denoted by $S$, and the areas of the projections on the coordinate planes $O y z, O x z$, and $O x y$ be denoted by $S_{x}, S_{y}$, and $S_{z}$ respectively ( $S_{z}=\sqrt{6}, S_{x}=\sqrt{7}$ ). According to the theorem of the area of the projection of a plane figure onto a plane,

$$
\sqrt{7}=S_{x}=S|\cos \alpha|, S_{y}=S|\cos \beta|, \sqrt{6}=S_{z}=S|\cos \gamma|
$$

Then

$$
7+6+S_{y} 2=S 2 \cos 2 \alpha+S 2 \cos 2 \beta+S 2 \cos 2 \gamma=S 2(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma)=S 2
$$

Therefore,

$$
S 2-S_{y} 2=13,\left(S-S_{y}\right)\left(S+S_{y}\right)=13
$$

Since both factors on the left side of the equation are positive integers, and the second factor is greater than the first, the equality is possible only in the case when

$$
\left\{\begin{array}{l}
S-S_{y}=1 \\
S+S_{y}=13
\end{array}\right.
$$

From this system, we find that $S_{y}=7$.

## Answer

On the edges $A 1 B 1, A B, A 1 D 1$, and $D D 1$ of the unit cube $A B C D A 1 B 1 C 1 D 1$, points $K, L, M$, and $N$ are taken respectively, such that $A 1 K=\frac{2}{3}, A L=\frac{1}{5}, A 1 M=\frac{1}{3}$. Determine which of the edges $A 1 D 1$ or $D 1 C 1$ is intersected by the plane parallel to the segment $M L$ and containing the segment $K N$. In what ratio does this edge divide the plane?

## Solution

Choose a coordinate system with the origin at point $A 1$. The $x$-axis is directed along the ray $A 1 D 1$, the $y$-axis along the ray $A 1 B 1$, and the $z$-axis along the ray $A 1 A$. Then the coordinates of points $K, L, M$, and $N$ are:

$$
K\left(0 ; \frac{2}{3} ; 0\right), L\left(0 ; \frac{1}{5} ; 1\right), M\left(\frac{1}{3} ; 0 ; 0\right), N(1 ; 0 ; t)
$$

where $0 \leqslant t \leqslant 1$. From point $L$, lay off the vector

$$
\overrightarrow{L P}=\overrightarrow{M K}=\left(0-\frac{1}{-3} ; 3-0 ; 0-0\right)=\left(-\frac{1}{3} ; \frac{2}{3} ; 0\right)
$$

Then point $P$ has coordinates $\left(-\frac{1}{3} ; \frac{13}{1} ; 1\right)$. The desired plane passes through points $K, P$, and $N$. We seek the equation of this plane in the form $a x+b y+c z=1$. Substituting the coordinates of points $K, P$, and $N$ into this equation, we find that

$$
c=\frac{1}{10(t+3)}, a=\frac{9+30}{10(t+3)}, b=\frac{3}{2}
$$

After obvious simplifications, we obtain the equation $(9 t+30) x+15(t+3) y+z=10(t+3)$. Substituting zeros for $y$ and $z$, we find the point of intersection of the plane with the line $A 1 D 1-x=\frac{10(t+3)}{3(3 t+10)}$, and since $0 \leqslant t \leqslant 1$, then $1 \leqslant x \leqslant \frac{40}{39}$. Therefore, the cutting plane does not intersect the edge $A 1 D 1$. Substituting $x=1$ and $z=0$ into the obtained equation, we find the point of intersection of the plane with the line $D 1 C 1-y=\frac{1}{15(t+3)}$, and since $0 \leqslant t \leqslant 1$, then $0 \leqslant y \leqslant \frac{1}{60}$. Therefore, the cutting plane intersects the edge $D 1 C 1$ and in this case, the point of intersection divides this edge in any ratio from 0 to $\frac{1}{59}$, counting from vertex $D 1$.

## Answer

$D 1 C 1$; in any ratio from 0 to $\frac{1}{59}$, counting from vertex $D 1$.