Task B-3.8. The lengths of the sides of the base of a right triangular prism are $6 \mathrm{~cm}, 8 \mathrm{~cm}$, and $4 \sqrt{6}$ $\mathrm{cm}$. A plane is drawn through the vertex of the largest angle of the base such that the intersection of the prism and the plane is an equilateral triangle. Calculate the area of this intersection.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-19.jpg?height=770&width=658&top_left_y=542&top_left_x=690)

Let $|BC|=8 \text{ cm}, |AC|=6 \text{ cm}, |AB|=4 \sqrt{6} \text{ cm}, |AM|=x \text{ cm}, |BN|=y \text{ cm}$.

Since $|MC|=|NC|=|MN|$, we have

$$
x^{2}+36=y^{2}+64=(y-x)^{2}+96
$$

which gives us the system of equations

$$
\left\{\begin{array}{l}
x^{2}-y^{2}=28 \\
2 x y - x^{2}=32
\end{array}\right.
$$

By substituting $\frac{y}{x}=z$, we get

$$
\left\{\begin{array}{l}
x^{2}-x^{2} z^{2}=28 \\
2 x^{2} z - x^{2}=32
\end{array}\right.
$$

Dividing the equations, we get

$$
\frac{1-z^{2}}{2 z-1}=\frac{7}{8}
$$

which simplifies to $8 z^{2}+14 z-15=0$, giving us $z_{1}=\frac{3}{4}$ and $z_{2}=-\frac{5}{2}$.

Since $x>0$ and $y>0$, we have $z=\frac{y}{x}>0$.

Thus, for $z=\frac{3}{4}$, we have

$$
\left\{\begin{array}{l}
y=\frac{3}{4} x \\
x^{2}-y^{2}=28
\end{array}\right.
$$

The positive solutions are $x=8 \text{ cm}, y=6 \text{ cm}$.

(1 point)

From triangle $AMC$, we have $|CM|^{2}=36+64$ or $|CM|=10 \text{ cm}$, so $P=25 \sqrt{3} \text{ cm}^{2}$.

(2 points)

## ZUPANIJSKO NATJECANJE IZ MATEMATIKE

## 4th grade - high school - B variant

## February 28, 2011

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.