5. Given is a triangle $\triangle A B C$ such that $|\angle B A C|+|\angle C B A|=135^{\circ}$. Point $N$ is the foot of the altitude from vertex $A$ such that $|N C|=6 \cdot|N B|$. Calculate the length of side $\overline{B C}$ if the area of triangle $\triangle A B C$ is $3024 \mathrm{~mm}^{2}$.

## COUNTY COMPETITION IN MATHEMATICS
5. Given is a triangle $\triangle A B C$ such that $|\angle B A C|+|\angle C B A|=135^{\circ}$. Point $N$ is the foot of the altitude from vertex $A$ for which $|N C|=6 \cdot|N B|$. Calculate the length of side $\overline{B C}$ if the area of triangle $\triangle A B C$ is 3024 $\mathrm{mm}^{2}$.

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5cfeb7f7e539b150a939g-15.jpg?height=565&width=596&top_left_y=543&top_left_x=730)

Sketch:

$1 \text{ POINT}$

$|\angle B A C|+|\angle C B A|=135^{\circ}$

$|\angle B A C|+|\angle C B A|+|\angle A C B|=180^{\circ}$

1 POINT

$|\angle A C B|=45^{\circ}$

1 POINT

The length $\overline{A N}$ is the altitude to side $\overline{B C}$, so $|\angle C N A|=90^{\circ}$.

It follows that $|\angle N A C|=45^{\circ}$ and $|A N|=|N C|$.

1 POINT

Let $|N B|=x$.

From $|N C|=6 \cdot|N B|$ it follows that $|A N|=|N C|=6 x,|B C|=7 x$.

$1 \text{ POINT}$

The area of triangle $\triangle A B C$ is:

$P=\frac{7 x \cdot 6 x}{2}$, or $P=(7 x \cdot 6 x): 2$.

1 POINT

$(7 x \cdot 6 x): 2=3024$

$42 x \cdot x=6048$ or $21 x \cdot x=3024 \quad 1$ POINT

$x \cdot x=144 \quad$ or $\quad x^{2}=144 \quad 1$ POINT

$144=2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3=(2 \cdot 2 \cdot 3) \cdot(2 \cdot 2 \cdot 3)=12 \cdot 12$

$x=12$

$1 \text{ POINT}$

$|B C|=7 x=84$

The length of side $\overline{B C}$ is $84 \mathrm{~mm}$.

$1 \text{ POINT}$

## COUNTY COMPETITION IN MATHEMATICS  March 29, 2021.