Task B-3.5. (20 points) The heights of triangle $ABC$ are in the ratio $v_{a}: v_{b}: v_{c}=$ $6: 2 \sqrt{3}: 3$, and the circumference of the circumscribed circle is $8 \pi \mathrm{cm}$. Determine the lengths of the sides and the measures of the angles of triangle $ABC$.
Solution. From $a v_{a}=b v_{b}$ we get $a: b=v_{b}: v_{a}=2 \sqrt{3}: 6=1: \sqrt{3}$, and from $b: c=v_{c}: v_{b}=$ $3: 2 \sqrt{3}=\sqrt{3}: 2$ so $a: b: c=1: \sqrt{3}: 2$. Let $a=k, b=\sqrt{3} k, c=2 k$.

(4 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_9c56249806d755c5106ag-12.jpg?height=611&width=606&top_left_y=368&top_left_x=728)

Now we have

$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3 k^{2}+4 k^{2}-k^{2}}{2 \cdot \sqrt{3} \cdot 2 k^{2}}=\frac{\sqrt{3}}{2}
$$

from which $\alpha=30^{\circ}$,

(4 points)

and from

$$
\cos \beta=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{k^{2}+4 k^{2}-3 k^{2}}{2 \cdot 1 \cdot 2 k^{2}}=\frac{1}{2}
$$

we get $\beta=60^{\circ}$.

(4 points)

From this we obtain $\gamma=90^{\circ}$, so the triangle is right-angled. The radius of the circumscribed circle is $r=\frac{c}{2}$, and since its circumference is $8 \pi=2 r \pi$ we get $r=4$ and $c=8$.

## COUNTY COMPETITION IN MATHEMATICS

## 4th grade - high school - B category

## February 23, 2009.