## Task A-1.6.

In triangle $A B C$ with centroid $T$, it is given that $|A T|=|B C|$ and $\varangle B C T=30^{\circ}$. Determine $|C T|$ : $|A T|$.
## Solution.

Let the midpoint of side $\overline{B C}$ be $P$, and let $a$ be the length of segment $\overline{B C}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_21f39afeb1dfc245d841g-05.jpg?height=880&width=649&top_left_y=1296&top_left_x=635)

From the statement of the problem, we have $|A T|=|B C|=a$.

Since $T$ is the centroid of the triangle, from $|A T|:|P T|=2: 1$ it follows that $|P T|=\frac{a}{2}$.

Since $P$ is the midpoint of side $\overline{B C}$, we have $|C P|=|B P|=\frac{a}{2}$. Since $|P T|=\frac{a}{2}$ as well, we conclude that the midpoint of side $\overline{B C}$ is the center of the circumscribed circle of triangle $TBC$, so this triangle is right-angled.

Furthermore, since $\varangle B C T=30^{\circ}$ and angle $\varangle C T B$ is a right angle, we get that $\varangle T B C=60^{\circ}$. 1 point Together with $|C P|=|T P|$, this means that triangle $B P T$ is equilateral, so $|B T|=\frac{a}{2}$. 2 points Applying the Pythagorean theorem in triangle $C T B$, we have

$$
|C T|=\sqrt{|B C|^{2}-|B T|^{2}}=\sqrt{a^{2}-\left(\frac{a}{2}\right)^{2}}=\frac{a \sqrt{3}}{2}
$$

Finally, we have $|A T|:|C T|=a: \frac{a \sqrt{3}}{2}=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$.

1 point