## Task B-4.6.

Let $A B C$ be a right triangle with acute angles $\varangle C A B=30^{\circ} \text{ and } \varangle A B C=60^{\circ}$, and let point $T$ be its centroid. Determine the cosine of the angle $\varangle A T B$. In what ratio are the area of triangle $A B C$ and the area of triangle $A T B$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-29.jpg?height=754&width=1145&top_left_y=597&top_left_x=410)

Let $|AB| = c$ and $\angle ATB = \varphi$. Then $|BC| = \frac{c}{2}$ and $|AC| = \frac{c \sqrt{3}}{2}$.

From the right triangle $ACD$, we get $t_{a}^{2} = \left(\frac{c \sqrt{3}}{2}\right)^{2} + \frac{c^{2}}{16} = \frac{13 c^{2}}{16}, t_{a} = \frac{c \sqrt{13}}{4}$.

From the right triangle $BCE$, we get $t_{b}^{2} = \left(\frac{c \sqrt{3}}{4}\right)^{2} + \frac{c^{2}}{4} = \frac{7 c^{2}}{16}, t_{b} = \frac{c \sqrt{7}}{4}$.

Applying the cosine rule to triangle $ATB$, we get:

$$
c^{2} = \left(\frac{2}{3} t_{a}\right)^{2} + \left(\frac{2}{3} t_{b}\right)^{2} - 2 \cdot \frac{2}{3} t_{a} \cdot \frac{2}{3} t_{b} \cos \varphi
$$

Substituting $(*)$ and $(**)$ into $(***)$, we get:

$$
\begin{aligned}
c^{2} & = \frac{4}{9} \cdot \frac{13 c^{2}}{16} + \frac{4}{9} \cdot \frac{7 c^{2}}{16} - 2 \cdot \frac{4}{9} \cdot \frac{c \sqrt{13}}{4} \cdot \frac{c \sqrt{7}}{4} \cos \varphi \\
c^{2} & = \frac{5}{9} c^{2} - c^{2} \cdot \frac{\sqrt{91}}{18} \cos \varphi \\
\cos \varphi & = -\frac{8}{\sqrt{91}}
\end{aligned}
$$

Then $P_{ABC} = \frac{c^{2} \sqrt{3}}{8}$ and

$$
\begin{aligned}
& P_{ATB} = \frac{1}{2} \cdot \frac{2}{3} t_{a} \cdot \frac{2}{3} t_{b} \sin \varphi = \frac{2}{9} \cdot \frac{c \sqrt{7}}{4} \cdot \frac{c \sqrt{13}}{4} \cdot \sqrt{1 - \left(-\frac{8}{\sqrt{91}}\right)^{2}} \\
& P_{ATB} = \frac{c^{2} \sqrt{91}}{72} \cdot \frac{3 \sqrt{3}}{\sqrt{91}} = \frac{c^{2} \sqrt{3}}{24}
\end{aligned}
$$

The desired ratio of areas is

$$
\frac{P_{ABC}}{P_{ATB}} = \frac{\frac{c^{2} \sqrt{3}}{8}}{\frac{c^{2} \sqrt{3}}{24}} = \frac{3}{1}
$$

Note: If a student uses the fact that the medians divide the triangle into 6 triangles of equal area, then the desired ratio is $6:2$, or $3:1$, assign 3 points.