Let's determine the angles of triangle $ABC$, given that the two trisection points of side $AB$ and the two quartering points of side $BC$ closer to the vertices form a cyclic quadrilateral, whose circumcircle is tangent to side $CA$.

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Determine the angles of triangle $ABC$ if we know that the two trisection points of side $AB$ and the two quartering points of side $BC$ closer to the vertices form a cyclic quadrilateral, whose circumcircle is tangent to side $CA$.
Solution. Using the notation of the figure, we denote the sides with the appropriate lowercase letters. The quadrilateral $D E F G$ is a cyclic quadrilateral according to the problem's conditions, such that the circumcircle touches the side $A C$ of the triangle. Therefore, the perpendicular bisectors of the sides of the quadrilateral intersect at a single point. Since $D$ and $E$ are points of trisection, $A D=E B=\frac{c}{3}$, so the midpoints of the segments $A B$ and $D E$ coincide. Similarly, the midpoints of the segments $B C$ and $F G$ also coincide. Therefore, the center $O$ of the circumcircle $(k)$ of the quadrilateral $D E F G$ is also the center of the circumcircle of the triangle $A B C$, as it lies on the perpendicular bisectors of the sides $A B$ and $B C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_cccaa9c722fccd9740cbg-1.jpg?height=520&width=553&top_left_y=480&top_left_x=775)

The circle $(k)$ touches the side $A C$ at point $H$. Then $O H \perp A C$. Since $O H$ passes through the center of the circumcircle of the triangle $A B C$ and is perpendicular to $A C$, $O H$ is the perpendicular bisector of the side $A C$, so $H$ is the midpoint of the side $A C$.

Let's write down the power of points $B$ and $C$ with respect to the circle $(k)$ in two ways.

For point $B$: $\frac{2 c^{2}}{9}=\frac{3 a^{2}}{16}, \quad$ for point $C$: $\frac{b^{2}}{4}=\frac{3 a^{2}}{16}$.

Rearranging: $b^{2}=\frac{3 a^{2}}{4}$ and $c^{2}=\frac{27 a^{2}}{32}$. From this, we can determine the ratio of the squares of the sides:

$$
a^{2}: b^{2}: c^{2}=1: \frac{3}{4}: \frac{27}{32}
$$

and the ratio of the sides:

$$
a: b: c=1: \frac{\sqrt{3}}{2}: \frac{3 \sqrt{3}}{4 \sqrt{2}}
$$

From this, considering the length of side $a$ as a unit, we can calculate the angles of the triangle using the cosine rule:

$$
\begin{aligned}
c^{2} & =a^{2}+b^{2}-2 a b \cdot \cos \gamma \\
\frac{27}{32} & =1+\frac{3}{4}-2 \cdot \frac{\sqrt{3}}{2} \cos \gamma \\
b^{2} & =a^{2}+c^{2}-2 a c \cdot \cos \beta \\
\frac{3}{4} & =1+\frac{27}{32}-2 \cdot \frac{3 \sqrt{3}}{4 \sqrt{2}} \cos \beta
\end{aligned}
$$

From these, the values of the angles of the triangle, rounded to two decimal places, are: $\gamma=58.45^{\circ}, \beta=53.46^{\circ}$ and $\alpha=180^{\circ}-$ $\gamma-\beta=60.09^{\circ}$.