A regular triangle's sides (in the same direction) were divided in the ratio $p: q$. By connecting the division points, a triangle is formed whose area is $\frac{19}{64}$ of the area of the original triangle. What is the ratio $p: q$? () Suggested by: Koncz Levente, Budapest
I. solution. The triangles cut off from the regular triangle $ABC$ are congruent (let their sides be $p$ and $q$, and one of their angles be $60^{\circ}-o s$), so their third sides are also equal, and thus the resulting triangle $EFG$ is also regular. Let the length of its side be denoted by $x$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b0dff191cd019b7d4fedg-1.jpg?height=430&width=439&top_left_y=300&top_left_x=840)

We apply the cosine rule in triangle $AEG$:

$$
x^{2}=p^{2}+q^{2}-2 p q \cdot \cos 60^{\circ}
$$

from which

$$
x^{2}=p^{2}+q^{2}-p q
$$

The area of triangle $ABC$ is $\frac{(p+q)^{2} \sqrt{3}}{4}$, and the area of the triangle formed by connecting the division points is $\frac{19}{64}$ of this, i.e.,

$$
\frac{x^{2} \sqrt{3}}{4}=\frac{19}{64}\left(\frac{(p+q)^{2} \sqrt{3}}{4}\right)
$$

Simplifying and substituting the previously obtained value of $x^{2}$:

$$
64\left(p^{2}+q^{2}-p q\right)=19(p+q)^{2}
$$

Performing the operations, we get

$$
45 p^{2}-102 p q+45 q^{2}=0
$$

Dividing by $3 q^{2} \neq 0$ and introducing the new variable $\frac{p}{q}=u$, we obtain the following quadratic equation:

$$
15 u^{2}-34 u+15=0
$$

From this, $u=\frac{p}{q}=\frac{5}{3}$, or $\frac{p}{q}=\frac{3}{5}$, depending on the direction in which we label the vertices of the inscribed triangle.

II. solution. The area of a triangle (with the usual notation) is $\frac{a b \cdot \sin \gamma}{2}$. In our case, the area of the "cut off" triangles $AEG$, $BFE$, and $CGF$ is $\frac{p}{p+q} \cdot \frac{q}{p+q}$ of the area of the original triangle, so $1-\frac{19}{64}=\frac{45}{64}=3 \frac{p q}{(p+q)^{2}}$, from which we obtain the equation $15 p^{2}-34 p q+15 q^{2}=0$ from the first solution. It can be seen that we get the same result for the ratio $p:q$ even if triangle $ABC$ is not regular.