What fraction of the plane is covered by those points which are closer to one vertex of an equilateral triangle and farther from the other two vertices than the side of the equilateral triangle?
I. Solution: Let the side length of the equilateral triangle $ABC$ be $d$. The points closer to vertex $C$ than $d$ are inside the circle $c$ with radius $d$ centered at $C$ (on the concave side of the circle), while the points farther from vertices $A$ and $B$ than $d$ are outside the circles $a$ and $b$ with radius $d$ centered at $A$ and $B$ (on the convex side of the circles).

![](https://cdn.mathpix.com/cropped/2024_05_02_32cad16e717cb0db44ebg-1.jpg?height=512&width=588&top_left_y=313&top_left_x=779)

Therefore, denoting the intersection points of $c$ with $a$ and $b$ (other than $B$ and $A$) as $E$ and $D$ respectively (see the figure), the points satisfying all three conditions fill the interior of the region bounded by the arcs $EC$, $CD$, and the arc $DE$ that does not contain point $A$.

The angles $ACE$ and $BCD$ are $60^\circ$ because the triangles $ACE$ and $BCD$ are equilateral, so $ED$ is a diameter of $c$. Therefore, the area of the mentioned region is obtained by subtracting the areas of the segments cut off by the chords $EC$ and $CD$ from the area of the semicircle above $ED$. These segments are congruent because their radii are equal and the angles subtended by the chords at $A$ and $B$ are $60^\circ$. The area of each segment is the difference between one-sixth of the area of the circle with radius $d$ and the area of the equilateral triangle with side length $d$:

$$
\frac{d^2 \pi}{6} - \frac{d^2 \sqrt{3}}{4}
$$

Now, the area of the desired region is:

$$
\frac{d^2 \pi}{2} - 2\left(\frac{d^2 \pi}{6} - \frac{d^2 \sqrt{3}}{4}\right) = d^2\left(\frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) \approx 1.3896 d^2
$$

Hild Erzsébet (Békéscsaba, Lorántffy Zsuzsanna Secondary School, 2nd grade)

II. Solution: By drawing the diameters $AA'$ and $BB'$ and the chords $EB'$ and $A'D$ of the circle $c$, the region bounded in the first solution is divided into five parts. Since $\angle B'CE = \angle BCD = \angle ACE = \angle A'CD = 60^\circ$, the segments cut off by $EB'$ and $A'D$ are mapped by a $60^\circ$ rotation around $E$ and $D$ respectively to the previously examined segments $EC$ and $CD$. Thus, the area of the desired region is equal to the combined area of the equilateral triangles $ECB'$ and $CDA'$ and the sector $A'B'C$:

$$
2 \cdot \frac{d^2 \sqrt{3}}{4} + \frac{d^2 \pi}{6}
$$

Pósch Margit (Budapest XIV., Teleki Blanka Secondary School, 2nd grade)

Note: The points on the arcs bounding the desired region do not satisfy at least one of the conditions, but this does not affect the measure of the area.