In a circle, three pairwise intersecting chords are drawn. Each chord is divided into three equal parts by the points of intersection.

Find the radius of the circle if one of the chords is equal to $a$.

#
Prove that all three chords are equal to each other.

## Solution

From the theorem of the products of the segments of intersecting chords, it follows that all three chords are equal to each other. Therefore, the points of intersection of the chords are the vertices of an equilateral triangle with side $\frac{a}{3}$. Since equal chords are equidistant from the center of the circle, the center $O$ of this triangle coincides with the center of the given circle. The distance from point $O$ to each chord is $\frac{a \sqrt{3}}{18}$. By the Pythagorean theorem, the square of the desired radius is $(a / 2)^{2} + 3 a^{2} / 18^{2} = 7 a^{2} / 27$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-42.jpg?height=523&width=512&top_left_y=822&top_left_x=773)

## Answer

$\frac{a \sqrt{7}}{3 \sqrt{3}}$.