<s>[INST] Geometry problem on circles from a competition Triangle $\triangle ABC$ is an equilateral triangle whose side is $16$. A circle meets the sides of the triangle at $6$ points:


*

*it intersects $AC$ at $G$ and $F$ and $|AG|=2$, $|GF|=13$, $|FC|=1$.

*it intersects $AB$ at $H$ and $J$ and $|AH|=3$, $|HJ|=7$, $|JB|=6$.

*it intersects $BC$ at $D$ and $E$.


Find $|DE|$ using plane geometry concepts.
 [/INST] Suppose the triangle is drawn with $A$ as the topmost point, $B$ as the left base point and $C$ as the right base point. 
The equation for the circle is $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0, y_0)$ is the centre coordinates and $r$ is the radius. Assume a coordinate system with the origin at $B$. Now the coordinates for the points $G$, $H$ and $J$ are found to be $(9,7 \sqrt{3})$, $(13/2, 13 \sqrt{3}/2)$ and $(3, 7 \sqrt{3})$ respectively. Putting these numbers in the equation for the circle gives you the following equation system:
$$\left\{\begin{array}{l}
  (7 \sqrt{3} - y_0)^2 + (9 - x_0)^2 = r^2 \\
  (13 \sqrt{3} / 2 - y_0)^2 + (13/2 - x_0)^2 = r^2 \\
  (3 \sqrt{3} - y_0)^2 + (3 - x_0)^2 = r^2
\end{array}\right.$$
The solution (with positive $r$) is $x_0 = 10$, $y_0 = 3 \sqrt{3}$ and $r=7$.
Note that circle is completely defined by three points only. By plugging in the coordinates of the point $F$ into the circle equation one can verify that the above solution is correct (assuming the problem makers have done their job). 
Now, $|DE|$ is simply the difference between the two solutions of the resulting circle equation when $y=0$, i.e.
$$|DE| = 2 \sqrt{r^2-y_0^2} = 2 \sqrt{22}$$
</s>