Triangle $\triangle ABC$ is an equilateral triangle whose side is $16$. A circle meets the sides of the triangle at $6$ points:
Find $|DE|$ using plane geometry concepts.
On
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.
$$|DE| = 2 \sqrt{r^2-y_0^2} = 2 \sqrt{22}$$
On
The above solutions are fine if the question is asking for $|DE|$ only.
However, this is not the end of the story because ....
After $|DE| = 2 \sqrt{22}$ has been found, we still need to find
(1) the correct value of $|BD|$ (which was found to be $10 \pm \sqrt{22}$)
---After testing,
----$|BD|$ shoud be $10 - \sqrt{22}$ only. [Not just simply take it for granted to accept the positive root.]
(2) the correct value of $|CE|$ (which was found to be $6 \pm \sqrt{22}$)
----$|CE|$ shoud be $6 - \sqrt{22}$ only.
The "Power of a Point" (with respect to a circle) Theorem ---in particular, its incarnation as the Secant-Secant Power Theorem--- comes in handy here. Relative to the six-point circle of this problem, the theorem states that $$\begin{align} |AH|\cdot|AJ| &= |AG|\cdot|AF| &(=\text{power of }A) \\ |BD|\cdot|BE| &= |BJ|\cdot|BH| &(=\text{power of }B) \\ |CF|\cdot|CG| &= |CE|\cdot|CD| &(=\text{power of }C) \end{align}$$
Substituting in known values ... $$\begin{align} 3\cdot(3+7) &= 2\cdot(2+13) &= 30\\ |BD|\cdot|BE| &= 6\cdot(6+7)&= 78\\ 1\cdot(1+13) &= |CE|\cdot|CD| &= 14 \end{align}$$ (The first equation isn't relevant to the problem at hand, but it's a nice check.)
Therefore, assuming the point ordering $B, D, E, C$, we have $$\begin{align} 78 &= |BD|\cdot\left(|BC|-|CE|\right) = |BD|\cdot\left(16-|CE|\right) \\ 14 &= |CE|\cdot\left(|BC|-|BD|\right) = |CE|\cdot\left(16-|BD|\right) \end{align}$$ Subtracting the second equation from the first gives $$64 = 16 \left( |BD|- |CE| \right) \qquad \implies \qquad |CE| = |BD| - 4$$ Thus, $$\begin{align} 78 = |BD|\cdot \left( 20 - |BD| \right) \qquad &\implies \qquad |BD| = 10 \pm \sqrt{22} \\ &\implies \qquad |CE| = \phantom{1}6 \pm \sqrt{22} \end{align}$$ so that $$\begin{align} |BD|+|DE|+|EC|=|BC| = 16 \qquad &\implies \qquad |DE| = 16 - \left( 16 \pm 2\sqrt{22} \right) \\ &\implies \qquad |DE| = 2\sqrt{22} \end{align}$$ (since we presumably want a non-negative $|DE|$).
I like that this solution begins with one of my favorite geometric theorems, but the ensuing algebra seems un-inspiringly mechanical. I suspect that there's a solution maintaining a more-geometric spirit throughout.