Find the value of $CE$.

Question

$ABCD$ is a square of side 1, the value of $FE$ is 1 and the points $A$, $C$, $E$ are collinear, as well as $B$, $F$, $E$. The question is to find the value of $CE$.

My teacher gave me this challenge but I can't solve this.

Answer 1

I would set up a coordinate system in which A is (0, 0), B is (0, 1), C is (1, 0), and D is (1, 1). Write E as (e, 0). The line from B to E, that is, through (0, 1) and (e, 0) can be written as y= 1- x/e. That line crosses the line x= 1 at (1, 1- 1/e) so F= (1, 1- 1/e). Now, we can calculate the length of FE: $\sqrt{(1-e)^2+ (0- 1+ 1/e)^2}= 1$.
So $(1- e)^2+ (1/e- 1)^2= e^2- 2e+ 1+ 1/e^2- 2e+ 1= e^2- 4e+ 1/e^2+ 2= 1$. We can write that as $e^2- 4e+ 1/e^2+ 1= 0$. Multiply through by $e^2$ to get $e^4- 4e^3+ e^2+ 1= 0$. Solve that quartic equation for e. (A quartic can have 4 complex roots but only a positive real root has a geometric meaning.)

Answer 2

Use the intercept theorem to find the length of FC and then apply the Pythagorean theorem for triangle FCE.

Answer 3

\begin{cases}\frac{CF}{AB}=\frac{CE}{AE}~................(1)\\CF^2+CE^2=FE^2~......(2)\end{cases} \begin{cases}\frac{CF}{1}=\frac{CE}{CE+1}~...............(1)\\CF^2+CE^2=1~..........(2)\end{cases}

\begin{equation}CE\approx0.883204\end{equation}