In the diagram, $A B C D$ is a square. Point $F$ is on $A B$ with $B F=2 A F$. Point $E$ is on $A D$ with $\angle F E C=\angle B C E$. If $0^{\circ}<\angle E C D<45^{\circ}$, what is the value of $\tan (\angle E C D)$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_45c89f07dc9351e0476cg-3.jpg?height=374&width=401&top_left_y=987&top_left_x=922)
Solution 1

Suppose that $\angle E C D=\theta$.

Without loss of generality, suppose that the square has side length 3.

Therefore, $A B=A D=D C=3$.

Since $B F=2 A F$ and $A F+B F=A B=3$, then $A F=1$ and $B F=2$.

Suppose that $E D=x$. Since $A D=3$, then $A E=3-x$.

Since $\angle E C D=\theta$, then $\angle B C E=90^{\circ}-\theta$.

Since $\angle F E C=\angle B C E$, then $\angle F E C=90^{\circ}-\theta$.

Since $\triangle E D C$ is right-angled at $D$, then $\angle D E C=90^{\circ}-\angle E C D=90^{\circ}-\theta$.

Since $A E D$ is a straight line, then

$$
\angle A E F=180^{\circ}-\angle F E C-\angle D E C=180^{\circ}-2\left(90^{\circ}-\theta\right)=2 \theta
$$

![](https://cdn.mathpix.com/cropped/2024_04_30_a838ccd738aa6ca69034g-14.jpg?height=393&width=461&top_left_y=812&top_left_x=905)

In $\triangle E D C$, we see that $\tan \theta=\frac{E D}{D C}=\frac{x}{3}$.

In $\triangle E A F$, we see that $\tan 2 \theta=\frac{A F}{A E}=\frac{1}{3-x}$.

Since $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$, then

$$
\begin{aligned}
\frac{1}{3-x} & =\frac{2 x / 3}{1-(x / 3)^{2}} \\
\frac{1}{3-x} & =\frac{6 x}{9-x^{2}} \\
9-x^{2} & =(3-x)(6 x) \\
5 x^{2}-18 x+9 & =0 \\
(5 x-3)(x-3) & =0
\end{aligned}
$$

Thus, $x=3$ or $x=\frac{3}{5}$.

If $x=3$, then $E C$ is a diagonal of the square, which would make $\angle E C D=45^{\circ}$, which is not allowed.

Therefore, $x=\frac{3}{5}$ and so $\tan (\angle E C D)=\tan \theta=\frac{3 / 5}{3}=\frac{1}{5}$.

## Solution 2

Without loss of generality, suppose that the square has side length 3.

Therefore, $A B=A D=D C=3$.

Since $B F=2 A F$ and $A F+B F=A B=3$, then $A F=1$ and $B F=2$.

Extend $E F$ and $C B$ to meet at $G$. Let $E F=t$.

![](https://cdn.mathpix.com/cropped/2024_04_30_a838ccd738aa6ca69034g-15.jpg?height=770&width=417&top_left_y=455&top_left_x=865)

Now, $\triangle E A F$ is similar to $\triangle G B F$, since each is right-angled and $\angle A F E=\angle B F G$.

Since $B F: A F=2: 1$, then $G F: E F=2: 1$ and so $G F=2 t$.

Since $\angle F E C=\angle B C E$, then $\triangle G E C$ is isosceles with $G E=G C$.

Thus, $G B=G C-B C=G E-3=3 t-3$.

Using the Pythagorean Theorem in $\triangle B G F$, we obtain

$$
\begin{aligned}
2^{2}+(3 t-3)^{2} & =(2 t)^{2} \\
4+9 t^{2}-18 t+9 & =4 t^{2} \\
5 t^{2}-18 t+13 & =0 \\
(5 t-13)(t-1) & =0
\end{aligned}
$$

and so $t=\frac{13}{5}$ or $t=1$.

If $t=1$, then $E F=A F$ which means that point $E$ is at $A$, which is not possible since we are told that $\angle E C D<45^{\circ}$.

Thus, $t=\frac{13}{5}$.

Since $A F: E F=1: \frac{13}{5}=5: 13$, which means that $\triangle F A E$ is similar to a $5-12-13$ triangle, and so $A E=\frac{12}{5}$.

Since $A D=3$, then $E D=A D-A E=\frac{3}{5}$.

Finally, this means that $\tan (\angle E C D)=\frac{E D}{D C}=\frac{3 / 5}{3}=\frac{1}{5}$.

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)