3. In Figure $1, \triangle A B C$ has points $D$ and $E$ on $A B$ and $A C$, respectively, so that $D E$ is parallel to $B C$. In this case, $\triangle A B C$ and $\triangle A D E$ are similar triangles because their corresponding angles are equal. These triangles thus have the property that $\frac{A D}{A B}=\frac{A E}{A C}=\frac{D E}{B C}$.
In Figure 2, $W X Y Z$ is a trapezoid with $W X$ parallel to $Z Y$. Also, points $M$ and $N$ and are on $W Z$ and $X Y$, respectively, with $M N$ parallel to $W X$ and $Z Y$.
(a) Suppose that, in Figure 1, $D E=6, B C=10, E C=3$, and $A E=x$. Determine the value of $x$.
(b) Suppose that, in Figure 2, $\frac{W X}{Z Y}=\frac{3}{4}$ and $\frac{W M}{M Z}=\frac{X N}{N Y}=\frac{2}{3}$. Determine the value of $\frac{W X}{M N}$.
(c) Suppose that, in Figure 2, $\frac{W X}{Z Y}=\frac{3}{4}$. Suppose also that $\frac{M Z}{W M}=\frac{N Y}{X N}$ and that this ratio is equal to a positive integer. If $W X+M N+Z Y=2541$ and the length of each of $W X, M N$ and $Z Y$ is an integer, determine all possible lengths of $M N$.
3. (a) In Figure 1, $\frac{A E}{A C}=\frac{D E}{B C}$.
Since $E C=3$ and $A E=x$, then $A C=A E+E C=3+x$.
Thus, $\frac{x}{3+x}=\frac{6}{10}$.
Therefore, $10 x=6(3+x)$ or $10 x=18+6 x$ and so $4 x=18$ which gives $x=\frac{9}{2}$.
(b) Since $\frac{W M}{M Z}=\frac{2}{3}$, then $W M=2 r$ and $M Z=3 r$ for some real number $r>0$.
Extend $Z W$ and $Y X$ to meet at point $P$ above $W X$. (We know that we can do this because $W X0$.

Note that each of $W X=3 k$ and $Z Y=4 k$ is an integer.
Since $k=Z Y-W X$, then $k$ itself is an integer.
Since $\frac{M Z}{W M}=\frac{N Y}{X N}$ and this ratio is an integer, then we can set $\frac{M Z}{W M}=t$ for some positive integer $t$.
If $W M=p$ for some real number $p>0$, then $M Z=p t$.
As in (b), we extend $Z W$ and $Y X$ to meet at $P$ above $W X$
We have again created a new version of the configuration in Figure 1.
Since $W X$ is parallel to $Z Y$, then $\triangle P W X$ and $\triangle P Z Y$ are similar, which means that $\frac{P W}{P Z}=\frac{W X}{Z Y}$.
Using the information determined so far,
$$
\begin{aligned}
\frac{P W}{P Z} & =\frac{W X}{Z Y} \\
\frac{P W}{P W+W+M Z} & =\frac{3}{4} \\
\frac{P W}{P W+p+p t} & =\frac{3}{4} \\
4 \times P W & =3(P W+p(t+1)) \\
4 \times P W & =3 \times P W+3 p(t+1) \\
P W & =3 p(t+1)
\end{aligned}
$$

We can also note that since $W X$ is parallel to $M N$, then $\triangle P W X$ and $\triangle P M N$ are similar, which means that $\frac{P W}{P M}=\frac{W X}{M N}$.
Since $p \neq 0$, then
$$
\frac{W X}{M N}=\frac{P W}{P W+W M}=\frac{3 p(t+1)}{3 p(t+1)+p}=\frac{p(3 t+3)}{p(3 t+4)}=\frac{3 t+3}{3 t+4}
$$

Since $W X=3 k$, then $\frac{3 k}{M N}=\frac{3 t+3}{3 t+4}$ and so $M N=\frac{(3 t+4) 3 k}{3 t+3}=\frac{(3 t+4) k}{t+1}$.
Now $M N=\frac{(3 t+4) k}{t+1}=\frac{3(t+1) k+k}{t+1}=3 k+\frac{k}{t+1}$.
Since $M N$ is an integer and $k$ is an integer, then $\frac{k}{t+1}$ is an integer, which means that the integer $k$ is a multiple of the integer $t+1$, say $k=q(t+1)$ for some positive integer $q$. We are given that $W X+M N+Z Y=2541$.
Substituting, we obtain successively the equivalent equations
$$
\begin{aligned}
3 k+\left(3 k+\frac{k}{t+1}\right)+4 k & =2541 \\
10 k+\frac{k}{t+1} & =2541 \\
10 q(t+1)+q & =2541 \\
q(10 t+10+1) & =2541 \\
q(10 t+11) & =2541
\end{aligned}
$$

Now $q$ and $t$ are positive integers.
Since $t \geq 1$, then $10 t+11 \geq 21$ and $10 t+11$ has units digit 1 .
So $q$ and $10 t+11$ form a positive integer factor pair of 2541 with each factor having units digit 1. (Since 2541 and $10 t+11$ have units digit 1 , then $q$ must also have units digit 1.) Factoring 2541, we obtain
$$
2541=3 \times 847=3 \times 7 \times 121=3 \times 7 \times 11^{2}
$$

The positive divisors of 2541 are
$$
1,3,7,11,21,33,77,121,231,363,847,2541
$$

The divisor pairs consisting of two integers each with units digit 1 are
$$
2541=1 \times 2541=11 \times 231=21 \times 121
$$

Since $10 t+1 \geq 21$, then we can have
- $q=1$ and $10 t+11=2541$, which gives $t=253$
- $q=11$ and $10 t+11=231$, which gives $t=22$
- $q=21$ and $10 t+11=121$, which gives $t=11$
- $q=121$ and $10 t+11=21$, which gives $t=1$

We are asked for the possible lengths of
$$
M N=3 k+\frac{k}{t+1}=3 q(t+1)+q=q(3 t+3+1)=q(3 t+4)
$$

Using the values of $q$ and $t$ above, we obtain
$$
\begin{aligned}
\text { - } q=1 \text { and } t=253: M N=763 \\
\text { - } q=11 \text { and } t=22: M N=770 \\
\text { - } q=21 \text { and } t=11: M N=777 \\
\text { - } q=121 \text { and } t=1: M N=847
\end{aligned}
$$

Therefore, the possible lengths for $M N$ are $763,770,777,847$,
It is indeed possible to construct a trapezoid in each of these cases. The corresponding lengths of $W X$ are $762,759,756,726$, respectively, and the correponding lengths of $Z Y$ are $1016,1012,1008,968$, respectively. One way to try to construct these is by putting right