<s>[INST] Geometry - Pentagon This is a tough program I have hard time to find the answer. Can anyone help me? Thank you very much in advance.
Problem - In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answer in simplest radical form.
This is a drawing for the problem:

 [/INST] Let $F$ be the intersection of $AC$ and $BE$. By angle chasing, we obtain the following picture:

Now $\triangle BAF$ and $\triangle BEA$ are similar, so 
$$BF:BA = BA:BE,$$
or 
$$\frac yx = \frac{x}{x+y}=\frac{1}{1+y/x}.\tag{1}$$
Solving that for $y$ in terms of $x$ we have 
$$\frac yx = \frac{\sqrt{5}-1}2.$$
Now apply Menelaus' Theorem to $\triangle AEF$ with $M$, $Z$, $B$ colinear we get
$$\frac{EM}{MA}\cdot \frac{ZA}{ZF}\cdot \frac{BF}{BE} = 1.\tag{2}$$
Note that $EM = MA$,
$$\frac{BF}{BE} = \frac{y}{x+y}=1-\frac{x}{x+y}=\frac{3-\sqrt{5}}2,$$
and
$$\frac{ZA}{ZF} = \frac{3}{y-3}.$$
Thus, (2) becomes
$$\frac{y-3}{3} = \frac{3-\sqrt{5}}2.$$
So
$$ y = \frac{9-3\sqrt{5}}{2}+3 = \frac{15-3\sqrt{5}}2.$$
And we get 
$$ x = \frac xy \cdot y = \frac{\sqrt{5}+1}2 \cdot \frac{15-3\sqrt{5}}2$$

Note: 


*

*I'm sure there are more elegant solutions.

*One can avoid Menelaus' Theorem by connecting $M$ with the midpoint $N$ of $AF$.

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