14. In $\triangle A B C$, $B D=D C$. There is a piece of cheese on side $A C$, located at the point closest to point $C$ among the four equal segments. On $A D$, there are three mirrors $W_{1}, W_{2}, W_{3}$, which divide $A D$ into four equal segments. A very suspicious mouse is crawling on $A B$ (from $A$ to $B$), with $A B=400$ meters. The mouse can observe the cheese when the mouse, a certain mirror, and the cheese are aligned. Due to its suspicion, the mouse hopes to see the cheese multiple times to ensure that the cheese has not been eaten by other mice before it reaches it. Therefore, in the first minute, it crawls forward 80 meters, in the second minute, it retreats 20 meters, in the third minute, it crawls forward 80 meters, in the fourth minute, it retreats 20 meters, and so on. When the mouse reaches point $B$, it directly rushes along $B C$ to eat the cheese. Question: How many times can the mouse see the cheese on segment $A B$?
【Answer】 5
【Solution】The key to this problem is to determine which points can see the cheese. Let the position of the cheese be $E$ (as shown in the figure below). Since $\frac{A E}{A C}=\frac{A W_{3}}{A D}=\frac{3}{4}$, it follows that $G_{3} E \parallel B C$, thus $\frac{A G_{3}}{A B}=\frac{3}{4}$. Given that $A B=400$ meters, we have $A G_{3}=300$ meters. The challenge lies in how to find the lengths of $A G_{1}$ and $A G_{2}$. We will first find $A G_{2}$, and two methods are provided:

Method 1: As shown in the figure, extend $G_{2} E$ to intersect the extension of $B C$ at point $H$. $\triangle A D C$ is intercepted by $W_{2} E H$, so $\frac{A W_{2}}{W_{2} D} \cdot \frac{D H}{H C} \cdot \frac{C E}{E A}=1 \Rightarrow 1 \cdot \frac{D H}{H C} \cdot \frac{1}{3}=1 \Rightarrow \frac{D H}{H C}=3$. Let $C H=a$, then $D C=2 a$, so $B D=D C=2 a$. Since $\triangle A B D$ is intercepted by $G_{2} W_{2} H$, we have $\frac{A G_{2}}{G_{2} B} \cdot \frac{B H}{H D} \cdot \frac{D W_{2}}{W_{2} A}=1 \Rightarrow \frac{A G_{2}}{G_{2} B} \cdot \frac{5}{3} \cdot 1=1 \Rightarrow \frac{A G_{2}}{G_{2} B}=\frac{3}{5} \Rightarrow \frac{A G_{2}}{A B}=\frac{3}{8}$. Given that $A B=400$ meters, we have $A G_{2}=150$ meters.

Method 2: As shown in the figure, draw $E I \parallel A K \parallel B C$, then $\frac{A J}{A D}=\frac{A E}{A C}=\frac{3}{4}$. Let $A D=4 k$, then $A J=3 k$. Since $W_{2}$ is the midpoint of $A D$, we have $A W_{2}=2 k$, thus $\frac{A K}{E J}=\frac{A W_{2}}{W_{2} J}=2 \Rightarrow A K=2 E J$. Since $\frac{E J}{D C}=\frac{A J}{A D}=\frac{I J}{B D}$ and $D C=B D$, we have $E J=I J \Rightarrow I E=2 E J$. Since $A K=2 E J$, we have $A K=I E$, so $\frac{A G_{2}}{G_{2} I}=\frac{A K}{I E}=1$. Since $\frac{A I}{A B}=\frac{A E}{A C}=\frac{3}{4}$, and combining $A G_{2}=G_{2} I$, we get $\frac{A G_{2}}{A B}=\frac{3}{8}$. Given that $A B=400$ meters, we have $A G_{2}=150$ meters.

Using the same method, we can find that $A G_{1}=60$ meters. Therefore, we have $A G_{1}=60$, $A G_{2}=150$, and $A G_{3}=300$. For the mouse, the distances it travels are as follows:

So, it sees the cheese five times.