A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.

Options

A. 4000 m
B. 4800 m
C. 5600 m
D. 6400 m

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Triangle
We have a right-angled triangle where the longest side (hypotenuse) is 80 m. Let's denote the vertices of the triangle as A, B, and C, where AB is the hypotenuse.

### Step 2: Divide the Hypotenuse
The Mayor wants to construct three paths from point C (the corner opposite the hypotenuse) such that they divide the hypotenuse AB into four equal segments. Since AB = 80 m, each segment will be:
\[ \text{Segment length} = \frac{80}{4} = 20 \text{ m} \]
So, the points dividing the hypotenuse are D, E, and F, where:
- AD = 20 m
- DE = 20 m
- EF = 20 m
- FB = 20 m

### Step 3: Identify the Paths
The paths from point C to points D, E, and F are CD, CE, and CF respectively. We need to find the sum of the squares of the lengths of these paths:
\[ \text{Sum} = CD^2 + CE^2 + CF^2 \]

### Step 4: Use the Apollonius Theorem
For triangle ACD, we can apply the Apollonius theorem, which states:
\[ AC^2 + AD^2 = 2(CD^2 + CE^2) \]
Here, AD = 20 m. Thus:
\[ AC^2 + 20^2 = 2(CD^2 + CE^2) \]

For triangle BCF, we similarly apply the theorem:
\[ BC^2 + DF^2 = 2(CF^2 + CD^2) \]
Here, DF = 20 m. Thus:
\[ BC^2 + 20^2 = 2(CF^2 + CD^2) \]

### Step 5: Relate the Sides
Since ABC is a right triangle, we can relate the sides using the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]
Given that AB = 80 m:
\[ 80^2 = AC^2 + BC^2 \implies 6400 = AC^2 + BC^2 \]

### Step 6: Substitute and Solve
Now we can substitute the values into the equations we derived from the Apollonius theorem:
1. From triangle ACD:
\[ AC^2 + 400 = 2(CD^2 + CE^2) \implies CD^2 + CE^2 = \frac{AC^2 + 400}{2} \]
2. From triangle BCF:
\[ BC^2 + 400 = 2(CF^2 + CD^2) \implies CF^2 + CD^2 = \frac{BC^2 + 400}{2} \]

### Step 7: Add the Equations
Now we can add the two equations:
\[ CD^2 + CE^2 + CF^2 + CD^2 = \frac{AC^2 + 400}{2} + \frac{BC^2 + 400}{2} \]
This simplifies to:
\[ CD^2 + CE^2 + CF^2 + CD^2 = \frac{AC^2 + BC^2 + 800}{2} \]
Substituting \(AC^2 + BC^2 = 6400\):
\[ CD^2 + CE^2 + CF^2 + CD^2 = \frac{6400 + 800}{2} = \frac{7200}{2} = 3600 \]

### Step 8: Final Calculation
Since we have \(CD^2\) appearing twice, we can rearrange:
\[ CD^2 + CE^2 + CF^2 = 3600 - CD^2 \]
Now, since \(CD\) is also 20 m, we can calculate:
\[ CD^2 = 20^2 = 400 \]
Thus:
\[ CE^2 + CF^2 = 3600 - 400 = 3200 \]
Finally, summing all squares:
\[ CD^2 + CE^2 + CF^2 = 400 + 3200 = 3600 \]

### Final Answer
The sum of the squares of the lengths of the three paths is:
\[ \boxed{3600} \]