If the vertices of a triangle have rational coordinates, then prove that the triangle cannot be equilateral.

Explanation

To prove that a triangle with vertices having rational coordinates cannot be equilateral, we will follow these steps:

### Step 1: Define the vertices of the triangle
Let the vertices of triangle ABC be given by the coordinates:
- A(x₁, y₁)
- B(x₂, y₂)
- C(x₃, y₃)

Here, \(x₁, y₁, x₂, y₂, x₃, y₃\) are all rational numbers.

### Step 2: Calculate the area of triangle ABC
The area \(A\) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \left| x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) \right| \]
Since \(x₁, y₁, x₂, y₂, x₃, y₃\) are rational numbers, the expression inside the absolute value will also yield a rational number. Therefore, the area \(A\) is a rational number.

### Step 3: Assume triangle ABC is equilateral
Assume that triangle ABC is equilateral with side length \(s\). The area \(A\) of an equilateral triangle can be expressed as:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Here, \(s^2\) is a positive rational number since the side length \(s\) is a rational number.

### Step 4: Analyze the area expression
Now, since \(s^2\) is rational, the area can be rewritten as:
\[ A = \frac{\sqrt{3}}{4} \cdot \text{(rational number)} \]
However, \(\frac{\sqrt{3}}{4}\) is an irrational number. The product of an irrational number and a rational number is always irrational.

### Step 5: Establish contradiction
Thus, if triangle ABC is equilateral, the area \(A\) would be irrational. This contradicts our earlier conclusion that the area of triangle ABC is rational.

### Conclusion
Since we have reached a contradiction, we conclude that a triangle with vertices having rational coordinates cannot be equilateral.

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