Show that the triangle, the coordinates of whose verticles are given by integers, can never be an equilateral triangle.

Answer

To prove that a triangle with vertices at integer coordinates cannot be an equilateral triangle, we will analyze the area of the triangle using two different methods: one based on the coordinates of the vertices and the other based on the length of the sides.

### Step-by-Step Solution:

1. **Define the Vertices**:
Let the vertices of the triangle be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \), where \( x_1, y_1, x_2, y_2, x_3, y_3 \) are all integers.

2. **Area of the Triangle using Coordinates**:
The area \( A \) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Since \( x_1, y_1, x_2, y_2, x_3, y_3 \) are integers, the expression inside the absolute value is an integer. Therefore, the area \( A \) is a half of an integer, which means \( A \) is a rational number.

3. **Area of the Equilateral Triangle**:
For an equilateral triangle with side length \( a \), the area is given by:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
Here, \( a \) is the length of the side of the triangle. If the triangle has vertices with integer coordinates, then the side length \( a \) must also be a rational number (since the distance between two points with integer coordinates is a rational number).

4. **Rational vs. Irrational**:
Since \( a \) is rational, \( a^2 \) is also rational. However, when we multiply a rational number \( a^2 \) by the irrational number \( \sqrt{3} \), the product \( \frac{\sqrt{3}}{4} a^2 \) becomes irrational. Thus, the area of the equilateral triangle is irrational.

5. **Conclusion**:
We have shown that the area of a triangle with integer coordinates is rational, while the area of an equilateral triangle (with integer coordinates) is irrational. Therefore, it is impossible for a triangle with vertices at integer coordinates to be an equilateral triangle.