Q.If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.

The core idea here involves Coordinate Geometry, specifically the properties of distance and area formulas when applied to points with integral coordinates.

1. Assume, for contradiction, that an equilateral triangle exists with all its vertices having integral coordinates, say $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, where $x_i, y_i \in \mathbb{Z}$.
2. The square of the side length, $a^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$, must be an integer, as the differences $(x_2 - x_1)$ and $(y_2 - y_1)$ are integers. Let $a^2 = K$, where $K \in \mathbb{Z}$. For a non-degenerate triangle, $K > 0$.
3. The area of a triangle with integral coordinates is given by $Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Since all $x_i, y_i$ are integers, this expression evaluates to an integer or a half-integer, meaning the area must be a rational number.
4. However, the area of an equilateral triangle with side length $a$ is also given by the formula $Area = \frac{\sqrt{3}}{4} a^2$. Substituting $a^2 = K$, we get $Area = \frac{\sqrt{3}}{4} K$. …

Let's explore why a triangle with all its vertices having integral coordinates cannot be equilateral. The core idea revolves around calculating the triangle's area in two different ways: once using the coordinates of its vertices, and again using the formula for an equilateral triangle's area based on its side length. We will then show that these two calculations lead to a contradiction.

When the coordinates of the vertices are integers, certain properties of the triangle, like the square of its side lengths and its area, become constrained.

1. Assume the opposite:

   Let's assume, for the sake of contradiction, that there exists an equilateral triangle $ABC$ whose vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ all have integral coordinates. This means $x_1, y_1, x_2, y_2, x_3, y_3 \in \mathbb{Z}$.

2. Determine the nature of the square of the side length:

   Let $s$ be the side length of the equilateral triangle. Using the distance formula between two vertices, say $A$ and $B$:

   $s^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

   Since $x_1, x_2, y_1, y_2$ are integers, the differences $(x_2 - x_1)$ and $(y_2 - y_1)$ are also integers. The square of an integer is an integer. Therefore, $(x_2 - x_1)^2$ and $(y_2 - y_1)^2$ are non-negative integers. Their sum, $s^2$, must also be a non-negative integer.

   Let $s^2 = k$, where $k \in \mathbb{Z}$ and $k \ge 0$.

   Since it's a triangle, its side length $s$ must be greater than $0$, so $k$ must be a positive integer ($k > 0$).

3. Calculate the area using coordinates:

   The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

   $Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

   Since all $x_i$ and $y_i$ are integers, the expression inside the absolute value, $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$, is a sum and product of integers, which results in an integer. Let this integer be $N$.

   So, the area of the triangle is $Area = \frac{1}{2} |N|$.

   This means the area of any triangle with integral coordinates must be a rational number (specifically, half an integer). For a non-degenerate triangle, $N \ne 0$, so the area is a positive rational number.

4. Calculate the area using the equilateral triangle formula:

   The area of an equilateral triangle with side length $s$ is given by:

   $Area = \frac{\sqrt{3}}{4} s^2$

   From Step 2, we established that $s^2 = k$, where $k$ is a positive integer. …