Animated Solution for Mathematics - Straight Lines: If the vertices $P,Q,R$ of a triangle $PQR$ are rational points, which of the following points of the triangle $PQR$ is (are) always rational point(s)?

My dear students, welcome to a beautiful exploration of coordinate geometry. Today, we are not just solving a problem; we are investigating the very nature of numbers within the geometric plane.

We are given a triangle $PQR$ where the vertices $P(x_1,y_1)$, $Q(x_2,y_2)$, and $R(x_3,y_3)$ are all rational points. This means every coordinate is a fraction of integers.

No square roots, no transcendental numbers—just the clean, structured world of $\mathbb{Q}$. Our mission is to determine which of the triangle's centers—the centroid, incentre, circumcentre, or orthocentre—inherit this rational nature.

Before we dive into the geometry, let us establish our mathematical foundation. The set of rational numbers, $\mathbb{Q}$, is a field.

This means it is closed under the four fundamental operations: addition, subtraction, multiplication, and division (by non-zero numbers). If you start with rational numbers and perform these operations, you will never, ever land on an irrational number.

This is our 'rational safety net.' If a center's coordinates can be expressed solely through these operations on the vertex coordinates, that center is guaranteed to be rational.

The Simple Average

Let us start with the centroid, $G$. Geometrically, it is the intersection of the medians. Algebraically, it is the most elegant of all centers.

Its coordinates are given by:

Look at this expression. We are taking three rational numbers, adding them, and dividing by three. By our golden rule, the result is undeniably rational. The centroid is safe.

Orthocentre and Circumcentre

Now, consider the orthocentre $H$ and the circumcentre $O$. These are slightly more complex, but they share a secret: they are defined by the intersection of lines.

For the orthocentre, we deal with altitudes. The slope of a side $QR$ is:

This value is rational. The altitude is perpendicular, so its slope is the negative reciprocal, $-\frac{1}{m_{QR}}$, which is also rational.

The equation of the altitude from $P$ is $y-y_1=m_P(x-x_1)$. This is a linear equation with rational coefficients.

When we solve for the intersection of two such altitudes, we are solving a system of linear equations. Since the coefficients are rational, the solution—the coordinates of $H$—must be rational.

The same logic applies to the circumcentre, which is the intersection of perpendicular bisectors. These bisectors also have rational slopes and pass through rational midpoints. Thus, both $H$ and $O$ are rational points.

The Irrational Trap

Finally, we arrive at the incentre, $I$. This is where the story takes a turn.

The formula for the incentre is:

Here, $a,b,c$ are the lengths of the sides opposite to vertices $P,Q,R$. To find a side length, say $a$, we use the distance formula:

Here is the danger: while the expression inside the square root is rational, the square root itself is not necessarily rational. For instance, the distance between $(0,0)$ and $(1,1)$ is $\sqrt{2}$.

Because the side lengths $a,b,c$ can be irrational, they can 'infect' the incentre formula, making the coordinates of $I$ irrational. Therefore, the incentre is not always a rational point.

We have peeled back the layers of these geometric centers. The centroid, orthocentre, and circumcentre rely only on linear operations that respect the closure of rational numbers.

The incentre, however, relies on distances, which introduce the potential for irrationality. Thus, the centroid, circumcentre, and orthocentre are the rational points we seek.

Keep this distinction in mind—it is the difference between a solid proof and a common trap!