Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equila

Q: Wait, why is the centroid the same as the incenter? I thought they were different points in a triangle.

That is a fantastic question! In a general triangle, the centroid, incenter, and circumcenter are indeed distinct. However, the equilateral triangle is a masterpiece of symmetry. Because all sides and angles are equal, the median, angle bisector, and perpendicular bisector from any vertex all coincide. This forces the centroid, incenter, and circumcenter to collapse into a single, beautiful point at the origin. You are working with a shape of perfect balance!

Q: Can I just use the distance from the origin to the line to get the circumradius directly?

Not directly! The distance from the origin to the side is the inradius $r$, because the incenter is the origin and the side is tangent to the incircle. The circumradius $R$ is the distance from the origin to the *vertices*. Since the origin is the circumcenter, the distance to the vertices is $R$. You must find $r$ first, then use the elegant relationship $R=2r$ to find the circumradius.

Q: Why is $R = 2r$ in an equilateral triangle? Is this always true?

Yes, it is always true for equilateral triangles! Think about the median. The centroid divides the median in a $2:1$ ratio. The part from the centroid to the vertex is the circumradius $R$, and the part from the centroid to the side is the inradius $r$. Thus, $R$ must be exactly twice $r$. It is one of the most useful shortcuts in geometry!

Question

Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x + y = 3$. If R and r be the radius of circumcircle and incircle respectively of $	riangle ABC$, then $(R + r)$ is equal to :

Options:

(A) $\frac{9}{\sqrt{2}}$

(B) $7\sqrt{2}$

(C) $2\sqrt{2}$

(D) $3\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Geometry of Perfection: Unlocking the Equilateral Triangle

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are exploring the profound symmetry of the equilateral triangle.

Imagine standing on a coordinate plane where the origin $(0,0)$ is the heart of a perfectly balanced triangle. This is not just any triangle; it is an equilateral one, a shape where every angle is $60^\circ$ and every side is equal.

When we place its centroid at the origin, we are aligning ourselves with the very center of its symmetry.

### Phase 1: The Heart of the Triangle

In the world of geometry, the equilateral triangle is unique. While other triangles have distinct centers—the centroid, the incenter, the circumcenter, and the orthocenter—the equilateral triangle brings them all together.

They collapse into a single point. When the problem tells us the centroid is at the origin, it is secretly telling us that the incenter and the circumcenter are also at the origin.

This is our first major breakthrough. We are not dealing with three separate points; we are dealing with one point of absolute stability at $(0,0)$.

### Phase 2: Finding the Inradius $r$

Now, let us look at the side of the triangle, defined by the line $x + y = 3$, or $x + y - 3 = 0$. The inradius $r$ is the radius of the circle inscribed within the triangle, touching all three sides.

Geometrically, the distance from the incenter to any side is exactly the inradius. Since our incenter is at $(0,0)$, we simply need to find the perpendicular distance from the origin to the line $x + y - 3 = 0$.

We reach for our trusty tool: the perpendicular distance formula. For a point $(x_1, y_1)$ and a line $ax + by + c = 0$, the distance $d$ is given by:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting our values, where $a=1$, $b=1$, $c=-3$, and $(x_1, y_1) = (0,0)$, we get:

$$r = \frac{|1(0) + 1(0) - 3|}{\sqrt{1^2 + 1^2}}$$

Simplifying this, we find $r = \frac{|-3|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$. This is the radius of our incircle. It is a simple, elegant result, but it holds the key to the entire problem.

### Phase 3: The Golden Ratio of $R$ and $r$

Now, we need the circumradius $R$. The circumcircle passes through the vertices of the triangle. In an equilateral triangle, the distance from the centroid to the vertex is the circumradius $R$.

We know that the centroid divides the median in a $2:1$ ratio. The distance from the centroid to the side is $r$, and the distance from the centroid to the vertex is $R$. Therefore, $R = 2r$.

This is a beautiful property that saves us from complex coordinate geometry calculations. We simply calculate:

$$R = 2 	imes \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}}$$

### Phase 4: The Final Synthesis

We have arrived at the final step. The question asks for the sum $(R + r)$. We have $R = \frac{6}{\sqrt{2}}$ and $r = \frac{3}{\sqrt{2}}$.

Adding these together is straightforward because they share the same denominator:

$$R + r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{6 + 3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$$

Look at that result. It is clean, precise, and derived from the fundamental symmetries of the triangle.

You have successfully navigated the problem by understanding the geometric soul of the equilateral triangle. Remember, in JEE Advanced, the math is often a reflection of the geometry. If you can visualize the symmetry, the equations will follow naturally.

Select Answer:

Visualized Solution

Visualizing the Setup

Property of Equilateral Triangles

Defining the Inradius $r$

Applying the Distance Formula

Substituting Values for $r$

Calculating the Inradius $r$

Relation between $R$ and $r$

Visualizing the Circumcircle

Calculating the Circumradius $R$

Final Sum: $R + r$

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

The Geometry of Perfection

Phase 1

Phase 2

Phase 3

Phase 4

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