Let the Centroid of an Equilateral Triangle ABC Be at the Origin and One of Its Sides Lie Along the Straight Line x + y = 3. If R and r Are the Radii of the Circumcircle and Incircle of Triangle ABC, Then R + r Is Equal To

JEE Main 2021 — 18 March, Shift 2. Previous Year Question.

Problem

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 $r$ are the radii of the circumcircle and incircle, respectively, of $\triangle ABC$, then $R+r$ is equal to:

Key insight. In an equilateral triangle the centroid, circumcenter and incenter all coincide at the same point, and the two radii are locked together by $R=2r$ — so the entire problem reduces to a single perpendicular-distance calculation from the origin to the given line.

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Approach

Because the triangle is equilateral, its centroid, incenter and circumcenter are all the same point — here, the origin. The inradius $r$ is then just the perpendicular distance from the origin to the given side, which is a one-line distance-from-a-point-to-a-line calculation. The circumradius follows immediately from the standard equilateral-triangle relation $R=2r$, so no coordinates for the triangle’s vertices are ever needed.

Solution

Step 1 — Identify the incenter and the side

Since $\triangle ABC$ is equilateral, its incenter coincides with its centroid, which is given as the origin $(0,0)$. The inradius $r$ is the perpendicular distance from the incenter to any side of the triangle — here, the side lying along the line $x+y=3$, or equivalently $x+y-3=0$.

Step 2 — Compute r as a perpendicular distance

Using the point-to-line distance formula for the origin and the line $x+y-3=0$:

$$r = \frac{|0+0-3|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt2}$$

Step 3 — Use R = 2r for an equilateral triangle

For any equilateral triangle, the circumradius is exactly twice the inradius, because both are measured from the same center and the ratio of the two radii depends only on the fixed angles of an equilateral triangle:

$$R = 2r = 2\cdot\frac{3}{\sqrt2} = \frac{6}{\sqrt2}$$

Step 4 — Add R and r

$$R+r = \frac{6}{\sqrt2}+\frac{3}{\sqrt2} = \frac{9}{\sqrt2}$$

Answer

Common mistakes

• Trying to find the triangle’s actual vertices first. It’s tempting to set up all three vertices of $\triangle ABC$ using the given side and the centroid, but this is unnecessary work — the perpendicular distance from the centroid to one side already gives $r$ directly.
• Using the wrong sign convention in the distance formula. The absolute value in $\left|\dfrac{ax_0+by_0+c}{\sqrt{a^2+b^2}}\right|$ matters — dropping it, or mishandling the constant $-3$, gives a negative or incorrect $r$.
• Forgetting that R=2r is specific to equilateral triangles. This relation does not hold for a general triangle, so it can only be used here because the problem explicitly states the triangle is equilateral.

Practise next

• An equilateral triangle has its centroid at $(1,1)$ and one side along the line $2x+y=5$. Find $R+r$ using the same perpendicular-distance-and-R=2r approach.