Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangles OPQ. The point R inside the triangle

Q: Wait, why does the centroid divide the triangle into equal areas? Is that always true?

Yes, it is a fundamental property! The centroid divides the triangle into three smaller triangles of equal area because the centroid lies on the medians, and medians divide triangles into equal areas. It is a beautiful geometric symmetry that makes this problem much simpler than it first appears.

Q: Can I solve this using the area formula for coordinates instead of the centroid property?

Absolutely! You could set the area of $\triangle OPR, \triangle PQR,$ and $\triangle OQR$ equal to one-third of the total area of $\triangle OPQ$. It involves more algebra, but it is a fantastic way to verify your intuition and practice your coordinate geometry skills.

Q: What if the point $R$ was outside the triangle? Would the area condition still hold?

The problem specifically places $R$ inside the triangle. If $R$ were outside, the signed areas might be equal, but the geometric interpretation of 'equal area' would change. For the purpose of this JEE problem, focus on the interior property!

Question

Let $O(0, 0), P(3, 4), Q(6, 0)$ be the vertices of the triangles $OPQ$. The point $R$ inside the triangle $OPQ$ is such that the triangles $OPR, PQR, OQR$ are of equal area. The coordinates of $R$ are

Options:

(A) $(4/3, 3)$

(B) $(3, 2/3)$

(C) $(3, 4/3)$

(D) $(4/3, 2/3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane with three vertices: $O(0, 0)$, $P(3, 4)$, and $Q(6, 0)$. Connecting these points forms the triangle $	riangle OPQ$.

We are looking for a mystery point $R$ located inside this triangle. When $R$ is connected to the vertices $O$, $P$, and $Q$, it creates three smaller triangles: $	riangle OPR$, $	riangle PQR$, and $	riangle OQR$.

The problem states that these three sub-triangles possess exactly the same area.

### The 'Aha!' Moment

Many students immediately jump into complex algebraic equations, attempting to set up area formulas for each of the three sub-triangles. While that is a valid path, it is the long way around.

In JEE Advanced, time is your most precious resource. The key here is to recognize the geometric soul of the problem.

What point inside a triangle divides it into three equal-area triangles? It is the **centroid**! The centroid is the intersection of the medians, and it is the unique point that balances the triangle perfectly.

### The Calculation

Now that we have identified $R$ as the centroid, we can use the elegant centroid formula:

$$R(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} ight)$$

Let us plug in our coordinates: $O(0, 0)$, $P(3, 4)$, and $Q(6, 0)$.

For the $x$-coordinate, we calculate:

$$x = \frac{0 + 3 + 6}{3} = \frac{9}{3} = 3$$

For the $y$-coordinate, we calculate:

$$y = \frac{0 + 4 + 0}{3} = \frac{4}{3}$$

And just like that, we have our answer: **$R(3, \frac{4}{3})$**.

### Final Thoughts

Geometry is not just about memorizing formulas; it is about seeing the underlying structure of the world. When you see a problem asking for equal area partitions, let your mind immediately jump to the centroid.

It is a powerful tool in your arsenal. Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of mathematics. You are doing great!

Animated Solution for Mathematics - Straight Lines: Let $O (0, 0), P (3, 4), Q (6, 0)$ be the vertices of the triangles $O P Q$ . The point $R$ inside the triangle $O P Q$ is such that the triangles $O P R, P QR, O QR$ are of equal area. The coordinates of $R$ are

Select Answer:

Visualized Solution

Vertices of $△ O P Q$

Forming $△ O P Q$

Locating Point $R$

Subdividing the Triangle

The Equal Area Condition

Identifying the Centroid

The Centroid Formula

Substituting x-coordinates

Calculating the x-coordinate

Substituting y-coordinates

Calculating the y-coordinate

Final Coordinates of $R$

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

Analyzing the Setup

The 'Aha!' Moment

The Calculation

Final Thoughts

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