Triangle PQR, Points A, B, C Divide Sides in Ratio 1:2, Find Area(PQR)/Area(ABC)

JEE Main 2023 — 24 January, Morning Shift. Previous Year Question.

Problem

Let $PQR$ be a triangle. The points $A$, $B$ and $C$ are on the sides $QR$, $RP$ and $PQ$ respectively such that

$$\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$$

Then $\dfrac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}$ is equal to?

Key insight. Writing $A$, $B$, $C$ as position vectors using the section formula turns the whole problem into pure vector algebra — the area ratio drops out of a cross product, with no need to ever compute an actual length or angle.

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Approach

Let $p, q, r$ be the position vectors of $P, Q, R$. Since $QA:AR = 1:2$, the section formula gives $A$’s position vector directly in terms of $q$ and $r$ — and likewise for $B$ and $C$. Once $A$, $B$, $C$ are all written as combinations of $p$, $q$, $r$, the area of $\triangle ABC$ can be compared to the area of $\triangle PQR$ using the identity $2 \cdot \text{Area}(\triangle PQR) = |p \times q + q \times r + r \times p|$.

Solution

Step 1 — Write A, B, C using the section formula

$A$ divides $QR$ with $QA:AR = 1:2$, so:

$$A = \frac{2q + r}{3}$$

By the same pattern (each point sits $1/3$ of the way from one vertex towards the next along its side):

$$B = \frac{2r + p}{3}, \qquad C = \frac{2p + q}{3}$$

Step 2 — Form two sides of triangle ABC

$$\vec{AB} = B – A = \frac{(2r+p) – (2q+r)}{3} = \frac{p – 2q + r}{3}$$

$$\vec{AC} = C – A = \frac{(2p+q) – (2q+r)}{3} = \frac{2p – q – r}{3}$$

Step 3 — Compute the cross product

Expanding $\vec{AB} \times \vec{AC}$ and collecting every term into the standard combination $S = p\times q + q\times r + r\times p$ (using $p\times p = 0$ and $x\times y = -y\times x$ throughout), every cross term collapses to a multiple of $S$:

$$\vec{AB} \times \vec{AC} = \frac{1}{9}\Big(3p\times q + 3q\times r + 3r\times p\Big) = \frac{S}{3}$$

Step 4 — Compare the two areas

Since $2\cdot\text{Area}(\triangle PQR) = |S|$, and $2\cdot\text{Area}(\triangle ABC) = |\vec{AB}\times\vec{AC}| = \dfrac{|S|}{3}$:

$$\text{Area}(\triangle ABC) = \frac{1}{3}\,\text{Area}(\triangle PQR)$$

$$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = 3$$

Answer

Common mistakes

• Mixing up which ratio goes with which vertex in the section formula. $QA:AR=1:2$ means $A$ is closer to $Q$, so $R$ (the farther point) gets the larger weight: $A = (2q+r)/3$, not $(q+2r)/3$.
• Forgetting that $p \times p = 0$ and $x\times y = -y\times x$ when expanding the cross product — without applying these consistently, the terms won’t collapse into the clean multiple of $S$.

See also

This same problem can also be worked by placing $Q$ at the origin, which reduces $A$, $B$, $C$ to simple multiples of just two position vectors instead of three. See Triangle PQR With QA/AR=RB/BP=PC/CQ=1/2 for that version.

Practise next

• Redo the same problem with $QA/AR = RB/BP = PC/CQ = 1/3$ instead of $1/2$, and find the new area ratio.