Let PQR Be a Triangle With A, B and C on the Sides QR, RP and PQ Such That QA/AR = RB/BP = PC/CQ = 1/2. Then Area(PQR)/Area(ABC) Is Equal To

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 $\dfrac{QA}{AR}=\dfrac{RB}{BP}=\dfrac{PC}{CQ}=\dfrac12$. Then $\dfrac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}$ is equal to:

Key insight. Placing one vertex of the triangle at the origin turns every other point’s position vector into a simple combination of just two vectors — the whole problem then reduces to a single cross-product calculation instead of tracking three separate points.

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Approach

Take $Q$ as the origin, and let $\vec P$ and $\vec R$ be the position vectors of $P$ and $R$. Since $A$, $B$, $C$ each divide a side of the triangle in a known ratio, their position vectors can be written directly using the section formula, entirely in terms of $\vec P$ and $\vec R$. From there, the area of $\triangle ABC$ comes from a cross product of two of its side vectors, and comparing it with the area of $\triangle PQR$ (itself just $\frac12|\vec P\times\vec R|$) gives the ratio directly.

Solution

Step 1 — Set up position vectors with Q as the origin

Let $Q$ be the origin, so its position vector is $\vec 0$. Let $\vec P$ and $\vec R$ be the position vectors of $P$ and $R$.

Step 2 — Find the position vector of A

$A$ lies on $QR$ with $QA:AR=1:2$, so $A$ divides $Q$ to $R$ in the ratio $1:2$:

$$\vec A = \frac{2\vec Q+1\vec R}{1+2} = \frac{\vec R}{3}$$

Step 3 — Find the position vector of B

$B$ lies on $RP$ with $RB:BP=1:2$, so $B$ divides $R$ to $P$ in the ratio $1:2$:

$$\vec B = \frac{2\vec R+1\vec P}{3} = \frac{2\vec R+\vec P}{3}$$

Step 4 — Find the position vector of C

$C$ lies on $PQ$ with $PC:CQ=1:2$, so $C$ divides $P$ to $Q$ in the ratio $1:2$:

$$\vec C = \frac{2\vec P+1\vec Q}{3} = \frac{2\vec P}{3}$$

Step 5 — Compute the side vectors of triangle ABC

$$\vec{AC} = \vec C-\vec A = \frac{2\vec P}3-\frac{\vec R}3 = \frac{2\vec P-\vec R}3$$

$$\vec{AB} = \vec B-\vec A = \frac{2\vec R+\vec P}3-\frac{\vec R}3 = \frac{\vec P+\vec R}3$$

Step 6 — Find the area of triangle ABC

$$\text{Area}(\triangle ABC) = \frac12\left|\vec{AC}\times\vec{AB}\right| = \frac12\cdot\frac19\left|(2\vec P-\vec R)\times(\vec P+\vec R)\right|$$

Expanding the cross product, using $\vec P\times\vec P=\vec 0$ and $\vec R\times\vec R=\vec 0$:

$$(2\vec P-\vec R)\times(\vec P+\vec R) = 2(\vec P\times\vec R)-(\vec R\times\vec P) = 2(\vec P\times\vec R)+(\vec P\times\vec R) = 3(\vec P\times\vec R)$$

since $\vec R\times\vec P=-\vec P\times\vec R$. So:

$$\text{Area}(\triangle ABC) = \frac12\cdot\frac19\cdot3\left|\vec P\times\vec R\right| = \frac16\left|\vec P\times\vec R\right|$$

Step 7 — Find the area of triangle PQR

Since $Q$ is the origin, $\vec{QR}=\vec R$ and $\vec{QP}=\vec P$:

$$\text{Area}(\triangle PQR) = \frac12\left|\vec R\times\vec P\right| = \frac12\left|\vec P\times\vec R\right|$$

Step 8 — Take the ratio

$$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \frac{\frac12\left|\vec P\times\vec R\right|}{\frac16\left|\vec P\times\vec R\right|} = \frac{1/2}{1/6} = 3$$

Answer

Common mistakes

• Not choosing a vertex as the origin. Working with three independent position vectors for $P$, $Q$, $R$ makes the algebra considerably messier — placing $Q$ at the origin (or any one vertex) is what makes every other position vector a clean combination of just two vectors.
• Mixing up the direction of each ratio. $QA:AR=1:2$ means $A$ is closer to $Q$, not the midpoint — using the section formula with the ratio reversed (getting $A=(2\vec R)/3$ instead of $\vec R/3$) silently flips the final answer to its reciprocal.
• Forgetting $\vec R\times\vec P=-\vec P\times\vec R$. This sign flip is exactly what turns $2(\vec P\times\vec R)-(\vec R\times\vec P)$ into $3(\vec P\times\vec R)$ rather than $(\vec P\times\vec R)$ — dropping it understates the area of $\triangle ABC$ by a factor of 3.

Practise next

• Repeat the same construction with $\dfrac{QA}{AR}=\dfrac{RB}{BP}=\dfrac{PC}{CQ}=\dfrac13$ instead of $\dfrac12$, and find $\dfrac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}$.