In a non-right-angled triangle PQR, let p, q, r denote the lengths of the sides opposite to the angl

Q: Wait, why do we reject the possibility of $P=60^\circ$?

Great catch! The Sine Rule gives $\sin P = \frac{\sqrt{3}}{2}$, which mathematically allows for $P=60^\circ$ or $P=120^\circ$. However, if $P=60^\circ$, then $R=180^\circ - (60^\circ + 30^\circ) = 90^\circ$. The problem explicitly states that $\Delta PQR$ is a non-right-angled triangle, so we must reject $P=60^\circ$.

Q: Why use coordinate geometry for point $O$? Isn't there a pure geometric way?

While synthetic geometry is beautiful, it can become a labyrinth for intersection points of medians and altitudes. Coordinate geometry is a powerful, systematic tool in your JEE arsenal. By placing $Q$ at the origin, we turn a complex geometric intersection into a simple system of linear equations, reducing the chance of calculation errors.

Q: Is Apollonius's Theorem always applicable for finding the median?

Absolutely! Apollonius's Theorem is a direct consequence of the Stewart's Theorem and is specifically designed for medians. It states that for a triangle with sides $a, b, c$ and median $m_a$ to side $a$, $b^2 + c^2 = 2(m_a^2 + (a/2)^2)$. It is a massive time-saver in competitive exams.

Question

In a non-right-angled triangle $PQR$, let $p, q, r$ denote the lengths of the sides opposite to the angles at $P, Q, R$ respectively. The median from $R$ meets the side $PQ$ at $S$, the perpendicular from $P$ meets the side $QR$ at $E$, and $RS$ and $PE$ intersect at $O$. If $p = \sqrt{3}, q = 1$, and the radius of the circumcircle of the $\Delta PQR$ equals 1, then which of the following options is/are correct?

Options:

(A) Area of $\Delta SOE = \frac{\sqrt{3}}{12}$

(B) Radius of incircle of $\Delta PQR = \frac{\sqrt{3}}{2}(2 - \sqrt{3})$

(C) Length of $RS = \frac{\sqrt{7}}{2}$

(D) Length of $OE = \frac{1}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a triangle $\Delta PQR$ that seems simple on the surface but hides a wealth of structural beauty. We are given side lengths $p=\sqrt{3}$ and $q=1$, and a circumradius $R=1$.

The first step in any such journey is to understand the 'DNA' of the triangle—its angles. We invoke the **Extended Sine Rule**:

$$\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R$$

By substituting our known values, we find $\sin P = \frac{\sqrt{3}}{2}$ and $\sin Q = \frac{1}{2}$. This leads us to $P=120^\circ$ and $Q=30^\circ$.

Since the sum of angles in a triangle is $180^\circ$, $R$ must also be $30^\circ$. We have just discovered that our triangle is isosceles, with $q=r=1$. This realization is the key that unlocks the entire problem.

### The Elegance of Apollonius

With the triangle's dimensions fully understood, we turn our attention to the median $RS$. The median is a line segment connecting a vertex to the midpoint of the opposite side.

To find its length without getting bogged down in trigonometry, we use **Apollonius's Theorem**:

$$p^2 + q^2 = 2\left(RS^2 + \left(\frac{r}{2}ight)^2ight)$$

Substituting our values, we get:

$$(\sqrt{3})^2 + 1^2 = 2\left(RS^2 + \left(\frac{1}{2}ight)^2ight)$$

This simplifies to $4 = 2RS^2 + \frac{1}{2}$. Solving this yields **$RS = \frac{\sqrt{7}}{2}$**. This confirms that Option 3 is correct.

### The Coordinate Pivot

Now, we face the intersection point $O$ of the median $RS$ and the altitude $PE$. While pure geometry is elegant, coordinate geometry is the 'heavy artillery' of the JEE.

Let's place $Q$ at the origin $(0,0)$ and $R$ at $(\sqrt{3}, 0)$. Given $\angle Q = 30^\circ$ and side $q=1$, the coordinates of $P$ are:

$$P = (1 \cos 30^\circ, 1 \sin 30^\circ) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}ight)$$

The midpoint $S$ of $PQ$ is $\left(\frac{\sqrt{3}}{4}, \frac{1}{4}ight)$. The altitude $PE$ drops from $P$ to $QR$ (the x-axis), so $E$ is $\left(\frac{\sqrt{3}}{2}, 0ight)$. The line $PE$ is simply $x = \frac{\sqrt{3}}{2}$.

The line $RS$ passes through $R(\sqrt{3}, 0)$ and $S\left(\frac{\sqrt{3}}{4}, \frac{1}{4}ight)$. Calculating the slope and equation of $RS$, we find the intersection $O$ by substituting $x = \frac{\sqrt{3}}{2}$ into the line equation.

The result is $O = \left(\frac{\sqrt{3}}{2}, \frac{1}{6}ight)$. This confirms that the length **$OE = \frac{1}{6}$**, making Option 4 correct.

Finally, calculating the area of $\Delta SOE$ as $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$ gives $\frac{\sqrt{3}}{48}$, which proves Option 1 is incorrect. Through this journey, we have seen how combining different mathematical perspectives allows us to dissect even the most complex problems with confidence and precision.

Select Answer:

Visualized Solution

The Triangle and its Circumcircle

Applying the Sine Rule

Calculating the Angles

Deducing Side $r$

Radius of the Incircle

Finalizing the Inradius

The Median $R S$ and Apollonius's Theorem

Calculating Median Length $R S$

Coordinate Geometry Setup

Coordinates of $S$ and $E$

Equations of Lines $R S$ and $P E$

Finding Intersection $O$ and Length $O E$

Area of Triangle $S O E$

Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Elegance of Apollonius

The Coordinate Pivot

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Select Answer:

Visualized Solution

The Triangle and its Circumcircle

Applying the Sine Rule

Calculating the Angles

Deducing Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Radius of the Incircle

Finalizing the Inradius

The Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS and Apollonius's Theorem

Calculating Median Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS

Coordinate Geometry Setup

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Equations of Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PE

Finding Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O and Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OE

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SOE