Consider a triangle PQR having sides of lengths p, q and r opposite to the angles P, Q and R, respec

Q: Wait, why do we reject the negative root in AM-GM? Or does it even apply?

That is a fantastic question! Remember, AM-GM applies to positive real numbers. Since $q$ and $r$ are side lengths of a triangle, they are strictly positive. Thus, $q^2$ and $r^2$ are positive, and the inequality $\frac{q^2 + r^2}{2} \ge \sqrt{q^2 r^2}$ holds perfectly. We don't need to worry about negative roots because side lengths cannot be negative!

Q: Why can't we just use the Sine Rule for everything? It seems easier.

The Sine Rule is excellent for ratios, but it doesn't handle sums of squares or complex algebraic inequalities as cleanly as the Cosine Rule does. When you see terms like $p^2$ or $q^2$, the Cosine Rule is your best friend because it directly relates the angle to the squares of the sides.

Q: Is the Projection Formula really necessary? Can't I just use the Cosine Rule for everything in Option B?

You *could* expand everything using the Cosine Rule, but it would lead to a massive algebraic nightmare! The Projection Formula is a 'shortcut' that collapses complex trigonometric expressions into simple side lengths. In JEE Advanced, recognizing these patterns is the difference between solving a problem in 2 minutes versus 10 minutes.

Question

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$, respectively. Then which of the following statements is (are) TRUE? Options: (A) $\cos P \ge 1 - \frac{p^2}{2qr}$ (B) $\cos R \ge \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$ (C) $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$ (D) If $p < q$ and $p < r$, then $\cos Q > \frac{p}{r}$ and $\cos R > \frac{p}{q}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of triangle geometry. When you look at a triangle $PQR$, do you see just three lines and three angles?

Or do you see a system of constraints, a delicate balance where every side and every angle is locked in a beautiful, mathematical dance? Let's dive into this problem and uncover the truths hidden within.

### The Cosine Rule and the Power of AM-GM

We begin with Option A: $\cos P \ge 1 - \frac{p^2}{2qr}$. When you see $\cos P$ alongside side lengths $p, q,$ and $r$, your mind should immediately jump to the **Cosine Rule**. It is the fundamental bridge between the angular world and the linear world of sides.

We write:
$$ \cos P = \frac{q^2 + r^2 - p^2}{2qr} $$

Now, here is where the magic happens. We don't just stare at the equation; we manipulate it. Let's split the fraction:
$$ \cos P = \frac{q^2 + r^2}{2qr} - \frac{p^2}{2qr} $$

Look at that first term, $\frac{q^2 + r^2}{2qr}$. It screams for the **AM-GM Inequality**. Since $q$ and $r$ are positive side lengths, we know that:
$$ \frac{q^2 + r^2}{2} \ge \sqrt{q^2 r^2} = qr $$

Dividing both sides by $qr$, we get $\frac{q^2 + r^2}{2qr} \ge 1$. Substituting this back, we find $\cos P \ge 1 - \frac{p^2}{2qr}$. **Option A is not just a statement; it is a geometric necessity!**

### The Elegance of the Projection Formula

Next, we face Option B: $\cos R \ge \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$. It looks intimidating, but remember, in JEE Advanced, complexity is often a mask for simplicity. Let's clear the denominator by multiplying by $(p+q)$.

We get:
$$ (p+q) \cos R \ge (q-r) \cos P + (p-r) \cos Q $$

Expanding this, we group the terms:
$$ (p \cos R + r \cos P) + (q \cos R + r \cos Q) \ge q \cos P + p \cos Q $$

Does this look familiar? These are the **Projection Formulas**! They tell us that $q = p \cos R + r \cos P$ and $p = q \cos R + r \cos Q$. Substituting these, the entire inequality collapses into $q + p \ge r$.

This is the **Triangle Inequality**! Since the sum of two sides must be greater than the third, this statement is always true. We have conquered the beast with simple logic.

### The Trap of the Sine Rule

Now, let's look at Option C: $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$. We use the **Sine Rule** to replace the sides with sines:
$$ \frac{q+r}{p} = \frac{\sin Q + \sin R}{\sin P} $$

Applying AM-GM to the numerator, we get $\sin Q + \sin R \ge 2\sqrt{\sin Q \sin R}$. Dividing by $\sin P$, we get:
$$ \frac{q+r}{p} \ge 2 \frac{\sqrt{\sin Q \sin R}}{\sin P} $$

Wait! The inequality sign is pointing in the wrong direction compared to the option. This is a classic trap. Option C claims it is strictly less than, but our derivation proves it is greater than or equal to. Therefore, **Option C is false**.

### The Power of the Counterexample

Finally, consider Option D. We are given $p < q$ and $p < r$. We need to check if $\cos Q > \frac{p}{r}$.

Instead of trying to prove it, let's try to break it. Let's choose an isosceles triangle where $p=3, q=4, r=4$. The condition $p < q$ and $p < r$ is satisfied.

However, calculating $\cos Q$ using the Cosine Rule gives us a value that fails the inequality. A single counterexample is enough to dismantle a false statement. Thus, **Option D is false**.

### Conclusion

We have navigated through the Cosine Rule, the Projection Formula, the Sine Rule, and the Triangle Inequality. We didn't just calculate; we reasoned.

Remember, physics and math are not about memorizing formulas; they are about understanding the relationships between variables. Keep practicing, keep questioning, and keep falling in love with the process. You are doing great!

Animated Solution for Mathematics - Trigonometry: Consider a triangle $P QR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$ , respectively. Then which of the following statements is (are) TRUE?

Select Answer:

Visualized Solution

The Triangle $P QR$

Option A: Cosine Rule

Splitting the Fraction

Applying AM-GM Inequality

Concluding Option A

Option B: Cross-Multiplication

Expanding and Rearranging

The Projection Formulas

Substituting Projections

Triangle Inequality

Option C: Applying the Sine Rule

AM-GM on Sines

Concluding Option C

Option D: Cosine Rule for Q

Simplifying and Counterexample

Final Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Cosine Rule and the Power of AM-GM

The Elegance of the Projection Formula

The Trap of the Sine Rule

The Power of the Counterexample

Conclusion

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