Animated Solution for Mathematics - Trigonometry: For a triangle $ABC$, the value of $\cos 2A+\cos 2B+\cos 2C$ is least. If its inradius is 3 and incentre is $M$, then which of the following is NOT correct?

Welcome, future engineers! Today, we are going to dissect a problem that seems to be about trigonometry, but is actually a beautiful dance of geometry and vector algebra.

We are given the expression $E=\cos 2A+\cos 2B+\cos 2C$ and told it is at its minimum. In the JEE Advanced arena, whenever you see a symmetric trigonometric expression involving triangle angles, your intuition should immediately scream: "Equilateral Triangle!"

Symmetry is the universe's way of minimizing energy and complexity. When $A=B=C=60^{\circ }$, the expression reaches its minimum value. We have just unlocked the secret identity of our triangle.

Connecting Inradius to Side Length

Now that we know $\Delta ABC$ is equilateral, we are given the inradius $r=3$. You might be tempted to dive into complex formulas, but keep it simple.

For an equilateral triangle, the relationship between the inradius $r$ and the side length $a$ is a fundamental result:

Substituting our known value, $3=\frac{a}{2\sqrt{3}}$, we find that $a=6\sqrt{3}$. This is the heartbeat of our problem. With $a$ in hand, the rest of the puzzle pieces fall into place like clockwork.

The Perimeter and the Sine Identity

Let's verify our findings. The perimeter $P$ is simply $3a$. Substituting $a=6\sqrt{3}$, we get $P=18\sqrt{3}$. Option 1 is correct!

Now, look at the sine identity: $\sin 2A+\sin 2B+\sin 2C=\sin A+\sin B+\sin C$. With $A=B=C=60^{\circ }$, the left side becomes:

The right side becomes $3\sin 60^{\circ }=3\times \frac{\sqrt{3}}{2}$. They match perfectly! Option 2 is also correct.

Dot Product and Geometry

This is where many students stumble. We need the dot product $\overrightarrow{MA}\cdot \overrightarrow{MB}$. Since $M$ is the incenter (and thus the circumcenter), the distance from $M$ to any vertex is the circumradius $R$.

In an equilateral triangle, $R=2r=6$. The vectors $\overrightarrow{MA}$ and $\overrightarrow{MB}$ originate from the center $M$. The angle between them is the angle subtended by the side $AB$ at the center, which is $120^{\circ }$.

Using the dot product formula $\overrightarrow{MA}\cdot \overrightarrow{MB}=|MA||MB|\cos 120^{\circ }$, we calculate:

Area Calculation

Finally, we check the area. The area of an equilateral triangle is $\Delta =\frac{\sqrt{3}}{4}{a}^2$. Plugging in $a=6\sqrt{3}$, we get $a^2=108$.

Thus, the area is:

The option claims the area is $\frac{27\sqrt{3}}{2}$. This is clearly incorrect! We have successfully navigated the traps and identified the false statement. Keep this level of precision in your practice, and no problem will ever be too daunting.