JEE Mathematics 2023 Question with Solution

Given: $\cos 2A + \cos 2B + \cos 2C$ is least, the inradius is $r = 3$, and the incentre is $M$.

Find: Which statement is NOT correct.

From the solution, the minimum value occurs when

$$A = B = C = 60^\circ$$

So $\triangle ABC$ is equilateral.

For an equilateral triangle of side $a$, the inradius is

$$r = \frac{a\sqrt{3}}{6}$$

Using $r = 3$,

$$3 = \frac{a\sqrt{3}}{6}$$ $$a = 6\sqrt{3}$$

Hence the perimeter is

$$3a = 18\sqrt{3}$$

So the perimeter statement is correct.

The area is

$$\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 108 = 27\sqrt{3}$$

Thus the area is $27\sqrt{3}$, not $\frac{27\sqrt{3}}{2}$.

Therefore, the incorrect statement is the area statement. The solution marks Option A, but this conflicts with the worked result and the listed options. Based on the actual working, the defensible incorrect option is D.

Since $A = B = C = 60^\circ$, we get

$$\sin 2A + \sin 2B + \sin 2C = 3\sin 120^\circ = 3 \cdot \frac{\sqrt{3}}{2}$$

and also

$$\sin A + \sin B + \sin C = 3\sin 60^\circ = 3 \cdot \frac{\sqrt{3}}{2}$$

So that statement is correct.

Also, in an equilateral triangle the incentre, circumcentre and centroid coincide. Hence

$$MA = MB = R = 2r = 6$$

So

$$MA \cdot MB = 6 \cdot 6 = 36$$

Therefore the expression written as $\overline{MA} \cdot \overline{MB} = -18$ cannot represent the product of lengths. The source solution does not discuss this option, so the answer is taken from the solution's own conclusion about the area statement.

Thus, from the extracted working, the intended NOT correct statement is the one about area.