Animated Solution for Mathematics - Trigonometry: A circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in this circle is ..................

Imagine you are standing before a perfect equilateral triangle, its sides stretching out with a length of $a$. There is a profound elegance in geometry, a sense of order that we, as students of physics and mathematics, are privileged to uncover.

Today, we are not just solving for an area; we are peeling back the layers of a nested system: a triangle, a circle, and a square, all locked in a dance of perfect symmetry. Let us begin by visualizing the heart of this system: the incenter.

When we place a circle inside an equilateral triangle, we are essentially defining the incenter, which we will call $I$. This point is equidistant from all three sides. To find the radius $r$ of this circle, we must construct a bridge.

Drop a perpendicular from $I$ to the base of the triangle, hitting at point $L$. This segment $IL$ is our radius $r$. Now, look at the triangle formed by the vertex $B$, the incenter $I$, and the point $L$.

Because the triangle is equilateral, the line $BI$ bisects the angle at $B$. Since the full angle is $60^{\circ }$, our angle $\angle IBL$ is exactly $30^{\circ }$. We know the base $BL$ is half the side length, so $BL=\frac{a}{2}$.

Now, we have a right-angled triangle $\Delta IBL$ where we know an angle and an adjacent side. Trigonometry is our best friend here. We use the tangent ratio:

Substituting our values, we get:

With a quick algebraic rearrangement, we find the inradius:

This is the fundamental constant of our system.

Now, we move to the next layer: the square inscribed within this circle. If a square is inscribed in a circle, its vertices must touch the circumference. By symmetry, the center of the square must coincide with the center of the circle $I$.

Consequently, the diagonal of the square must pass through the center, making the diagonal equal to the diameter of the circle, which is $2r$. Let the side of our square be $s$.

We know from the Pythagorean theorem that the diagonal of a square is $s\sqrt{2}$. Therefore, we set up our equation:

Solving for $s$, we get:

This is a beautiful, simple relationship. The side of the square is simply the radius scaled by $\sqrt{2}$.

We are at the finish line. We need the area of the square, which is $s^2$. Substituting our expression for $s$, we get:

Now, we bring back our value for $r$ from the first phase. Substituting $r=\frac{a}{2\sqrt{3}}$, the equation becomes:

Let us expand this carefully. The numerator becomes $a^2$. The denominator becomes $(2\sqrt{3}{)}^2$, which is $4\times 3=12$. So, we have:

Simplifying the fraction, we arrive at our final, elegant result:

Look at that result. It is clean, it is simple, and it perfectly encapsulates the relationship between these three shapes. You have successfully navigated the nesting of geometry. Remember, in JEE Advanced, the complexity is often just a mask for fundamental principles.