Animated Solution for Mathematics - Differentiation: $AB$ is a diameter of a circle and $C$ is any point on the circumference of the circle. Then

Imagine you are standing on a vast, perfectly circular field. You have a straight path, a diameter $AB$, marked out across the center of this field.

Now, pick any point $C$ anywhere on the edge of this circle. By connecting $A$ to $C$ and $B$ to $C$, you have just carved out a triangle, $\Delta ABC$, from the landscape.

This is the setup for our problem, and it is a classic in the world of JEE geometry. The beauty of this problem lies in the hidden constraints imposed by the circle itself.

Before we dive into the algebra, we must acknowledge a powerful geometric truth. Because $AB$ is a diameter, the angle $\angle ACB$ is locked in at exactly $90^{\circ }$.

This is Thales' Theorem, and it transforms our arbitrary triangle into a right-angled triangle. This is our first major simplification.

We are no longer dealing with a generic triangle; we are dealing with a right-angled triangle where the hypotenuse is fixed at length $d$.

Now, let us introduce a variable to describe the shape of this triangle. Let $\angle CAB=\alpha$.

Because we are in a right-angled triangle, we can express the other two sides, the base $AC$ and the height $BC$, using the hypotenuse $d$ and our angle $\alpha$. Using basic trigonometry, we find:

We have now successfully parameterized the entire triangle using only the diameter $d$ and the angle $\alpha$. This is the bridge between geometry and algebra.

We want to maximize the area of this triangle. The formula for the area of a right-angled triangle is $A=\frac{1}{2}\times \text{base}\times \text{height}$.

Substituting our expressions, we get:

This expression looks good, but we can make it even more elegant. By multiplying and dividing by $2$, we can invoke the double-angle identity: $\sin 2\alpha =2\sin \alpha \cos \alpha$.

Our area function becomes:

Finding the Maximum

Now, look at this equation. The diameter $d$ is a constant. The only thing that changes as we move point $C$ around the circle is the angle $\alpha$.

To maximize the area $A$, we must maximize the term $\sin 2\alpha$. We know from trigonometry that the maximum value of the sine function is $1$, which occurs when the argument is $90^{\circ }$.

Therefore, we set $2\alpha =90^{\circ }$, which gives us $\alpha =45^{\circ }$.

When $\alpha =45^{\circ }$, the other angle $\angle ABC$ must also be $45^{\circ }$ (since the sum of angles in a triangle is $180^{\circ }$). A triangle with two $45^{\circ }$ angles is, by definition, an isosceles right-angled triangle.

We started with a simple circle and a point, and through the language of trigonometry, we discovered that the area is maximized when the triangle is perfectly balanced—an isosceles right-angled triangle.

This is the elegance of JEE mathematics: taking a complex, moving system and finding the point of perfect symmetry. Remember this result, for it appears in many guises throughout your physics and math journey.