The triangle of maximum area that can be inscribed in a given circle of radius 'r' is: (1) A right a

Q: Wait, why do we assume it's an equilateral triangle? Can't it be an isosceles triangle that isn't equilateral?

That is a fantastic question! While all equilateral triangles are isosceles, not all isosceles triangles are equilateral. If you were to use calculus to solve this, you would define the area as a function of the angles or the position of the vertices. You would find that the derivative of the area function is zero only when the angles are equal. Symmetry is nature's way of optimizing! In the JEE, recognizing this symmetry saves you precious minutes.

Q: Why is the circumcenter $O$ dividing the altitude in a $2:1$ ratio?

In an equilateral triangle, the altitude, median, and angle bisector are all the same line. The centroid divides the median in a $2:1$ ratio. Since the circumcenter, centroid, and orthocenter all coincide in an equilateral triangle, this property holds true. It is a powerful geometric shortcut that turns a complex problem into a simple ratio calculation!

Q: I calculated $BD = \frac{\sqrt{3}r}{2}$, but the answer is $\sqrt{3}r$. Did I miss something?

You are very close! You have correctly identified the length of the segment from the altitude to one vertex. Remember that $BD$ is only half of the base $BC$. In your right-angled triangle $\triangle OBD$, $BD$ represents the distance from the altitude to one vertex. To get the full side length, you must multiply by $2$. Always visualize the full triangle to avoid this common trap!

Question

The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:

Options:

(A) A right angle triangle having two of its sides of length $2r$ and $r$.

(B) An equilateral triangle of height $\frac{3r}{2}$.

(C) An isosceles triangle with base equal to $2r$.

(D) An equilateral triangle having each of its side of length $\sqrt{3}r$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the center of a circle with radius $r$. You hold a piece of string, and you want to form a triangle inside this circle that encloses the largest possible space.

This is not just a geometry problem; it is a quest for optimization. In the world of JEE Advanced, we don't just solve for $x$; we understand the physical soul of the problem.

### The Intuition of Symmetry

Why does the triangle of maximum area have to be equilateral? If you have a scalene triangle, you can always 'push' one of the vertices to increase the height or the base without leaving the circle.

The only configuration where you cannot improve the area by shifting a vertex is when the triangle is perfectly balanced. This is the principle of symmetry.

In any optimization problem involving a circle, symmetry is your best friend. We conclude that the triangle must be equilateral. This realization is the first step in our journey.

### The Anatomy of the Altitude

Now, let us get technical. We draw an altitude from the top vertex $A$ down to the base $BC$. Let us call this line $AD$.

Because our triangle is equilateral, the center of the circle $O$ must lie on this altitude. This is a beautiful consequence of the triangle's symmetry.

Here is where the JEE toolkit comes in. We know that in an equilateral triangle, the circumcenter $O$ is also the centroid.

The centroid divides the median (which is also our altitude $AD$) in a $2:1$ ratio. This means the segment from the vertex to the center, $OA$, is twice the length of the segment from the center to the base, $OD$.

Mathematically, we express this as:

$$\frac{OA}{OD} = \frac{2}{1}$$

Since $OA$ is the radius of the circle, we know $OA = r$. Substituting this into our ratio, we get:

$$\frac{r}{OD} = 2 \implies OD = \frac{r}{2}$$

This is a profound moment. We have just determined the exact position of the base relative to the center of the circle. The total height of our triangle, $AD$, is simply the sum of these two segments:

$$AD = OA + OD = r + \frac{r}{2} = \frac{3r}{2}$$

### The Trigonometric Bridge

We have the height, but we need the side length. Let us focus our attention on the right-angled triangle $	riangle OBD$.

We know the hypotenuse $OB$ is the radius $r$. We also know that the angle $\angle BOC$ is $120^\circ$ because the circle is divided into three equal arcs by the vertices of the equilateral triangle.

Therefore, the angle $\angle BOD$ is half of that, which is $60^\circ$. Using basic trigonometry, we can find the length of $BD$:

$$\sin(60^\circ) = \frac{	ext{Opposite}}{	ext{Hypotenuse}} = \frac{BD}{OB}$$

Substituting our known values:

$$\frac{\sqrt{3}}{2} = \frac{BD}{r}$$

Solving for $BD$, we get:

$$BD = \frac{\sqrt{3}r}{2}$$

### The Final Reveal

Do not stop here! Remember, $BD$ is only half of the base $BC$. The full side length of our equilateral triangle is twice $BD$:

$$BC = 2 	imes BD = 2 	imes \left( \frac{\sqrt{3}r}{2} ight) = \sqrt{3}r$$

There it is. The side length of the triangle that maximizes the area is **$\sqrt{3}r$**.

We started with a simple circle and a vague goal, and through the power of symmetry, the properties of the centroid, and a touch of trigonometry, we have arrived at a precise, elegant result. This is the essence of mathematics—taking a complex, seemingly infinite set of possibilities and narrowing them down to the one perfect configuration.

Animated Solution for Mathematics - Differentiation: The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:

Select Answer:

Visualized Solution

The Given Circle

Condition for Maximum Area

Drawing the Altitude

Properties of the Circumcenter

Setting up the Ratio

Calculating $O D$

Total Height of the Triangle

Focusing on $△ O B D$

Angle at the Center

Trigonometry in $△ O B D$

Calculating Half-Base $B D$

Total Side Length $B C$

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Intuition of Symmetry

The Anatomy of the Altitude

The Trigonometric Bridge

The Final Reveal

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