In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4 sin² (A/2). If

Q: Why can't we just use the Sine Rule immediately to convert everything to side lengths?

That is a brilliant instinct! While the Sine Rule is powerful, applying it directly to $\cos B + \cos C$ would introduce square roots and complex algebraic expressions involving $a, b,$ and $c$ immediately. In JEE Advanced problems, the 'path of least resistance' is often to simplify the trigonometric expression first using identities. By reducing the equation to a clean ratio of cosines, we make the subsequent algebra much more manageable.

Q: How do I know when to use Componendo and Dividendo?

Great question! Look for the structure. Whenever you have a ratio of two terms equal to a constant, like $\frac{X}{Y} = k$, and you suspect that the sum or difference of those terms will lead to a standard identity (like $\cos(x-y) + \cos(x+y)$), that is your cue. In this problem, we had $\frac{\cos \frac{B-C}{2}}{\cos \frac{B+C}{2}} = 2$. Applying Componendo and Dividendo allowed us to transform the numerator and denominator into products of sines and cosines, which is exactly what we needed to reach the tangent half-angle formulas.

Q: Wait, why is the locus of point $A$ an ellipse? I thought it was just a calculation.

The calculation $b + c = 2a$ is the algebraic proof, but the geometric interpretation is the 'soul' of the problem. By definition, an ellipse is the locus of a point such that the sum of its distances from two fixed points (the foci) is constant. Here, the fixed points are $B$ and $C$. Since $b$ is the distance $AC$ and $c$ is the distance $AB$, the condition $b + c = 2a$ (where $2a$ is constant) perfectly matches the definition of an ellipse with foci at $B$ and $C$.

Question

In a triangle $ABC$ with fixed base $BC$, the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$. If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$, respectively, then

Options:

(A) $b + c = 4a$

(B) $b + c = 2a$

(C) locus of point $A$ is an ellipse

(D) locus of point $A$ is a pair of straight lines

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not merely solving a trigonometry problem; we are embarking on a journey to uncover a hidden geometric truth. In the world of JEE Advanced, problems are rarely just about plugging in numbers.

They are about recognizing patterns, understanding the 'why' behind the algebra, and visualizing the motion of points in space. Let us dissect the problem of the moving vertex $A$ in triangle $ABC$.

### The Trigonometric Dance

We start with a triangle $ABC$ where the base $BC$ is fixed. This means the side length $a$ is a constant. The vertex $A$ is constrained by the equation $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$.

At first glance, this equation looks intimidating. The key is to simplify the left-hand side using the sum-to-product identity: $\cos X + \cos Y = 2 \cos \frac{X+Y}{2} \cos \frac{X-Y}{2}$.

Applying this to our equation, we get:

$$2 \cos \frac{B+C}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$$

Now, recall the fundamental property of any triangle: $A + B + C = \pi$. This implies that $\frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2}$.

When we take the cosine of both sides, we find a beautiful substitution: $\cos \frac{B+C}{2} = \sin \frac{A}{2}$. Substituting this back into our equation, the terms align:

$$2 \sin \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$$

Assuming $\sin \frac{A}{2} 
eq 0$ (since $A$ is an angle of a triangle), we divide both sides by $2 \sin \frac{A}{2}$ to obtain the relation:

$$\cos \frac{B-C}{2} = 2 \sin \frac{A}{2}$$

### The Algebraic Bridge

We need to relate this to the side lengths $b$ and $c$. Substituting $\sin \frac{A}{2}$ back with $\cos \frac{B+C}{2}$, we get:

$$\cos \frac{B-C}{2} = 2 \cos \frac{B+C}{2}$$

Rearranging this into a ratio, we have $\frac{\cos \frac{B-C}{2}}{\cos \frac{B+C}{2}} = 2$. We now employ the technique of **Componendo and Dividendo**:

$$\frac{\cos \frac{B-C}{2} + \cos \frac{B+C}{2}}{\cos \frac{B-C}{2} - \cos \frac{B+C}{2}} = \frac{2+1}{2-1} = 3$$

Using the identities $\cos(x-y) + \cos(x+y) = 2 \cos x \cos y$ and $\cos(x-y) - \cos(x+y) = 2 \sin x \sin y$, the left side simplifies to $\cot \frac{B}{2} \cot \frac{C}{2} = 3$. Taking the reciprocal, we arrive at:

$$	an \frac{B}{2} 	an \frac{C}{2} = \frac{1}{3}$$

### The Geometric Revelation

In any triangle, the product of the tangents of the half-angles is given by $	an \frac{B}{2} 	an \frac{C}{2} = \frac{s-a}{s}$, where $s$ is the semi-perimeter $\frac{a+b+c}{2}$. Equating this to our result:

$$\frac{s-a}{s} = \frac{1}{3}$$

Cross-multiplying gives $3s - 3a = s$, or $2s = 3a$. Since $2s = a + b + c$, we substitute this to find $a + b + c = 3a$, which simplifies to:

$$b + c = 2a$$

This is the final piece of the puzzle. We have proven that the sum of the distances from the moving vertex $A$ to the fixed points $B$ and $C$ is constant.

By the definition of an ellipse—the locus of a point where the sum of distances to two fixed foci is constant—we conclude that the path of $A$ is an **ellipse**.

Animated Solution for Mathematics - Trigonometry: In a triangle $A B C$ with fixed base $B C$ , the vertex $A$ moves such that $cos B + cos C = 4 sin^{2} \frac{A}{2}$ . If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$ , respectively, then

Select Answer:

Visualized Solution

The Given Condition

Applying Sum-to-Product Formula

Using Angle Sum Property

Substituting Back into Equation

Simplifying the Expression

Relating B and C Angles

Applying Componendo and Dividendo

Simplifying to Tangents

Half-Angle Formula Identity

Solving for Semi-perimeter

Finding the Side Relation

The Locus is an Ellipse

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Trigonometric Dance

The Algebraic Bridge

The Geometric Revelation

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