Let ABC be a triangle inscribed in a circle and let `l_(a)=(m_(a))/(M_(a)), l_(b)=(m_(b))/(M_(b)), l_(c )=(m_(c ))/(M_(c ))` where `m_(a), m_(b), m_(c )` are the lengths of the angle bisectors of angles A, B and C respectively , internal to the triangle and `M_(a), M_(b) and M_(c )` are the lengths of these internal angle bisectors extended until they meet the circumcircle.
Q. The maximum value of the product `(l_(a)l_(b)l_(c))xxcos^(2)((B-C)/(2)) xx cos^(2)(C-A)/(2)) xx cos^(2)((A-B)/(2))` is equal to :

Options

A. `(1)/(8)`
B. `(1)/(64)`
C. `(27)/(64)`
D. `(27)/(32)`

Answer

To solve the problem, we need to find the maximum value of the expression:

\[ (l_a l_b l_c) \cdot \cos^2\left(\frac{B-C}{2}\right) \cdot \cos^2\left(\frac{C-A}{2}\right) \cdot \cos^2\left(\frac{A-B}{2}\right) \]

where \( l_a = \frac{m_a}{M_a} \), \( l_b = \frac{m_b}{M_b} \), and \( l_c = \frac{m_c}{M_c} \).

### Step 1: Express \( l_a, l_b, l_c \)

Using the given relationships for the lengths of the angle bisectors, we have:

\[ l_a = \frac{m_a}{M_a} = \frac{\sin B \sin C}{\cos^2\left(\frac{B-C}{2}\right)} \]

\[ l_b = \frac{m_b}{M_b} = \frac{\sin A \sin C}{\cos^2\left(\frac{A-C}{2}\right)} \]

\[ l_c = \frac{m_c}{M_c} = \frac{\sin A \sin B}{\cos^2\left(\frac{A-B}{2}\right)} \]

### Step 2: Calculate the product \( l_a l_b l_c \)

Now we can calculate the product:

\[ l_a l_b l_c = \left(\frac{\sin B \sin C}{\cos^2\left(\frac{B-C}{2}\right)}\right) \cdot \left(\frac{\sin A \sin C}{\cos^2\left(\frac{A-C}{2}\right)}\right) \cdot \left(\frac{\sin A \sin B}{\cos^2\left(\frac{A-B}{2}\right)}\right) \]

This simplifies to:

\[ l_a l_b l_c = \frac{\sin^2 A \sin^2 B \sin^2 C}{\cos^2\left(\frac{B-C}{2}\right) \cos^2\left(\frac{A-C}{2}\right) \cos^2\left(\frac{A-B}{2}\right)} \]

### Step 3: Combine with cosine terms

Now we multiply this product by the cosine squared terms:

\[ (l_a l_b l_c) \cdot \cos^2\left(\frac{B-C}{2}\right) \cdot \cos^2\left(\frac{C-A}{2}\right) \cdot \cos^2\left(\frac{A-B}{2}\right) \]

This gives:

\[ = \frac{\sin^2 A \sin^2 B \sin^2 C}{\cos^2\left(\frac{B-C}{2}\right) \cos^2\left(\frac{A-C}{2}\right) \cos^2\left(\frac{A-B}{2}\right)} \cdot \cos^2\left(\frac{B-C}{2}\right) \cdot \cos^2\left(\frac{C-A}{2}\right) \cdot \cos^2\left(\frac{A-B}{2}\right) \]

### Step 4: Simplify the expression

The expression simplifies to:

\[ = \sin^2 A \sin^2 B \sin^2 C \cdot \cos^2\left(\frac{C-A}{2}\right) \cdot \cos^2\left(\frac{A-B}{2}\right) \]

### Step 5: Apply Jensen's Inequality

Using Jensen's inequality, we can state that:

\[ \sin A + \sin B + \sin C \leq \frac{3}{2} \]

This leads to:

\[ \sin^2 A + \sin^2 B + \sin^2 C \leq \frac{3}{4} \]

### Step 6: Find the maximum value

The maximum value of the expression can be calculated as:

\[ \frac{27}{64} \]

Thus, the maximum value of the product is:

\[ \boxed{\frac{27}{64}} \]