Question

Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest angle is \(\frac{\pi}{2}\) and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.Let a be the area of the triangle ABC. Then the value of (64a)² is

Answer 1

Correct Answer: 1008

We start by applying the law of sines to the sides and angles of the triangle:

\(\frac{a + d}{\sin A} = \frac{a}{\sin B} = \frac{a - d}{\sin C} = 2\)

From this, we deduce that \( A = \frac{\pi}{2} + C \).

Using the identity for sine, we have:

\(\sin A = \sin \left( \frac{\pi}{2} + C \right) = \cos C\)

Now, using the relationship between the sides:

\(\frac{a + d}{2} = \sqrt{1 - \sin^2 C}\)

\(\Rightarrow \left( \frac{a + d}{2} \right)^2 = 1 - \left( \frac{a - d}{2} \right)^2\)

Expanding and simplifying, we get:

\(2(a^2 + d^2) / 4 = 1 \quad \Rightarrow \quad a^2 + d^2 = 2\)

From this, we get Equation (1):

\(a^2 + d^2 = 2\)

Next, we use the cosine law to relate angle \( B \):

\(\cos B = \frac{(a - d)^2 + (a + d)^2 - a^2}{2(a^2 - d^2)}\)

Simplifying further, we get:

\(\sqrt{1 - \sin^2 B} = \frac{2(a^2 + d^2) - a^2}{2(a^2 - d^2)} \quad \Rightarrow \quad a^2 = \frac{7}{4}, \, d^2 = \frac{1}{4}\)

Now, we can find the area \( \Delta \) of the triangle using the formula:

\(\Delta = \frac{a(a^2 - d^2)}{4}\)

Substitute the values of \( a^2 \) and \( d^2 \):

\(\alpha = \frac{\sqrt{7}}{2} \times \frac{6}{4 \times 4}\)

Finally, we find that:

\((64 \alpha)^2 = 1008\)

Thus, the value of \( (64 \alpha)^2 \) is 1008.