Question

Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals:

• A. 18
• B. 8
• C. 20
• D. 6

Hint:

Review properties of centroids and binomial coefficients to simplify the problem-solving process.

Answer 1

The Correct Option is C

nC_r-1 = 28, nC_r = 56

The first equation:

\( \frac{nC_{r-1}}{nC_r} = \frac{28}{56} \)

\( \frac{n}{(n-r+1)} = \frac{1}{2} \)

This simplifies to:

\( \frac{1}{(n-r+1)} = \frac{1}{2} \)

3r = n + 1 ........ (i)

The second equation:

\( \frac{nC_{r-1}}{nC_r} = \frac{56}{70} \)

By solving (i) & (ii):

(r = 3), (n = 8)

A(4cos t, 4sin t) B(2sin t, -2cos t) C(3r - n, r² - n - 1)

A(4cos t, 4sin t) B(2sin t, -2cos t) C(1, 0)

\( (3x - 1)^2 + (3y)^2 = (4 \cos t + 2 \sin t)^2 + (4 \sin t - \cos t)^2 \)

\( (3x - 1)^2 + (3y)^2 = 20 \)