Key Concepts and Formulas

• Combinations: ${ }^n C_r = \frac{n!}{r!(n-r)!}$, where $n!$ denotes the factorial of $n$.
• Ratio of Combinations: $\frac{{ }^n C_r}{{ }^n C_{r+1}} = \frac{r+1}{n-r}$ and $\frac{{ }^n C_{r-1}}{{ }^n C_r} = \frac{r}{n-r+1}$.
• Centroid of a Triangle: If the vertices of a triangle are $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, then the centroid is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
• Trigonometric Identity: $\cos^2 t + \sin^2 t = 1$.

Step-by-Step Solution

Step 1: Find the relationship between $n$ and $r$ using the given combination values. We are given ${ }^n C_{r-1}=28$ and ${ }^n C_r=56$. We can form a ratio to eliminate the factorials and relate $n$ and $r$. $\frac{{ }^n C_{r-1}}{{ }^n C_r} = \frac{28}{56} = \frac{1}{2}$ Using the formula for the ratio of combinations, we have: $\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}} = \frac{r}{n-r+1} = \frac{1}{2}$ $2r = n-r+1$ $3r = n+1 \quad \text{..... (i)}$

Step 2: Find another relationship between $n$ and $r$. We are also given ${ }^n C_r=56$ and ${ }^n C_{r+1}=70$. We can form another ratio. $\frac{{ }^n C_r}{{ }^n C_{r+1}} = \frac{56}{70} = \frac{4}{5}$ Using the formula for the ratio of combinations, we have: $\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}} = \frac{r+1}{n-r} = \frac{4}{5}$ $5(r+1) = 4(n-r)$ $5r+5 = 4n-4r$ $9r = 4n-5 \quad \text{..... (ii)}$

Step 3: Solve for $n$ and $r$. Now we have a system of two linear equations with two variables: $3r = n+1 \quad \text{..... (i)}$ $9r = 4n-5 \quad \text{..... (ii)}$ Substitute $n = 3r-1$ from (i) into (ii): $9r = 4(3r-1)-5$ $9r = 12r - 4 - 5$ $3r = 9$ $r = 3$ Now, substitute $r=3$ into equation (i): $3(3) = n+1$ $9 = n+1$ $n = 8$ Therefore, $n=8$ and $r=3$.

Step 4: Determine the coordinates of vertex C. Given $C(3r-n, r^2-n-1)$, substitute $n=8$ and $r=3$ into the coordinates: $C(3(3)-8, (3)^2-8-1) = C(9-8, 9-8-1) = C(1, 0)$

Step 5: Determine the coordinates of the centroid. The coordinates of the vertices are $A(4 \cos t, 4 \sin t)$, $B(2 \sin t, -2 \cos t)$, and $C(1, 0)$. The centroid $(x, y)$ is given by: $x = \frac{4 \cos t + 2 \sin t + 1}{3}$ $y = \frac{4 \sin t - 2 \cos t + 0}{3}$ Thus, $3x = 4 \cos t + 2 \sin t + 1$ and $3y = 4 \sin t - 2 \cos t$. Therefore, $3x - 1 = 4 \cos t + 2 \sin t$ and $3y = 4 \sin t - 2 \cos t$.

Step 6: Find the locus of the centroid. Square both equations and add them together: $(3x-1)^2 + (3y)^2 = (4 \cos t + 2 \sin t)^2 + (4 \sin t - 2 \cos t)^2$ $(3x-1)^2 + (3y)^2 = 16 \cos^2 t + 16 \sin^2 t + 4 \sin^2 t + 4 \cos^2 t + 16 \cos t \sin t - 16 \sin t \cos t$ $(3x-1)^2 + (3y)^2 = 16(\cos^2 t + \sin^2 t) + 4(\sin^2 t + \cos^2 t)$ $(3x-1)^2 + (3y)^2 = 16(1) + 4(1) = 20$ We are given that $(3x-1)^2 + (3y)^2 = \alpha$. Therefore, $\alpha = 20$.

Common Mistakes & Tips

• Be careful with the ratio of combinations. Ensure you are dividing them in the correct order and applying the formula correctly.
• When squaring and adding the equations to eliminate $t$, make sure to expand correctly and use the trigonometric identity $\sin^2 t + \cos^2 t = 1$.
• Double-check your algebra when solving the system of equations for $n$ and $r$.

Summary

We first used the given combination equations to find the values of $n$ and $r$. Then, we used these values to find the coordinates of vertex $C$. Using the coordinates of vertices $A$, $B$, and $C$, we found the coordinates of the centroid. Finally, we manipulated the centroid equations to eliminate the parameter $t$ and find the locus of the centroid, which allowed us to determine the value of $\alpha$.

Final Answer

The final answer is \boxed{20}, which corresponds to option (C).