Let A(4 t, 4 t), B(2 t, -2 t) and C(3r-n, r^2-n-1) be the vertices of a triangle ABC, wher…

Question

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

Accepted Answer

From $\dfrac{^nC_{r-1}}{^nC_r} = \dfrac{28}{56}=\dfrac12$: $\dfrac{r}{n-r+1}=\dfrac12 \Rightarrow 2r = n-r+1 \Rightarrow 3r=n+1$ \ldots (i)

From $\dfrac{^nC_r}{^nC_{r+1}} = \dfrac{56}{70}$: $\dfrac{r+1}{n-r}=\dfrac{56}{70}=\dfrac45 \Rightarrow 5(r+1)=4(n-r) \Rightarrow 9r=4n-5$ \ldots (ii)

Solving (i),(ii): $r=3,\ n=8$. So $C = (3(3)-8,\ 9-8-1) = (1,0)$.

Centroid $G = \left(\dfrac{4\cos t+2\sin t+1}{3}, \dfrac{4\sin t-2\cos t}{3}\right)$, so $3x-1 = 4\cos t+2\sin t$, $3y=4\sin t-2\cos t$.

$$(3x-1)^2+(3y)^2 = (4\cos t+2\sin t)^2+(4\sin t-2\cos t)^2 = 16+4 = 20$$

Solutions · 1