Question

Consider a triangle ABC whose vertices are A(0, \( \alpha \), \( \alpha \)), B(\( \alpha \), 0, \( \alpha \)) and C(\( \alpha \), \( \alpha \), 0), \( \alpha \) > 0. Let D be a point moving on the line x + z \( - \) 3 = 0 = y and G be the centroid of \( \Delta \)ABC. If the minimum length of GD is \( \sqrt {{{57} \over 2}} \), then \( \alpha \) is equal to ____________.

Hint:

Find the centroid \(G\); the minimum distance from \(G\) to the line is the perpendicular distance, obtained by minimising \(GD^2\) over the line parameter.

Answer 1

Correct Answer: 6

Step 1: Centroid \(G=\left(\tfrac{2\alpha}3,\tfrac{2\alpha}3,\tfrac{2\alpha}3\right)\); put \(a=\tfrac{2\alpha}3\). The line \(y=0,\ x+z=3\) has points \(D(t,0,3-t)\).
Step 2: \(GD^2=(t-a)^2+a^2+(3-t-a)^2\). Minimising over \(t\): derivative \(2(t-a)-2(3-t-a)=0\Rightarrow t=\tfrac32\).
Step 3: \(GD^2_{\min}=2\left(\tfrac32-a\right)^2+a^2=\tfrac92-6a+3a^2\).
Step 4: Set equal to \(\tfrac{57}2\): \(3a^2-6a-24=0\Rightarrow a^2-2a-8=0\Rightarrow a=4\) (or \(-2\), rejected since \(\alpha>0\)).
Step 5: \(\alpha=\tfrac{3a}2=6\).
Answer: 6