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Offline Centres

A line through the centroid G of an equilateral triangle ABC cuts the sides of the triangle AC at P, AB at N and extended side BC at M. Let the side of the triangle be $l$ then

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Sum of perpendicular distances of A, C from the line is equal to perpendicular distance of B from the line.

If $B M < C M$ then $B N < A N$

If X is foot of perpendicular on AC from G then $G X = \frac{l}{\sqrt{3}}$

If X is foot of perpendicular on AC from G then $G X = \frac{l}{2 \sqrt{3}}$

answer is A, C.

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Detailed Solution

Let $∠ B M G = θ, ∠ B N M = 60 - θ, ∠ G P A = 60 + θ$

$G X = G Z = G Y = \frac{l}{2 \sqrt{3}}$ ( For triangle ABC all centres coincide)

$G P = \frac{G X}{S i n ∠ G P A} = \frac{l / 2 \sqrt{3}}{S i n (60 + θ)}, G N = \frac{l}{2 \sqrt{3} s i n (60 - θ)} a n d G M = \frac{l}{2 \sqrt{3} \sin θ}$

$\sin^{2} θ + \sin^{2} (60 - θ) + \sin^{2} (60 + θ) = 3 / 2$

$\sin^{2} θ + {(\frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ)}^{2} + {(\frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ)}^{2} = \frac{3}{2}$

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