Question

If G is the centroid of triangle ABC , then find the value of
\frac{AB^2+BC^2+CA^2}{AG^2+BG^2+CG^2}

Answer 1

Answer :

:Answer is 3

Step-by-step explanation :

Let A (x1, y1), B(x2, y2) and C(x3, y3), be the vertices of ∆ABC.

Without the loss of Generality, assume the centroid of the ΔABC to be at the origin, i.e. G = (0, 0).

Centroid of the ∆ABC

⇒ x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0

Squaring on both sides, we get

x1² + x2² + x3² + 2x1x2 + 2x2x3 + 2x3x1 = 0 and y1² + y2² + y3² + 2y1y2 + 2y2y3 + 2y3y1 = 0 … (1)

AB² + BC² + CA²

= [(x2 – x1)² + (y2 – y1)²] + [(x3 – x2)² + (y3 – y2)²] + [(x1 – x3)² + (y1 – y3)²]

= [(x1² + x2² – 2x1x2 + y1² + y2² – 2y1y2) + (x2² + x3² – 2x2x3 + y2² + y3² – 2y2y3) + (x1² + x3² – 2x1x3 + y1² + y3² – 2y1y3)

= (2x12 + 2x22 + 2x32 – 2x1x2 – 2x2x3 – 2x1x3) + (2y12 + 2y22 + 2y32 – 2y1y2 – 2y2y3 – 2y1y3)

= (3x12 + 3x22 + 3x32) + (3y12 + 3y22 + 3y32) (From (1))

= 3(x12 + x22 + x32) + 3(y12 + y22 + y32) … (2)

3(GA² + GB² + GC²)

= 3 [(x1 – 0)2 + (y1 – 0)2 + (x2 – 0)2 + (y2 – 0)2 + (x3 – 0)2 + (y3 – 0)2]

= 3 (x12 + y12 + x22 + y22 + x32 + y32)

= 3 (x12 + x22 + x32) + 3(y12 + y22 + y32) … (3)

From (2) and (3), we get

AB² + BC² + CA² = 3(GA² + GB² + GC²)

(AB² + BC² + CA²)/(GA² + GB² + GC²) = 3

Answer 2

Step-by-step explanation:

Consider AG² + BG²+ CG²

If AD, BE and CF are the medians , we know that centroid divides the median in the ratio 2:1, thus

= (2/3*AD)² + (2/3*BE)² + (2/3*CF)²,

= 4/9[ AD² + BE² + CF²]

Now we know length of median formula  

Similarly BE² = AD² = 1/4*[2*(AB²+BC²)-AC²] , and

CF² = 1/4*[2*(BC²+AC²)-AB²]

Therefore,

AG² + BG²+ CG² = 4/9*1/4*3*(AB²+BC²+AC²)

=1/3*(AB²+BC²+AC²)

Thus, AG² + BG²+ CG²/AB²+BC²+AC² = 1/3.