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}
Answers
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
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.