Coordinate geometry proving

Question

Let $G$ be the centroid of $\Delta ABC$. Prove that: $$AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2).$$

how to do this sum? Plzzz help me out. I don't know how to do it

Answer 1

Let $a$, $b$ and $c$ be sides-lengths of our triangle.

Thus, sides-lengths of medians are $m_a=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$ and similar, which gives $GA=\frac{2}{3}m_a=\frac{1}{3}\sqrt{2b^2+2c^2-a^2}$.

Id est, $$3(GA^2+GB^2+GC^2)=3\sum_{cyc}\left(\frac{1}{3}\sqrt{2b^2+2c^2-a^2}\right)^2=$$ or $$=\frac{1}{3}\sum_{cyc}(2b^2+2c^2-a^2)=a^2+b^2+c^2=AB^2+AC^2+BC^2.$$ Done!

Answer 2

To find the centroid of a triangle with coordinates: (a,b) (c,d) (e,f) is simply the midpoint of the two and then from 2/3 of the other vertex to the midpoint.

Midpoint has the following coordinates: ((a+e)/2, (b+f)/2).

Centroid has the following coordinates:

{1/2 ((a + e)×4/3 + c/3), 1/2 ((b + f)×4/3 + d/3)}

After you find the coordinates of the centroid find the distances from each vertex, and pairwise distances among vertices. Then you can prove it with setting equality.

Answer 3

Let $\vec{AB}=\vec{a}$ and $\vec{AC}=\vec{b}$.

Thus, $$\vec{AG}=\frac{2}{3}\cdot\frac{1}{2}(\vec{a}+\vec{b})=\frac{1}{3}(\vec{a}+\vec{b}),$$ $$\vec{BG}=\frac{1}{3}(\vec{BA}+\vec{BC})=\frac{1}{3}(\vec{b}-\vec{a}-\vec{a})=\frac{1}{3}(\vec{b}-2\vec{a})$$ and $$\vec{CG}=\frac{1}{3}(\vec{CB}+\vec{CA})=\frac{1}{3}(\vec{a}-\vec{b}-\vec{b})=\frac{1}{3}(\vec{a}-2\vec{b}).$$ Thus, $$3(AG^2+BG^2+CG^2)=\frac{1}{3}((\vec{a}+\vec{b})^2+(\vec{b}-2\vec{a})^2+(\vec{a}-2\vec{b})^2)=$$ $$=2((\vec{a})^2+(\vec{b})^2-\vec{a}\vec{b})=(\vec{a})^2+(\vec{b})^2+(\vec{b}-\vec{a})^2=AB^2+AC^2+BC^2.$$ Done!

Answer 4

Let $A(a_1,a_2)$, $B(b_1,b_2)$ and $C(c_1,c_2)$.

Thus, $$G\left(\frac{a_1+b_1+c_1}{3},\frac{a_2+b_2+c_2}{3}\right)$$ and $$3(GA^2+GB^2+GC^2)=3\sum_{cyc}\left(\sqrt{\left(a_1-\frac{a_1+b_1+c_1}{3}\right)^2+\left(a_2-\frac{a_2+b_2+c_2}{3}\right)^2}\right)^2=$$ $$=\frac{1}{3}\sum_{cyc}((2a_1-b_1-c_1)^2+(2a_2-b_2-c_2)^2)=2\sum_{cyc}(a_1^2-a_1b_1+a_2^2-a_2b_2)=$$ $$=\sum_{cyc}((a_1-b_1)^2+(a_2-b_2)^2)=AB^2+AC^2+BC^2.$$ Done!