The circum centre of a triangle lies at the origin and its centroid is the mid point of th…

Question

The circum centre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points $(a^2+1,a^2+1)$ and $(2a,-2a),\ a
e0$. Then for any $a$, the orthocentre of this triangle lies on the line:

Accepted Answer

Circumcentre $O=(0,0)$. Centroid:
$$G = \left(\frac{(a^2+1)+2a}2,\frac{(a^2+1)-2a}2\right) = \left(\frac{(a+1)^2}2,\frac{(a-1)^2}2\right)$$

Since $O,G,H$ (orthocentre) are collinear (Euler line) with $HG:GO=2:1$:
$$H = 3G-2O = \left(\frac{3(a+1)^2}2,\frac{3(a-1)^2}2\right)$$

These coordinates satisfy $(a-1)^2x-(a+1)^2y=0$: substitute
$$(a-1)^2\cdot\frac{3(a+1)^2}2-(a+1)^2\cdot\frac{3(a-1)^2}2 = 0$$  (identically zero)

So the orthocentre lies on $(a-1)^2x-(a+1)^2y=0$.

Solutions · 1