If the points P and Q are respectively the circumcenter and the orthocentre of a , then PA…

Question

If the points $P$ and $Q$ are respectively the circumcenter and the orthocentre of a $	riangleABC$, then $\overset{	o \leftarrow }{PA}+\overset{	o \leftarrow }{PB}+\overset{	o \leftarrow }{PC}$ is
equal to :

Accepted Answer

1. Circumcenter $P$:

The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.

2. Orthocenter $Q$:

The orthocenter of a triangle is the point of intersection of its three altitudes. An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).

3. Centroid $G$:

The centroid of a triangle is the point of intersection of its medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid always divides each median in a 2:1 ratio, with the larger segment being closer to the vertex.

With the above definitions, it's known that the centroid divides the line segment joining the circumcenter and orthocenter in the ratio 2 : 1, meaning :

$\overset{	o \leftarrow }{PG}=\frac{2}{3}\overset{	o \leftarrow }{PQ}$

$\overset{	o \leftarrow }{PQ}=3\overset{	o \leftarrow }{PG}$

The position vector of the centroid $G$ in terms of the vertices $A,B,$ and $C$ is :

$\overset{	o }{G}=\frac{\overset{	o }{A}+\overset{	o }{B}+\overset{	o }{C}}{3}$

Because $P$ is the circumcenter and is at the origin in this problem:

$\overset{	o \leftarrow }{PA}=\overset{	o }{A}$

$\overset{	o \leftarrow }{PB}=\overset{	o }{B}$

$\overset{	o \leftarrow }{PC}=\overset{	o }{C}$

Substituting these into the equation for $G$ :

$\overset{	o \leftarrow }{PG}=\frac{\overset{	o \leftarrow }{PA}+\overset{	o \leftarrow }{PB}+\overset{	o \leftarrow }{PC}}{3}$

Now, using the relationship between $PQ$ and $PG$ established earlier :

$\overset{	o \leftarrow }{PQ}=3\overset{	o \leftarrow }{PG}=\overset{	o \leftarrow }{PA}+\overset{	o \leftarrow }{PB}+\overset{	o \leftarrow }{PC}$

Solutions · 1