Home > Affine Geometry, Geometry > Five points on a circle, centroids and perpendiculars

## Five points on a circle, centroids and perpendiculars

Suppose you have ${5}$ different points on a circle. For each triangle formed by three of these points, consider the line passing through its centroid, perpendicular to the line determined by the other two.

Proof: Pick one point ${P}$ among the five points on the circle, and consider the homothety of center ${P}$ and ratio ${3/2}$. This homothety maps the centroid of a triangle ${PAB}$ on the midpoint of ${AB}$. Thus, if ${A,B,C,D}$ are the remaining vertices, the line passing through the centroid of ${AB}$, perpendicular on ${CD}$ is mapped to a line passing through the midpoint of ${AB}$, perpendicular on ${CD}$.

Now, we have the following result: Suppose you have ${4}$ points ${A,B,C,D}$ on a circle. Draw the lines which pass through the midpoint of a segment determined by these, perpendicular to the line determined by the other two. Then these ${6}$ lines are concurrent. Furthermore, if ${X}$ is the point of concurrency, ${O}$ the center of the circle, and ${G}$ the centroid of ${ABCD}$, then ${G}$ is the midpoint of ${OX}$.

To prove this secondary result, note that if we take the midpoints ${M,N}$ of ${AB}$ and ${CD}$, for example, and consider the intersection ${X}$ point of the perpendicular from ${M}$ to ${CD}$ and the perpendicular from ${N}$ to ${AB}$, then ${MXNO}$ is a parallelogram. Thus, the midpoint of ${OX}$ is the same as the midpoint of ${MN}$, which is the centroid of ${ABCD}$.

Thus, among the ${10}$ lines, there are ${5}$ families of six lines which are concurrent. Each such two families have at least two common lines, which means the intersection point is unique.

Take a look at this MathOverflow discussion, where you can find more informations and generalizations. Here you can find a Geogebra interactive picture.