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This essential trains for: AMC-10, AMC-12, AIME, Math Kangaroo 9-10.

Mass point geometry is useful when we have to find ratios of segment lengths, primarily in triangles. It is an easy strategy that can replace, in most cases, the application of the Menelaus and Ceva theorems.

The center of mass of a system of two masses m and M is on the line that connects them. The position of the center of gravity is the same as that of the center of mass. Imagine we connect the two masses by a weighless rod and suspend the rod at a point P. The rod with remain horizontal in a gravity field if and only if the point P is the center of mass.

If the rod is in equilibrium then the torques created by the gravity of the two masses are equal:

equation

figure

From the value of the masses, we can obtain the ratio of the segments l and L. Conversely, information about the lengths of the segments helps assign masses to the two points.

If three mass points are placed non-collinearly, their center of mass is unique. To find the position of the center of mass of the system of three mass points:

figure

In problems, we often have a drawing in which we know certain lengths or ratios of lengths and we must find other lengths or ratios. We start by assigning masses to points according to the lengths of the segments that separate them.

Example: The intersection of the bisector of angle A and the median from vertex B in triangle ABC is denoted with P. We know that AB=7 and AC=5. What is the ratio CP:PF, where F is the intersection of AB and extension of CP?

Solution: Denote with M and N the intersections of the median and the bisector with the sides of the triangle.

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We use the properties of the median to assign masses to the points A and C:

equation

Since M is the midpoint of AC:

equation

The masses at points A and C must be equal. For now, we will make them both equal to 1:

equation

We use the angle bisector theorem to assign masses to the points B and C:

equation

The masses must be such as to balance the segment lengths:

equation

equation

equation

We can, therefore, set the mass at B to be 5 and the mass at C to be 7. But the the mass at C had been set to 1! Not to worry. Let us set the mass at A be 7 as well. Now we have the masses:

figure

From the masses we can now derive a number of ratios of segment lengths:

equation

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Multiple intersection points:

We have used the fact that the center of mass is unique. As a consequence, the cevians AN, BM, and CF in the example, intersected at the same point.

If the cevians do not intersect at the same point, then the resulting intersections are not centers of mass of the whole system of three mass points. In this case, we have to use several sets of mass points, one for each center of mass (intersection) that we need.