Mass points

Mass points is a technique in Euclidean geometry that can greatly simplify the proofs of many theorems concerning polygons, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local coordinate system to identify points by the ratios into which they divide line segments. Mass points are generalized by barycentric coordinates.

Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).

The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Any line passing this central point will balance the figure. If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. If two points are balanced, the point on the balancing line used to balance them has a mass of the sum of the masses of the two points.

For a full explanation with images, go to this link: http://old.mathtalentquest.org/images/Mass_Point.pdf

Examples

Consider a triangle $ABC$ with its three medians drawn, with the intersection points being $D, E, F,$ corresponding to $AB, BC,$ and $AC$ respectively. Thus, if we label point $A$ with a weight of $1$, $B$ must also have a weight of $1$ since $A$ and $B$ are equidistant from $D$. By the same process, we find $C$ must also have a weight of 1. Now, since $A$ and $B$ both have a weight of $1$, $D$ must have a weight of $2$ (as is true for $E$ and $F$). Thus, if we label the centroid $P$, we can deduce that $DP:PC$ is $1:2$ - the inverse ratio of their weights.

Problems

2013 AMC 10B Problems/Problem 16

2004 AMC 10B Problems/Problem 20

2009 AIME I Problems/Problem 5

2009 AIME I Problems/Problem 4

2001 AIME I Problems/Problem 7

2011 AIME II Problems/Problem 4

1992 AIME Problems/Problem 14

1988 AIME Problems/Problem 12

1989 AIME Problems/Problem 15

1985 AIME Problems/Problem 6

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