Mass point geometry, colloquially known as mass points, is a problem-solving technique in geometry which applies the physical principle of the center of mass to geometry problems involving triangles and intersecting cevians. All problems that can be solved using mass point geometry can also be solved using either similar triangles, vectors, or area ratios, but many students prefer to use mass points. Though modern mass point geometry was developed in the 1960s by New York high school students, the concept has been found to have been used as early as 1827 by August Ferdinand Möbius in his theory of homogeneous coordinates.

Definitions

The theory of mass points is defined according to the following definitions:

Methods

Concurrent cevians

First, a point is assigned with a mass (often a whole number, but it depends on the problem) in the way that other masses are also whole numbers. The principle of calculation is that the foot of a cevian is the addition (defined above) of the two vertices (they are the endpoints of the side where the foot lie). For each cevian, the point of concurrency is the sum of the vertex and the foot. Each length ratio may then be calculated from the masses at the points. See Problem One for an example.

Splitting masses

Splitting masses is the slightly more complicated method necessary when a problem contains transversals in addition to cevians. Any vertex that is on both sides the transversal crosses will have a split mass. A point with a split mass may be treated as a normal mass point, except that it has three masses: one used for each of the two sides it is on, and one that is the sum of the other two split masses and is used for any cevians it may have. See Problem Two for an example.

Other methods

Examples

Problem One

Problem. In triangle ABC, E is on AC so that CE = 3AE and F is on AB so that BF = 3AF. If BE and CF intersect at O and line AO intersects BC at D, compute and. Solution. We may arbitrarily assign the mass of point A to be 3. By ratios of lengths, the masses at B and C must both be 1. By summing masses, the masses at E and F are both 4. Furthermore, the mass at O is 4 + 1 = 5, making the mass at D have to be 5 - 3 = 2 Therefore = 4 and. See diagram at right.

Problem Two

Problem. In triangle ABC, D, E, and F are on BC, CA, and AB, respectively, so that, BD = CE = 3, and BF = 5. If DE and CF intersect at O, compute and. Solution. As this problem involves a transversal, we must use split masses on point C. We may arbitrarily assign the mass of point A to be 15. By ratios of lengths, the mass at B must be 6 and the mass at C is split 10 towards A and 9 towards B. By summing masses, we get the masses at D, E, and F to be 15, 25, and 21, respectively. Therefore and.

Problem Three

Problem. In triangle ABC, points D and E are on sides BC and CA, respectively, and points F and G are on side AB with G between F and B. BE intersects CF at point O_1 and BE intersects DG at point O_2. If FG = 1,, and BG = CE = 3, compute. Solution. This problem involves two central intersection points, O_1 and O_2, so we must use multiple systems.