Mass point geometry

Mass point geometry, colloquially known as mass points, is a geometry problem-solving technique which applies the physical principle of the center of mass to geometry problems involving triangles and intersecting cevians.^[1] All problems that can be solved using mass point geometry can also be solved using either similar triangles, vectors, or area ratios^[2], but mass point geometry is far quicker than those methods^[3] and thus is used more often on math competitions in which time is an important factor. Though modern mass point geometry was developed in the 1960s by New York high school students,^[4] the concept has been found to have been used as early as 1827 by August Ferdinand Möbius in his theory of homogenous coordinates.^[5]

Definitions

Example of mass point addition

The theory of mass points is rigorously defined according to the following definitions:^[6]

• Mass Point - A mass point is a pair (m,P), also written as mP, including a mass, m, and an ordinary point, P on a plane.
• Coincidence - We say that two points mP and nQ coincide if and only if m = n and P = Q.
• Addition - The sum of two mass points mP and nQ has mass m + n and point R where R is the point on PQ such that PR:RQ = n:m. In other words, R is the fulcrum point that perfectly balances the points P and Q. An example of mass point addition is shown at right. Mass point addition is closed, idempotent, commutative, and associative.
• Scalar Multiplication - Given a mass point mP and a positive real scalar k, we define multiplication to be k(m,P) = (km,P). Mass point scalar multiplication is distributive over mass point addition.

Methods

Concurrent cevians

The method of using mass point geometry to solve problems with concurrent cevians is quite simple, and does not require much knowledge of the theory behind it. To begin with, a single point is assigned an arbitrary mass, usually one that allows the other masses in the problem to be integral. The masses at the other points are calculated so that the feet of cevians are the sum of the two mass point vertices they are between. For each cevian, the point of concurrency is the sum of the mass point vertex and foot of that cevian. 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

• Routh's theorem - Many problems involving triangles with cevians will ask for areas, and mass points does not provide a method for calculating areas. However, Routh's theorem, which goes hand in hand with mass points, uses ratios of lengths to calculate the ratio of areas between a triangle and a triangle formed by three cevians.
• Special cevians - When given cevians with special properties, like an angle bisector or an altitude, other theorems may be used alongside mass point geometry that determine length ratios. One very common theorem used likewise is the angle bisector theorem.
• Stewart's theorem - When asked not for the ratios of lengths but for the actual lengths themselves, Stewart's theorem may be used to determine the length of the entire segment, and then mass points may be used to determine the ratios and therefore the necessary lengths of parts of segments.
• Higher dimensions - The methods involved in mass point geometry are not limited to two dimensions; the same methods may be used in problems involving tetrahedra, or even higher-dimensional shapes, though it is rare that a problem involving four or more dimensions will require use of mass points.

Examples

Diagram for solution to Problem One

Diagram for solution to Problem Two

Diagram for Problem Three

Diagram for Problem Three, System One

Diagram for Problem Three, System Two

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 $\tfrac{OB}{OE}$ and $\tfrac{OD}{OA}$.

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 $\tfrac{OB}{OE}$ = 4 and $\tfrac{OD}{OA} = \tfrac{3}{2}$. See diagram at right.

Problem Two

Problem. In triangle ABC, D, E, and F are on BC, CA, and AB respectively so that AE = AF = CD = 2, BD = CE = 3, and BF = 5. If DE and CF intersect at O, compute $\tfrac{OD}{OE}$ and $\tfrac{OC}{OF}$.

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 $\tfrac{OD}{OE} = \tfrac{25}{15} = \tfrac{5}{3}$ and $\tfrac{OC}{OF} = \tfrac{21}{10 + 9} = \tfrac{21}{19}$.

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 O₁ and BE intersects DG at point O₂. If FG = 1, AE = AF = DB = DC = 2, and BG = CE = 3, compute $\tfrac{O_1O_2}{BE}$.

Solution. This problem involves two central intersection points, O₁ and O₂, so we must use multiple systems.

• System One. For the first system, we will choose O₁ as our central point, and we may therefore ignore segment DG and points D, G, and O₂. We may arbitrarily assign the mass at A to be 6, and by ratios of lengths the masses at B and C are 3 and 4 respectively. By summing masses, we get the masses at E, F, and O₁ to be 10, 9, and 13 respectively. Therefore, $\tfrac{EO_1}{BO_1} = \tfrac{3}{10}$ and $\tfrac{EO_1}{BE} = \tfrac{3}{13}$.

• System Two. For the second system, we will choose O₂ as our central point, and we may therefore ignore segment CF and points F and O₁. As this system involves a transversal, we must use split masses on point B. We may arbitrarily assign the mass at A to be 3, and by ratios of lengths, the mass at C is 2 and the mass at B is split 3 towards A and 2 towards C. By summing masses, we get the masses at D, G, and O₂ to be 4, 6, and 10 respectively. Therefore, $\tfrac{BO_2}{EO_2} = \tfrac{5}{3 + 2} = 1$ and $\tfrac{BO_2}{BE} = \tfrac{1}{2}$.

• Original System. We now know all the ratios necessary to put together the ratio we are asked for. The final answer may be found as follows:

$\tfrac{O_1O_2}{BE} = \tfrac{BE - BO_2 - EO_1}{BE} = 1 - \tfrac{BO_2}{BE} - \tfrac{EO_1}{BE} = 1 - \tfrac{1}{2} - \tfrac{3}{13} = \tfrac{7}{26}.$

See also

• Cevian
• Ceva's theorem
• Menelaus's theorem
• Stewart's theorem
• Angle bisector theorem
• Routh's theorem
• Barycentric coordinates

Notes

1. ^ Rhoad, R., Milauskas, G., and Whipple, R. Geometry for Enjoyment and Challenge. McDougal, Littell & Company, 1991.
2. ^ http://mathcircle.berkeley.edu/archivedocs/2007_2008/lectures/0708lecturesps/MassPointsBMC07.ps
3. ^ http://www.artofproblemsolving.com/Wiki/index.php/Mass_Point_Geometry
4. ^ Rhoad, R., Milauskas, G., and Whipple, R. Geometry for Enjoyment and Challenge. McDougal, Littell & Company, 1991
5. ^ D. Pedoe Notes on the History of Geometrical Ideas I: Homogeneous Coordinates. Math Magazine (1975), 215-217.
6. ^ H. S. M. Coxeter, Introduction to Geometry, pp. 216-221, John Wiley & Sons, Inc. 1969