Mass point geometry

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. ^[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 many students prefer to use mass points. Though modern mass point geometry was developed in the 1960s by New York high school students, ^[3] the concept has been found to have been used as early as 1827 by August Ferdinand Möbius in his theory of homogeneous coordinates. ^[4]

Definitions

Example of mass point addition

The theory of mass points is defined according to the following definitions: ^[5]

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

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

• 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 ${\displaystyle ABC}$, ${\displaystyle E}$ is on ${\displaystyle AC}$ so that ${\displaystyle CE=3AE}$ and ${\displaystyle F}$ is on ${\displaystyle AB}$ so that ${\displaystyle BF=3AF}$. If ${\displaystyle BE}$ and ${\displaystyle CF}$ intersect at ${\displaystyle O}$ and line ${\displaystyle AO}$ intersects ${\displaystyle BC}$ at ${\displaystyle D}$, compute ${\displaystyle {\tfrac {OB}{OE}}}$ and ${\displaystyle {\tfrac {OD}{OA}}}$.

Solution. We may arbitrarily assign the mass of point ${\displaystyle A}$ to be ${\displaystyle 3}$. By ratios of lengths, the masses at ${\displaystyle B}$ and ${\displaystyle C}$ must both be ${\displaystyle 1}$. By summing masses, the masses at ${\displaystyle E}$ and ${\displaystyle F}$ are both ${\displaystyle 4}$. Furthermore, the mass at ${\displaystyle O}$ is ${\displaystyle 4+1=5}$, making the mass at ${\displaystyle D}$ have to be ${\displaystyle 5-3=2}$ Therefore ${\displaystyle {\tfrac {OB}{OE}}}$${\displaystyle =4}$ and ${\displaystyle {\tfrac {OD}{OA}}={\tfrac {3}{2}}}$. See diagram at right.

Problem Two

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

Solution. As this problem involves a transversal, we must use split masses on point ${\displaystyle C}$. We may arbitrarily assign the mass of point ${\displaystyle A}$ to be ${\displaystyle 15}$. By ratios of lengths, the mass at ${\displaystyle B}$ must be ${\displaystyle 6}$ and the mass at ${\displaystyle C}$ is split ${\displaystyle 10}$ towards ${\displaystyle A}$ and ${\displaystyle 9}$ towards ${\displaystyle B}$. By summing masses, we get the masses at ${\displaystyle D}$, ${\displaystyle E}$, and ${\displaystyle F}$ to be ${\displaystyle 15}$, ${\displaystyle 25}$, and ${\displaystyle 21}$, respectively. Therefore ${\displaystyle {\tfrac {OD}{OE}}={\tfrac {25}{15}}={\tfrac {5}{3}}}$ and ${\displaystyle {\tfrac {OC}{OF}}={\tfrac {21}{10+9}}={\tfrac {21}{19}}}$.

Problem Three

Problem. In triangle ${\displaystyle ABC}$, points ${\displaystyle D}$ and ${\displaystyle E}$ are on sides ${\displaystyle BC}$ and ${\displaystyle CA}$, respectively, and points ${\displaystyle F}$ and ${\displaystyle G}$ are on side ${\displaystyle AB}$ with ${\displaystyle G}$ between ${\displaystyle F}$ and ${\displaystyle B}$. ${\displaystyle BE}$ intersects ${\displaystyle CF}$ at point ${\displaystyle O_{1}}$ and ${\displaystyle BE}$ intersects ${\displaystyle DG}$ at point ${\displaystyle O_{2}}$. If ${\displaystyle FG=1}$, ${\displaystyle AE=AF=DB=DC=2}$, and ${\displaystyle BG=CE=3}$, compute ${\displaystyle {\tfrac {O_{1}O_{2}}{BE}}}$.

Solution. This problem involves two central intersection points, ${\displaystyle O_{1}}$ and ${\displaystyle O_{2}}$, so we must use multiple systems.

• System One. For the first system, we will choose ${\displaystyle O_{1}}$ as our central point, and we may therefore ignore segment ${\displaystyle DG}$ and points ${\displaystyle D}$, ${\displaystyle G}$, and ${\displaystyle O_{2}}$. We may arbitrarily assign the mass at ${\displaystyle A}$ to be ${\displaystyle 6}$, and by ratios of lengths the masses at ${\displaystyle B}$ and ${\displaystyle C}$ are ${\displaystyle 3}$ and ${\displaystyle 4}$, respectively. By summing masses, we get the masses at ${\displaystyle E}$, ${\displaystyle F}$, and ${\displaystyle O_{1}}$ to be 10, 9, and 13, respectively. Therefore, ${\displaystyle {\tfrac {EO_{1}}{BO_{1}}}={\tfrac {3}{10}}}$ and ${\displaystyle {\tfrac {EO_{1}}{BE}}={\tfrac {3}{13}}}$.
• System Two. For the second system, we will choose ${\displaystyle O_{2}}$ as our central point, and we may therefore ignore segment ${\displaystyle CF}$ and points ${\displaystyle F}$ and ${\displaystyle O_{1}}$. As this system involves a transversal, we must use split masses on point ${\displaystyle B}$. We may arbitrarily assign the mass at ${\displaystyle A}$ to be ${\displaystyle 3}$, and by ratios of lengths, the mass at ${\displaystyle C}$ is ${\displaystyle 2}$ and the mass at ${\displaystyle B}$ is split ${\displaystyle 3}$ towards ${\displaystyle A}$ and 2 towards ${\displaystyle C}$. By summing masses, we get the masses at ${\displaystyle D}$, ${\displaystyle G}$, and ${\displaystyle O_{2}}$ to be 4, 6, and 10, respectively. Therefore, ${\displaystyle {\tfrac {BO_{2}}{EO_{2}}}={\tfrac {5}{3+2}}=1}$ and ${\displaystyle {\tfrac {BO_{2}}{BE}}={\tfrac {1}{2}}}$.
• 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:
  $${\displaystyle {\tfrac {O_{1}O_{2}}{BE}}={\tfrac {BE-BO_{2}-EO_{1}}{BE}}=1-{\tfrac {BO_{2}}{BE}}-{\tfrac {EO_{1}}{BE}}=1-{\tfrac {1}{2}}-{\tfrac {3}{13}}={\tfrac {7}{26}}.}$$

See also

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

Notes

1. Rhoad, R., Milauskas, G., and Whipple, R. Geometry for Enjoyment and Challenge. McDougal, Littell & Company, 1991.
2. "Archived copy". Archived from the original on 2010-07-20. Retrieved 2009-06-13.
3. Rhoad, R., Milauskas, G., and Whipple, R. Geometry for Enjoyment and Challenge. McDougal, Littell & Company, 1991
4. D. Pedoe Notes on the History of Geometrical Ideas I: Homogeneous Coordinates. Math Magazine (1975), 215-217.
5. H. S. M. Coxeter, Introduction to Geometry, pp. 216-221, John Wiley & Sons, Inc. 1969