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.

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:
  $${\displaystyle {\tfrac {O_{1}O_{2}}{BE}}={\tfrac {BE-BO_{2}-EO_{1}}{BE}}=1-{\tfrac {BO_{2}}{BE}}-{\tfrac {EO_{1}}{BE}}=1-{\tfrac {1}{2}}-{\tfrac {3}{13}}={\tfrac {7}{26}}.}$$