Center of Gravity and Motion of Material Points

What is this about?


Let $A,$ $B,$ $C,$ $D$ be points in a plane and $\lambda$ and $\mu$ two real numbers. Define $E$ and $F$ on $AB$ and $CD$ by $BE/AB=DF/CD=\lambda.$ Define $K$ and $L$ on $AD$ and $BC$ by $DK/AD=BL/BC=\mu.$ Finally, define $G,$ $H,$, $I,$ $M$ - the midpoints of $AC,$ $EF,$ $BD,$ and $KL,$ respectively.

Four material points

Prove that $G,$ $H,$ $I,$ and $M$ are collinear


Take a look into center of gravity (barycenter) and barycentric coordinates of material points.

Solution 1

Consider $A,$ $B,$ $C,$ $D$ as material points, placing masses $\lambda$ at $A$ and $C$ and $1-\lambda$ at $B$ and $D.$ Then, with $E=\lambda A + (1-\lambda)B,$ the material point $(E,1)$ is the barycenter of points $(A,\lambda)$ and $(B,1-\lambda).$ Similarly, $(F,1)$ is the barycenter of points $(C,\lambda)$ and $(D,1-\lambda).$ The midpoint $(H,2)=((E+F)/2,2)$ is the barycenter of all four points.

Four material points - solution 1

Now, proceed a little differently, first finding $(G,2\lambda)$ and $(I,2(1-\lambda)$ is the barycenters of pairs $(A,\lambda),$ $(C,\lambda)$ and $(B,1-\lambda),$ $(D,1-\lambda).$ Then $(\lambda (A+C)/2 + (1-\lambda)(B+D)/2, 2)$ is the barycenter of all four points and needs to coincide with $(H,2).$ Or, differently, since the barycenter of two points is on the line between these points, the barycenter of the four points $A,$ $B,$ $C,$ $D$ is on the lines $EF$ as well as $GI$ and, hence, at their intersection, implying that $G,$ $H,$ and $I,$ are collinear.

Now, with $\mu$ instead of $\lambda$ find first barycenters of $A$ and $D$ and also of $B$ and $C$ and then combine the two into the barycenter of all four points, with the same result as above. This will show that $M$ is collinear with $G$ and $I$ and, therefore, also with $H.$

Solution 2

The parametric equations of lines $AB$ and $CD$ are, respectively, $f(t)=A+t\mathbf{AB}$ and $g(t)=C+t\mathbf{CD},$ where the two vectors are in bold (e.g., $\mathbf{AB}=B-A)$ and $t$ is a real parameter such that $f(0)=A,$ $f(1)=B,$ $g(0)=C,$ $g(1)=D.$ The function $h=(f+g)/2$ is also linear such that $h(0)=G$ and $h(1)=I.$ Furthermore, for any $t,$ $h(t)$ is the midpoint between $f(t)$and $g(t),$ as well as being on the line between $h(0)$ and $h(1).$ The given problem corresponds to choosing $t=1-\lambda.$

Four material points - solution 2

The derivation for point $M$ is analogous.

We may imagine one traveler walking along $AB$ and another along $CD$ with constant speeds such that they are at $A$ and $C$ at time $t=0$ and at $B$ and $D$ at time $T=1.$ If there is an elastic band stretched between them, its midpoint will also move along a straight line with a fixed speed.

Finally, note that, instead of talking of a midpoint, we could have chosen any fixed ratio $r$ and points it defines between the pairs $A,$ $C$ and $B,$ $D.$


The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.

This is a clear generalization of a property of bimedians and Newton's line in a quadrilateral.

Finally, the statement is actually a reformulation of Hjelmslev's Theorem.

Barycenter and Barycentric Coordinates

  1. 3D Quadrilateral - a Coffin Problem
  2. Barycentric Coordinates
  3. Barycentric Coordinates: a Tool
  4. Barycentric Coordinates and Geometric Probability
  5. Ceva's Theorem
  6. Determinants, Area, and Barycentric Coordinates
  7. Maxwell Theorem via the Center of Gravity
  8. Bimedians in a Quadrilateral
  9. Simultaneous Generalization of the Theorems of Ceva and Menelaus
  10. Three glasses puzzle
  11. Van Obel Theorem and Barycentric Coordinates
  12. 1961 IMO, Problem 4. An exercise in barycentric coordinates
  13. Centroids in Polygon
  14. Center of Gravity and Motion of Material Points
  15. Isotomic Reciprocity
  16. An Affine Property of Barycenter
  17. Problem in Direct Similarity
  18. Circles in Barycentric Coordinates
  19. Barycenter of Cevian Triangle
  20. Concurrent Chords in a Circle, Equally Inclined

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