Vectors, center of mass, medians and cevians

When reading these notes, please draw pictures. They will help. If you see anymistakes, please let me know.

We always regard the plane as a set of points. Thus the plane is the same thingas the set of all points in the plane. We can introduce a coordinate system in theplane. Then any point is represented by a pair of coordinates (x, y).

Definition 1. A vector is an oriented segment defined up to parallel translation.

This definition requires some explanation. First, a segment AB with endpointsA and B is said to be oriented if an order of the endpoints is fixed. We will write~AB for the oriented segment with the order of the endpoints as indicated: A comes

first and B comes second. An oriented segment ~AB is also called a vector. We willidentify two vectors ~AB and ~CD if the segments AB and CD are two parallel sidesof a parallelogram, whose other parallel sides are AC and BD or if they lie on thesame line, have equal lengths and the same direction.

Problem 1. For any vector ~x and a point A, there is a unique point B such that~AB = ~x.

We will usually denote vectors by letters with arrows: ~x, ~y, etc. Vectors can beidentified with points as follows. Given a point A, we define the corresponding vectorjust as OA, where O is the origin (the point with coordinates (0, 0)). Conversely,given a vector ~x, we define the corresponding point as the unique point A such that~OA = ~x. We will use this identification between points and vectors sometimes.The coordinates of a vector are defined as the coordinates of the corresponding

point. The coordinates of the vector ~x are usually written as x1 and x2. We willalso write ~x = (x1, x2) thus identifying the vector ~x with the pair of its coordinates.

The sum of two vectors ~x and ~y is defined geometrically as follows. Suppose that~x = ~AB and ~y = ~BC for some points A, B and C. Then we set ~x + ~y = ~AC. Inother words, if the end of the first vector coincides with the beginning of the secondvector (and we can always arrange that), then the sum of the two vectors is thevector going from the beginning of the first vector to the end of the second vector.

Problem 2. The sum of two vectors ~x and ~y is independent of the choice of pointsA, B and C such that ~x = ~AB and ~y = ~BC.

Problem 3. In coordinates, the sum of two vectors ~x = (x1, x2) and ~y = (y1, y2) isgiven by the formula

~x + ~y = (x1 + y1, x2 + y2).

The zero vector is the vector ~0 = ~AA. Since vectors are only defined up to aparallel translation, this definition does not depend on the choice of the point A. Ifwe add the zero vector to any vector ~x, then we obtain ~x.

Let ~x = ~AB be a vector and a real number. Choose a point C on the line ABsuch that

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the length of the segment AC is times the length of AB, if > 0, then the points B and C are on the same side of A, if < 0, then the points B and C are on different sides of A.

Then the product ~x is defined as the vector ~AC.In particular, if we multiply the zero vector by any real number, then we obtain

the zero vector. If we multiply any vector by zero, then we obtain the zero vector.It is not hard to see that if we multiply a vector ~AB by 1, then we obtain thevector ~BA, which is called the opposite vector for the vector ~AB. The sum of twopositive vectors is always the zero vectors. All these statements are very simple, butplease make sure you understand their geometric meaning.

Problem 4. The product ~x is independent of the choice of points A and B suchthat ~x = ~AB.

Problem 5. In coordinates, the product of a real number and a vector ~x = (x1, x2)is given by the formula

~x = (x1, x2).

Hint: use similar triangles.The advantage of using vectors is that one can perform essentially the same alge-

braic manipulations on vectors as on numbers. One only needs to always remember,that in general the division of vectors is undefined.

Let us discuss a geometric way to define the center of mass. Physically, anyparticle has mass. Mathematically, we just assign a positive real number to a pointand call it the mass of the point. We will sometimes write (A,m) for a point A withmass m. Of course, we can put any positive mass to any point.

Let (A,m) and (B, ) be two points (so that m is the mass of A and is the massof B). A point G in the plane is called a center of mass of the system A,B if thefollowing equality holds:

m ~GA + ~GB = ~0.

In particular, we see that the vectors ~GA and ~GB are proportional, therefore, theygo along the same line. In other words, the point G belongs to the line AB.

We can say more than that. The vector ~GB can be expressed as (m/) ~GA,the coefficient being negative. Since the coefficient is negative, the points A and Bare on different sides of the point G. We can rephrase this as follows: the point Galways belongs to the segment AB. How to find G on the segment AB? The rule isthe following: the point G divides the segment AB in the ration : m, i.e.

|AG|

|GB|=

m.

Problem 6. Prove this.

In particular, since we have a unique way to find the point G, it follows that thepoint G is unique. Still, let us prove uniqueness differently, by using vector algebra.

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Suppose that we have two points G and G satisfying the same property, namely,

m ~GA + ~GB = 0, m ~GA + ~GB = 0.

We will prove that G = G in this case. This is the same as to prove that the vector~GG is equal to the zero vector. Subtract the equation on G from the equation on

G:

m( ~GA ~GA) + ( ~GB ~GB) = 0.

By the properties of the vector addition, we have

~GA ~GA = ~GG, ~GB ~GB = ~GG.

Therefore, we have

(m + ) ~GG = 0.

It follows that ~GG = 0, therefore, G = G.Thus for any pair of points with positive masses assigned, there is a uniquely

defined center of mass. However, it would be nice to have an explicit formula forthe center of mass. Up to now, we only have the defining property, which does notlook like an explicit expression. Take any point O in the plane. Define a point G bythe following explicit formula (this is a formula defining the vector ~OG, but this isthe same thing as to define the point G, provided that we know the exact positionof O):

~OG =m ~OA + ~OB

m + .

Here we divide a vector by the sum of masses. By definition, this is the same thingas to multiply the vector by the number 1/(m + ).

It turns out that the point G thus defined is the center of mass for (A,m) and(B, ). In particular, this point is independent of the choice of the point O (althoughour explicit expression uses this choice). To prove that G is the center of mass, justcheck that G satisfies the property defining the center of mass (we already knowthat a point with this property is unique). Indeed, we have

~GA = ~OA ~OG, ~GB = ~OB ~OG.

Therefore,

m ~GA + ~GB = m( ~OA ~OG) + ( ~OB ~OG) = m ~OA + ~OB (m + ) ~OG = ~0.

We provided an explicit way of finding the center of mass for any pair of points.In particular, it follows that the center of mass always exists.

Problem 7. Suppose that points A and B have equal masses. Then their center ofmass is the midpoint of the segment AB.

Consider a triangle ABC. Assign equal masses to the vertices of this triangle.We can assume without loss of generality that the masses are equal to 1. Define

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the center of mass for the three points A,B,C as a point G in the plane with thefollowing property:

~GA + ~GB + ~GC = ~0.

To prove the existence of such point G, we can choose any point O in the plane,and define the vector ~OG by the following formula:

~OG =~OA + ~OB + ~OC

3.

This is very similar to what we did for a pair of points. Note, however, that allmasses are equal to 1 in our case. This is why all coefficients in the numerator areequal to 1 and the denominator is equal to 3 = 1 + 1 + 1.

Problem 8. Check that the point G defined by this formula is indeed the center ofmass for points A, B and C.

Problem 9. Prove that the center of mass of points A, B and C is unique. Inparticular, whatever point O we choose, the formula we gave defines the same pointG.

We can now choose the point O to be A. Then we have the following formula:

~AG =~AB + ~AC

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We already know that the midpoint A of the side BC is given by the formula:

~AA =~AB + ~AC

2.

Problem 10. Explain this.

We see that the vectors ~AG and ~AA are proportional (they are both equal to the

vector ~AB + ~AC times certain real numbers), which means that the three points A,A and G lie on the same line. In other words, the point G lies on the line AA.But AA is the median of the triangle ABC connecting A with the midpoint of theopposite side. Thus G belongs to the median AA. Similarly, we can prove that thepoint G belongs to any other median of the triangle ABC. It follows that all threemedians of the triangle intersect at one point, which is the center of mass for thesystem of vertices (provided that we assigned equal masses to all vertices).

Let us now discuss a more general notion of the center of mass. Let A1, . . . , Anbe points in the plane and m1, . . . ,mn positive numbers called masses. Assign to apoint Ak the mass mk. Then we can think of Ak as a physical particle of mass mk.However, our construction is purely mathematical.

By definition, a center of mass of points A1, . . . , An is defined as a point G suchthat

m1 ~GA1 + m2 ~GA2 + + mn ~GAn = 0.4

Problem 11. Let O be any point in the plane (which may be one of the pointsA1, . . . , An or may be different from all of them). Set m = m1 + + mn. Provethat the point

~OG =m1m

~OA1 + +mnm

~OAn

is a center of mass of points (A1,m1), . . . , (An,mn).

This is very similar to what we did for a pair of points and for the three verticesof a triangle. The only difference is that now we have an arbitrary number of points.

Problem 12. Prove that the center of mass is unique. In other words, if G andG are points, both of which satisfy the definition of a center of mass, then G = G.Hint: show that the vector ~GG is the zero vector.

Finally, there is a useful rule for finding the center of mass for a system of points.Divide our system of points into two groups (we can divide in any way we want).Then compute the center of mass for each of the group. Replace each group witha single point, namely, its center of mass. The mass of this point will be the totalmass of the group (i.e. the sum of masses of all points from the group). We nowhave only two points with certain masses assigned. We can take the center of massfor this pair of points. It turns out that this is the same thing as the center of massfor the whole system.

The rule we just described can be formalized as the following mathematical state-ment:

Problem 13. Let a point P be the center of mass of points (A1,m1), . . . , (An,mn), apoint Q the center of mass of points (B1, 1), . . . , (Bk, k), and a point G the centerof mass of points (P,m) and (Q,), where m = m1 + +mn and = 1 + +k.Prove that G is the center of mass of points

(A1,m1), . . . , (An,mn), (B1, 1), . . . , (Bk, k).

Hint: first prove this statement for the case n = 1 and k = 2 (i.e. the first groupcontains just one point, and the second group consists of two points)

A good illustration to this rule is again the centroid of a triangle (i.e. the centerof mass of all vertices). Suppose that the vertices of a triangle ABC have equalmasses. Then their center of mass can be found as follows. We divide the tripleof points A,B,C into two groups. One group will contain the point A. The othergroup will contain the pair of points B and C.

For the first group, we do not need to do anything. The point A is the center ofmass of itself. Its mass is equal to 1.

For the second group, the center of mass is the midpoint A of the segment BC.We need to assign mass 2 to this point. This is the total mass of the second group.

Finally, to find the centroid of the triangle, we need to compute the center ofmass for points (A, 1) and (A, 2). In this way, we obtain yet another proof of thestatement that the centroid G divides each median in the ratio 2 : 1.

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Let us now place arbitrary masses to the vertices of the triangle ABC. Denoteby mA, mB and mC the masses of points A, B and C, respectively. The center ofmass G now satisfies the following property

mA ~GA + mB ~GB + mC ~GC = ~0

and can be computed by the following formula:

~OG =mA ~OA + mB ~OB + mC ~OC

mA + mB + mC,

where O is an arbitrary point in the plane.

Problem 14. Let a point A be the center of mass for B and C, a point B thecenter of mass for A and C, and C the center of mass for A and B. Using the rulefrom problem 13, show that the lines AA, BB and CC all pass through a point G.In particular, they intersect at one point.

This observation can help to solve the following problem. Let A, B and C bepoints on sides BC, AC and AB, respectively. What are the conditions on pointsA, B and C necessary and sufficient for the three lines AA, BB and CC tointersect at one point. If these lines do intersect at one point, then they are calledcevians.

We saw that if A, B and C are centers of mass for systems of points {B,C},{A,C} and {A,B}, respectively, for some values of masses, then the lines AA, BB

and CC indeed intersect at one point.

Problem 15. Prove the Ceva theorem (see the textbook) using the center of mass.

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