The Centroid

http://www.mathwithmccarthy.com/1/post/2012/01/balancing-act.html

Outline of Investigation• Investigate the intersection of the medians of

a given triangle.• Six parts

1. Draw one example on graph paper.2. Investigate using GeoGebra3. Establish a conjecture4. Prove the conjecture using areas of triangles5. Prove the conjecture using coordinates6. Prove the conjecture using dilations (later)

Using Quadrille Paper

• Construct a triangle using most of one sheet of quadrille graph paper. It helps if the vertices occur at intersections of vertical and horizontal lines an even number of lines apart.

• Indicate the midpoints of each of the sides.• Draw the medians – lines that connect the

midpoints to the opposite vertex.• What do you notice?

Using GeoGebra• Construct points A,B,C and triangle ABC.• Construct midpoints D,E on sides AB, AC.• Construct medians CD, BE and their intersection point

F.• Construct ray AF and its intersection (G) with side BC.• Compare lengths CG, BG.• Compare lengths AF, FG.• Formulate conjectures

based on observations. • Find the point X that

minimizes the sum of the

squares of the distances

to the vertices A, B, C.

A Proof—Using Areas• Given triangle ABC.• Construct TWO medians BE and CD.• Indicate F as the intersection of these TWO medians.• Construct ray AF, intersecting side BC at point G.• Let A1, … A6 denote the areas of the six triangles.• Denote p=CG and q=BG• Objective: prove that AG is the THIRD median, i.e., p=q

A Proof (cont’d)• A1=A6 and A2=A3

Reason: equal base lengths and heights.• A1+A2+A3=A6+A1+A2

Reason: each equals half the area of triangle ABC since E and D are midpoints.

• Therefore, A3=A6 and A1=A2=A3=A6Let T be this common value

A Proof (cont’d)• A4/A5=p/q and q(A4)=p(A5).

REASON: Triangles CFG and BFG have the same altitudes• (A2+A3+A4)/(A1+A5+A6)=p/q => q(2T+A4)=p(2T+A5)

REASON: Triangles AGC and AGB have the same altitudes • 2Tq + q(A4) = 2Tp + p(A5) =>p=q

A Proof Using Coordinates

Conjecture: The medians of a triangle ABC intersect at a point that is 2/3 of the way from each vertex to the midpoint of the opposite side.Let the vertices of triangle ABC be represented as

A: (a1,a2), B: (b1,b2), C: (c1,c2)

1. Determine the coordinates of the midpoint (G) of side BC.

ANSWER: ( (b1+c1)/2, (b2+c2)/2 )

A Proof Using Coordinates (cont’d)

2. Determine the coordinates (x,y) of the point that is 2/3 of the way from A to G.ANSWER: x= (1/3)a1 + (2/3)(1/2)(b1+c1)

(x,y) = ( (1/3)(a1+b1+c1), (1/3)(a2+b2+c2) ).

3. What is the corresponding result for the other medians? Explain.ANSWER: All that we need do is permute the symbols a,b,c. The result is the same point (x,y).

Interpretations & Applications

• CENTROID is the point – for which the sum of the squares of the distances

from the three vertices is as small as possible.– through which the best (least squares) linear fit

to three data points is guaranteed to pass.– on which one can balance the triangle.

An Extension to 3D• Given 4 points A, B, C, D in 3D space.• Connect the points to form a tetrahedron (triangular

pyramid).• Construct line segments from each vertex to the

centroid of the opposite face.• These 4 segments intersect at a point that is 75% of the

way from each vertex to the opposite face.• The coordinates of this point of concurrency is the

average of the coordinates of the vertices, e.g., x=(1/4)(a1 + b1 + c1 + d1), and serve as the center of mass of a solid, uniform density tetrahedron.