CENTERS OF GRAVITY OF POLYGONS





The center of gravity of an object is its balance point. The balance point of a two dimensional figure or a three dimensional figure of constant height and density can be thought of a balance point for the area of the figure. One way of locating such a point is finding the intersection of lines that divide the figure into regions of equal area.

1. Triangles

The median of a triangle is one such line (segment). To find the center of gravity of a triangle (called the centroid), find the intersection of its medians. However, every triangle has three medians. We need to establish that the medians of a triangle are always concurrent. Then we can officially conjecture that "the" intersection of the medians is "the" centroid of a triangle. We will also need to prove this conjecture using our regions of equal area claim.

Following the argument discussed in Rethinking Proof, p. 54-55, we will use one proof to establish our two propositions.

Proposition 1: The three medians of a triangle always intersect in one point, and this point of

concurrency is the centroid. Proposition 2: The six small triangles formed by the sides of the triangle and subsets of its medians have equal areas.

Given: Triangle ABC Medians
            AD and CE intersecting at point G

Show: Median BX passes through G (X =dummy variable for midpoint of AC)
            Triangles AXG, CGX, GDB, BDG, GEB, and AGE have equal areas.

Proof:

Set up the argument by making some constructions. Construct segment BG. Then extend BG to H so that GH = GB, and G is between H and B. Let F be the intersection of GH and AC. Construct HA and HC.


 
Statements
Reasons
Triangle ABH ~ Triangle EBG  EB = 1/2AB, <ABH=<EBG, GB= 1/2HB
EG || AH <BAH & <BEG congruent (by similarity) & corresponding with transversal AB
Triangle CBH ~ Triangle DBG  BD=1/2BC, <CBH=<DBG, GB= 1/2HB
DG || CH <BCH & <BDG congruent (by similarity) & corresponding with transversal AB
GC || AH EG || AH; C collinear with E,G
AG || CH DG || CH; A collinear with D,G
Quadrilateral AHCG is a parallelogram Two pairs of || sides from previous two statements
Segments AC and GH bisect each other Diagonals of parallelogram
F is the midpoint of AC, ie. F=X Definition of bisect
*Therefore, BF is median  Definition of median
*Therefore, median passes through G G lies on BF
Area (AFG ) = Area (CFG)

Area (CDG ) = Area (BDG)

Area (BEG ) = Area (AEG)

 Congruent bases (F,D,E are midpoints of AC,CB,BA, resp)

Equal height because of shared altitudes

(altitudes from G to AC, G to CB, and G to BA)

Area = 1/2(Base)(Height)
Area (AFB ) = Area (CFB) Congruent bases (F is midpoint of AC)

Equal height because of shared altitude from G to AC

Area = 1/2(Base)(Height)
Area (ABG ) = Area (CBG) Area (ABG) = Area (AFB) - Area (AFG) 

Area (CBG) = Area (CFB) - Area (CFG) 

Differences of equal areas by previous two statements
Area (CDG ) = Area (BDG) = Area (BEG ) = Area (AEG) 2Area (CDG) = Area(CGB) = Area (ABG) = 2 Area (BEG) follows from above

Transitivity
Area (AFG) = Area (CDG)  
*Therefore, all six triangles are congruent. From preceding statements by transitivity
*G is centroid, or center of gravity, of triangle ABC. G is the common vertex of 6 adjacent triangles of equal area that make up ABC.

Now that we have gone to the trouble of developing this proof, we want to reuse this result in locating the centers of gravity of other polygons rather than starting from scratch.

2. Convex Quadrilaterals


Given a convex quadrilateral (the ordinary, familiar type), we can decompose the quadrilateral as a pair of triangles two different ways using each of the two diagonals.
 
 

Construct a diagonal of a convex quadrilateral and consider the two triangles formed. We can easily find the centroid of each triangle using our result in 1. Then if we think of each triangle's area as a point mass located as the centroid, we have only to find the balance point of the segment connecting the two triangles' centroids. Since these two triangles will generally have different areas, the segment's balance point is generally not the midpoint.


However, if we then decompose the quadrilateral into two triangles using the other diagonal , the quadrilateral's center of gravity must also lie on the segment connecting these two triangles' centroids.


So we can locate the center of gravity by finding the intersection of the two segments connecting the centroids of the paired triangles.

 
 

3. Concave Quadrilaterals

The general procedure used above for convex quadrilaterals does not work for concave quadrilaterals because the two segments connecting the pairs of centroids do not intersect.


In this case, the balance point of quadrilateral ABDC would not lie on the green SEGMENT. The green centroids that are endpoints of this segment are not balance points of subregions of the quadrilateral. Instead, the balance point of triangle ADC accounts for the region outside the quadrilateral that is balance by the centroid of triangle ABC . Quadrilateral ABCD's balance point should still lie on the LINE containing the centroids of triangles ADC and ABC.

So we can find the balance point of ABCD ( or any quadrilateral, convex or concave) by generalizing our convex solution to find the intersection of lines connecting centroids of pairs of triangles formed by the diagonals.