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Calculus and the centroid of a triangle
The goal of this document is to verify the following physical assertion, which is presented
immediately after the proof of Theorem I I I.4.1 :
If X is the solid triangular region in RRRR
of uniform density whose vertices are the noncollinear points A, B and C, then the center of mass for X is
given by (^1 / 3 ) · [A + B + C].
There are two parts to our argument, the first of which is the following “physically obvious” statement about the effect of rigid motions (or Galilean transformations) on centroids:
Theorem 1. Let X be a subset of RRRR
which is measurable in the sense that one can define
its area by some reasonable method , let z denote the centroid of X defined by the usual sorts
of formulas from integral calculus , and let G be a Galilean transformation of RRRR
. Then G(z)
is the centroid of the subset G[X].
(Source: http://www.math.cornell.edu/~mec/Summer2008/youssef/Groups/images/triangle_rotation.jpg)
Although this statement corresponds to everyday experience with physical objects, writing up a full mathematical proof is considerably more complicated than one might expect (among other things, it is necessary to be precise about a reasonable method for defining area). Details will be given at the end of this document.
If G is a Galilean transformation of RRRR
, then the results of Unit I I show that G sends
(^1 / 3 ) · [A + B + C] to (^1 / 3 ) · [G(A) + G(B) + G(C)], and therefore the if the centroid
formula is true for X then it is also true for G[X]. Consequently, it will suffice to prove the
centroid formula for a class K of triangles DEF such that every triangle is congruent to a
triangle in F. The second part of the verification is to show that the formula is true for a suitable family of this type.
CLAIM. Every triangle ABC in RRRR
is congruent to a triangle whose vertices are ( p , 0),
(0, h ) and ( – q , 0), where p and h are positive real numbers and q is a nonnegative real
number.
Proof of the claim. We know that the triangle has at least two acute angles, and without loss of generality we may as well assume the vertices of two such angles are A and C. In this
case Corollary I I I.4.1 implies that the foot D of the perpendicular from B to AC lies between A and C (see the drawing above). Then, as the drawing suggests, there is a
Galilean transformation G be a Galilean transformation sending line AC to the x – axis such that D is sent to the origin, the first coordinate of A is negative, the first coordinate of C is positive, and the second coordinate of D is positive (the explicit construction of G is left to the reader as an exercise).
By the preceding discussion, the verification of the centroid formula reduces to doing so for all triangles like those in the drawing; in other words, we need to show that the standard integral
calculus formulas yield the value (^1 / 3 ) · ( p – q , h ) for the centroid of the closed triangular
region bounded by ABC.
For the sake of simplifying the algebra, we shall first consider the special case where q = 0.
By construction the line AB is defined by the equation y = h – ( hx / p ) , and the standard centroid formulas from integral calculus immediately yield the moment of the solid triangular
region with respect to the y – axis. Similarly, if we rewrite the equation for the line in the form
x = p – ( py / h ) , we get the moment of the triangular region with respect to the x – axis:
Formula 2. The moments of the solid triangular region with respect to the x – and y – axes
are equal to (^1 / 6 ) · p^2 h and (^1 / 6 ) · ph^2 respectively.
The derivation of this formula is a routine exercise in integral calculus.
Derivation of the centroid formula for the triangles in the Claim. Let M(x) and M(y)
denote the moments for the solid triangular region of ABC with respect to the y – and
x – axes, let M++++(x) and M++++(y) denote the corresponding moments for the solid triangular
region of ABD, and let M–––– (x) and M–––– (y) denote the corresponding moments for the solid
triangular region of ACD. The moments M++++(x) and M++++(y) are given by Formula 2, and a
similar argument shows that M–––– (x) and M–––– (y) are given by – (^1 / 6 ) · q
2 h and (^1 / 6 ) · q h
2
respectively. Therefore the total moments are given as follows:
M(x) = M++++(x) + M–––– (x) = (^1 / 6 ) · p
2 h – (^1 / 6 ) · q
2 h
M(y) = M++++(y) + M–––– (y) = (^1 / 6 ) · p h
2
2
The area of the solid triangular region for ABC is given by S = ½ ( p + q ) h , and
therefore the coordinates of the centroid are given by x* = M(x)/S = ( p + q ) / 3 and y* =
M(y)/S = h / 3 , which is what we wanted to prove.
Proof. We shall verify the assertions in the stated order.
Verification of (1). By definition M 1 (P) = x m and M 2 (P) = y m , and for the same reasons
M 1 (Q) = ( x + a ) m and M 2 (Q) = ( y + b ) m.
Verification of (2). Once again M 1 (P) = x m and M 2 (P) = y m , and for the same reasons
M 1 (Q) = x m and M 2 (Q) = – y m.
Verification of (3). This case is slightly less trivial. As the drawing below suggests, a rotation
through an angle θθθθ sends P = ( x , y ) to Q = ( x cos θθθθ – y sin θθθθ , y cos θθθθ + x sin θθθθ).
By definition the moments at Q are equal to m times its coordinates, and this yields the formula in the remaining case.
Proof of Theorem 1
Since Galilean transformations preserve areas, we know that the areas of X and G[X] are equal to the same value which we shall call C. The next step is to compute the moments of
G[X] with respect to the standard coordinate axes, and if we apply Theorem 3 we see that these moments given by the integrals on the next page. In analogy with previous notation, if W is a closed region of uniform density in the plane, then M 1 (W) and M 2 (W) will are its moments with
respect to the y – and x – axes.
The integral formulas defining the moments of W with respect to the y− and x−axes
are given by
M 1 (W ) =
W
u du dv , M 2 (W ) =
W
v du dv
and the centroid coordinates are given by
(u, v) =
M 1 (W )
C
M 2 (W )
C
Similarly, the moments of G[W ] with respect to these axes are given as follows in
the three basic cases. Throughout this discussion we adopt the notational convention
G(u, v) = (x, y).
Translations. For translations we have x = u + a and y = v + b, so that
M 1 (G[W ]) =
W
u + a du dv = M 1 (W ) + aC ,
M 2 (G[W ]) =
W
v + b du dv = M 2 (W ) + bC.
If we divide both of these equations by the area C, we see that the centroid of G[W ] has
coordinates (u + a, v + b).
Reflections. For the reflection in the second case we have x = u and y = −v, so that
M 1 (G[W ]) =
W
u du dv = M 1 (W ) ,
M 2 (G[W ]) =
W
−v du dv = −M 2 (W ).
If we divide both of these equations by the area C, we see that the centroid of G[W ] has
coordinates (u, −v).
Rotations. For the rotations in the second case we have x = u cos θ − v sin θ and
y = v cos θ + u sin θ, so that
M 1 (G[W ]) =
W
u cos θ − v sin θ du dv = M 1 (W ) cos θ − M 2 (W ) sin θ ,
M 2 (G[W ]) =
W
v cos θ + u sin θ du dv = M 2 (W ) cos θ + M 1 (W ) sin θ.
If we divide both of these equations by the area C, we see that the centroid of G[W ] has
coordinates
(u cos θ − v sin θ, v cos θ + u sin θ) = G(u, v).
