Difference between revisions of "Centroid"
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''Readers unfamiliar with [[homothety]] should consult the second proof.'' | ''Readers unfamiliar with [[homothety]] should consult the second proof.'' | ||
| − | Let | + | Let <math>\displaystyle D,E,F</math> be the respective midpoints of sides <math>\displaystyle BC, CA, AB</math> of triangle <math>\displaystyle ABC</math>. We observe that <math>\displaystyle DE, EF, FE</math> are parallel to (and of half the length of) <math>\displaystyle AB, BC, CA</math>, respectively. Hence the triangles <math>\displaystyle ABC, DEF</math> are homothetic with respect to some point <math>\displaystyle G</math> with dilation factor <math>\displaystyle -\frac{1}{2}</math>; hence <math>\displaystyle AD, BE, CF</math> all pass through <math>\displaystyle G</math>, and <math>\displaystyle AG = 2 GD; BG = 2 GE; CG = 2 GF</math>. Q.E.D. |
=== Proof 2 === | === Proof 2 === | ||
| − | Let | + | Let <math>\displaystyle ABC</math> be a triangle, and let <math>\displaystyle D,E,F</math> be the respective [[midpoint]]s of the segments <math>\displaystyle BC, CA, AB</math>. Let <math>\displaystyle G</math> be the intersection of <math>\displaystyle BE</math> and <math>\displaystyle CF</math>. Let <math>\displaystyle E',F'</math> be the respective midpoints of <math>\displaystyle BG, CG </math>. We observe that both <math>\displaystyle EF </math> and <math>\displaystyle E'F'</math> are [[parallel]] to <math>\displaystyle CB </math> and of half the length of <math>\displaystyle CB </math>. Hence <math>\displaystyle EFF'E' </math> is a [[parallelogram]]. Since the diagonals of parallelograms [[bisect]] each other, we have <math>\displaystyle GE = E'G = BE'</math>, or <math>\displaystyle BG = 2GE </math>. Hence each median passes through a similar trisection point of any other median; hence the medians concur. Q.E.D. |
We note that both of these proofs give the result that the distance of a vertex of a point of a triangle to the centroid of the triangle is twice the distance from the centroid of the traingle to the midpoint of the opposite side. | We note that both of these proofs give the result that the distance of a vertex of a point of a triangle to the centroid of the triangle is twice the distance from the centroid of the traingle to the midpoint of the opposite side. | ||
Revision as of 19:36, 25 August 2006
This article is a stub. Help us out by expanding it.
The centroid of a triangle is the point of intersection of the medians of the triangle and is conventionally denoted 
. The centroid has the special property that, for each median, the distance from a vertex to the centroid is twice that of the distance from the centroid to the midpoint of the side opposite that vertex. Also, the three medians of a triangle divide it into six regions of equal area.
The centroid is the center of mass of the triangle; in other words, if you connected a string to the centroid of a triangle and held the other end of the string, the triangle would be level.
The coordinates of the centroid of a coordinatized triangle is (a,b), where a is the arithmetic average of the x-coordinates of the vertices of the triangle and b is the arithmetic average of the y-coordinates of the triangle.
Proof of concurrency of the medians of a triangle
Note: The existance of the centroid is a trivial consequence of Ceva's Theorem. However, there are many interesting and elegant ways to prove its existance, such as those shown below.
Proof 1
Readers unfamiliar with homothety should consult the second proof.
Let 
be the respective midpoints of sides 
of triangle 
. We observe that 
are parallel to (and of half the length of) 
, respectively. Hence the triangles 
are homothetic with respect to some point 
with dilation factor 
; hence 
all pass through , and 
. Q.E.D.
Proof 2
Let be a triangle, and let be the respective midpoints of the segments . Let be the intersection of 
and 
. Let 
be the respective midpoints of 
. We observe that both 
and 
are parallel to 
and of half the length of . Hence 
is a parallelogram. Since the diagonals of parallelograms bisect each other, we have 
, or 
. Hence each median passes through a similar trisection point of any other median; hence the medians concur. Q.E.D.
We note that both of these proofs give the result that the distance of a vertex of a point of a triangle to the centroid of the triangle is twice the distance from the centroid of the traingle to the midpoint of the opposite side.
