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Difference between revisions of "Centroid"

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== Proof of concurrency of the medians of a triangle ==
 
== Proof of concurrency of the medians of a triangle ==
  
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''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.''
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''Note: The existence of the centroid is a trivial consequence of [[Ceva's Theorem]].  However, there are many interesting and elegant ways to prove its existence, such as those shown below.''
  
 
=== Proof 1 ===
 
=== Proof 1 ===
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Let <math>ABC</math> be a triangle, and let <math>D,E,F</math> be the respective [[midpoint]]s of the segments <math>BC, CA, AB</math>.  Let <math>G</math> be the intersection of <math>BE</math> and <math>CF</math>.  Let <math>E',F'</math> be the respective midpoints of <math>BG, CG </math>.  We observe that both <math>EF </math> and <math>E'F'</math> are [[parallel]] to <math>CB </math> and of half the length of <math>CB </math>.  Hence <math>EFF'E' </math> is a [[parallelogram]].  Since the diagonals of a parallelogram [[bisect]] each other, we have <math>GE = E'G = BE'</math>, or <math>BG = 2GE </math>.  Hence each median passes through an appropriate trisection point of each other median and the medians concur.
 
Let <math>ABC</math> be a triangle, and let <math>D,E,F</math> be the respective [[midpoint]]s of the segments <math>BC, CA, AB</math>.  Let <math>G</math> be the intersection of <math>BE</math> and <math>CF</math>.  Let <math>E',F'</math> be the respective midpoints of <math>BG, CG </math>.  We observe that both <math>EF </math> and <math>E'F'</math> are [[parallel]] to <math>CB </math> and of half the length of <math>CB </math>.  Hence <math>EFF'E' </math> is a [[parallelogram]].  Since the diagonals of a parallelogram [[bisect]] each other, we have <math>GE = E'G = BE'</math>, or <math>BG = 2GE </math>.  Hence each median passes through an appropriate trisection point of each other median and the medians concur.
  
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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.
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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 triangle to the midpoint of the opposite side.
  
 
== See also ==
 
== See also ==

Revision as of 13:05, 9 July 2009

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 $G$. 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 are $(a,b)$ where $a$ is the arithmetic mean of the $x$-coordinates of the vertices of the triangle and $b$ is the arithmetic mean of the $y$-coordinates of the triangle.


Centroid.PNG


Proof of concurrency of the medians of a triangle

Note: The existence of the centroid is a trivial consequence of Ceva's Theorem. However, there are many interesting and elegant ways to prove its existence, such as those shown below.

Proof 1

Readers unfamiliar with homothety should consult the second proof.

Let $D,E,F$ be the respective midpoints of sides $BC, CA, AB$ of triangle $ABC$. We observe that $DE, EF, FE$ are parallel to (and of half the length of) $AB, BC, CA$, respectively. Hence the triangles $ABC, DEF$ are homothetic with respect to some point $G$ with dilation factor $-\frac{1}{2}$; hence $AD, BE, CF$ all pass through $G$, and $AG = 2 GD; BG = 2 GE; CG = 2 GF$.

Proof 2

Let $ABC$ be a triangle, and let $D,E,F$ be the respective midpoints of the segments $BC, CA, AB$. Let $G$ be the intersection of $BE$ and $CF$. Let $E',F'$ be the respective midpoints of $BG, CG$. We observe that both $EF$ and $E'F'$ are parallel to $CB$ and of half the length of $CB$. Hence $EFF'E'$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, we have $GE = E'G = BE'$, or $BG = 2GE$. Hence each median passes through an appropriate trisection point of each other median and the medians concur.

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 triangle to the midpoint of the opposite side.

See also

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