Difference between revisions of "Centroid"
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| − | The '''centroid''' of a [[triangle]] is the point of intersection of the [[median of a triangle |medians]] of the triangle and is | + | The '''centroid''' of a [[triangle]] is the [[point]] of [[intersection]] of the [[median of a triangle | medians]] of the triangle and is conventionally denoted <math>G</math>. 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 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. | ||
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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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| + | === Proof 1 === | ||
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| + | ''Readers unfamiliar with [[homothety]] should consult the second proof.'' | ||
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| + | Let $\displaystyle D,E,F$ be the respective midpoints of sides <math>\displaystyle BC, CA, AB</math> of triangle $\displaystyle ABC$. We observe that $\displaystyle DE, EF, FE$ are parallel (and of half the length of) $\displaystyle AB, BC, CA$, respectively. Hence the triangles $ABC$, $DEF$ are homothetic with respect to some point $\displaystyle G$ with dilation factor $\displaystyle -\frac{1}{2}$; hence $AD, BE, CF$ all pass through $\displaystyle G$, and $\displaystyleAG = 2 GD; BG = 2 GE; CG = 2 GF$. Q.E.D. | ||
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| + | === Proof 2 === | ||
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| + | Let $\displaystyle ABC$ be a triangle, and let $\displaystyle D,E,F$ be the respective [[midpoint]]s of the segments $\displaystyle BC, CA, AB$. Let $\displaystyle G$ be the intersection of $\displaystyle BE$ and $\displaystyle $CF$. Let $\displaystyle E',F'$ be the respective midpoints of $\displaystyle BG, CG $. We observe that both $\displaystyle EF $ and $\displaystyle E'F'$ are [[parallel]] to $\displaystyle CB $ and of half the length of $\displaystyle CB $. Hence $\displaystyle EFF'E' $ is a [[parallelogram]]. Since the diagonals of parallelograms [[bisect]] each other, we have $\displaystyle GE = E'G = BE'$, 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. | ||
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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. | ||
== See also == | == See also == | ||
* [[Cevian]] | * [[Cevian]] | ||
* [[Geometry]] | * [[Geometry]] | ||
Revision as of 20:23, 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 $\displaystyle D,E,F$ be the respective midpoints of sides 
of triangle $\displaystyle ABC$. We observe that $\displaystyle DE, EF, FE$ are parallel (and of half the length of) $\displaystyle AB, BC, CA$, respectively. Hence the triangles $ABC$, $DEF$ are homothetic with respect to some point $\displaystyle G$ with dilation factor $\displaystyle -\frac{1}{2}$; hence $AD, BE, CF$ all pass through $\displaystyle G$, and $\displaystyleAG = 2 GD; BG = 2 GE; CG = 2 GF$. Q.E.D.
Proof 2
Let $\displaystyle ABC$ be a triangle, and let $\displaystyle D,E,F$ be the respective midpoints of the segments $\displaystyle BC, CA, AB$. Let $\displaystyle G$ be the intersection of $\displaystyle BE$ and $\displaystyle $CF$. Let $\displaystyle E',F'$ be the respective midpoints of $\displaystyle BG, CG $. We observe that both $\displaystyle EF $ and $\displaystyle E'F'$ are parallel to $\displaystyle CB $ and of half the length of $\displaystyle CB $. Hence $\displaystyle EFF'E' $ is a parallelogram. Since the diagonals of parallelograms bisect each other, we have $\displaystyle GE = E'G = BE'$, 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.
