the first answer works other solutions: I midpoint of AD;J midpoint of BC G midpoint of IJ or K midpoint of AC,L midpoint of BD G midpoint of KL and the 3 ways give the same point G! anonymous1

theorem the centroid of the vertices of a quadrilateral is the midpoint of the lines joining the midpoint of opposite sides,it is also the midpoint of the line joining the midpoint of the diagonals anonymous 1

hmmm ok! there are three possible interpretations of problem statement: I1) centroid of 4 equal 0-dimensional masses at quadrilateral vertices I2) centroid of 4 homogeneously massive 1-dimensional lines forming the given quadrilateral I3) centroid of a homogeneously massive 2-dimensional quadrilateral.

@Anonymous 1: your solution solves I1 only

I highly suspect that initial 8foxes intention was I3. Anyway, all three interpretations have a construction solution...

I2 the centroid of a massive line is the midpoint with mass proportional to the length l let be A',B',C',D' the midpoints of the sides a,b,c,d we must find K center of the point masses (A',a),(B',b) A'K=a/(a+b)A'B' this can be done with parallel lines L center of (C',c)and (D',d) G2 is the centroid of (K,a+b),(L,c+d)

I3 the center of gravity of an homogenious triangle is the centroid with a mass proportional to the aera each diagonal divides ABCD into 2 triangles with centroids C1,C2(first diagonal),C3,C4(second diagonal) G3 lies on the lines C1C2 and C3C4 G3 is the intersection of this lines (thank you for the hints) anonymous1

The centroid of the vertices of a quadrilateral occurs at the point of intersection of the bimedians (i.e., the lines M(AB)-M(CD) and M(AD)-M(BC) joining pairs of opposite midpoints) (Honsberger 1995, pp. 36-37). In addition, it is the midpoint of the line M(AC)-M(BD) connecting the midpoints of the diagonals AC and BD (Honsberger 1995, pp. 39-40).

this simple construction of G3 M midpoint of diagonal BD O intersection of the diagonals on diagonal AC,AN=CO then MG=1/3 MN Essai sur la determination des centres de gravité Hyacinthe Celestin Gaubert (1839)p27 anonymus1

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ABCD the quadrilaterl

ReplyDeleteE midpoint of AB;F midpoint of CD

G midpoint of EF

No, this won't work.

ReplyDeletethe first answer works

ReplyDeleteother solutions:

I midpoint of AD;J midpoint of BC

G midpoint of IJ

or

K midpoint of AC,L midpoint of BD

G midpoint of KL

and the 3 ways give the same point G!

anonymous1

theorem

ReplyDeletethe centroid of the vertices of a quadrilateral

is the midpoint of the lines joining the midpoint of opposite sides,it is also the midpoint of the line joining the midpoint of the diagonals

anonymous 1

hmmm ok! there are three possible interpretations of problem statement:

ReplyDeleteI1) centroid of 4 equal 0-dimensional masses at quadrilateral vertices

I2) centroid of 4 homogeneously massive 1-dimensional lines forming the given quadrilateral

I3) centroid of a homogeneously massive 2-dimensional quadrilateral.

@Anonymous 1: your solution solves I1 only

I highly suspect that initial 8foxes intention was I3. Anyway, all three interpretations have a construction solution...

"I highly suspect that initial 8foxes intention was I3. Anyway, all three interpretations have a construction solution..."

ReplyDeleteCorrect. Slight coloring of the quadrilateral implies that.

I2

ReplyDeletethe centroid of a massive line is the midpoint

with mass proportional to the length l

let be A',B',C',D' the midpoints of the sides a,b,c,d

we must find K center of the point masses (A',a),(B',b)

A'K=a/(a+b)A'B'

this can be done with parallel lines

L center of (C',c)and (D',d)

G2 is the centroid of (K,a+b),(L,c+d)

I3

the center of gravity of an homogenious triangle is the centroid with a mass proportional to the aera

each diagonal divides ABCD into 2 triangles

with centroids C1,C2(first diagonal),C3,C4(second

diagonal)

G3 lies on the lines C1C2 and C3C4

G3 is the intersection of this lines

(thank you for the hints)

anonymous1

The centroid of the vertices of a quadrilateral occurs at the point of intersection of the bimedians (i.e., the lines M(AB)-M(CD) and M(AD)-M(BC) joining pairs of opposite midpoints) (Honsberger 1995, pp. 36-37). In addition, it is the midpoint of the line M(AC)-M(BD) connecting the midpoints of the diagonals AC and BD (Honsberger 1995, pp. 39-40).

ReplyDeletehttp://mathworld.wolfram.com/Quadrilateral.html

There is a nice animation in Joe Wilson's page.

ReplyDeleteSee here:

http://jwilson.coe.uga.edu/emt668/EMT668.Folders.F97/Patterson/EMT%20669/centroid%20of%20quad/Centroid.html

The name of the guy is Jim Wilson. Sorry about that...

ReplyDeletethis simple construction of G3

ReplyDeleteM midpoint of diagonal BD

O intersection of the diagonals

on diagonal AC,AN=CO

then MG=1/3 MN

Essai sur la determination des centres de gravité

Hyacinthe Celestin Gaubert (1839)p27

anonymus1

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ReplyDelete