0

I know how to construct the centroid of a quadrilateral as mentioned here.

But my question is different from that.

We know that if points B,C,G are given in geometry plane and for locating point A such that $G(∆ABC)=G$ we have to do these following things:

(a)Take m(B,C) as shown in figure:Centroid of ∆

(b) then we reflect m(B,C) in G to get m'(B,C) .

(c) we reflect G in m'(B,C) to get G' and from there we observed that $G(∆G'BC)=G$ so we mentioned $G'=A$.

But if we do such things in case of Quadrilateral then it becomes a challenging task to construct .

So my Question is "If points A,B,C,G are Given then how to locate D such that $G(ABCD)=G$ ?"

My Attempt:

I take $G_b$=$G(∆ABC)$ and placed point $G_d$ on line $GG_b$ then I construct Point D such that $G_d$=$G(∆ADC)$ as shown in figure:

Quadrilateral centroid

Then Construct point $G_a$ and $G_c$ such that $G_a=G(∆ABD)$ and $G_c=G(∆BCD)$ as shown in figure: Centroid of quadrilateral But I observed that Line $G_aG_c$ intersect Line $G_bG_d$ at different points other than G.

So the construction becomes wrong so please help to locate point D such that $G=G(ABCD)$.

Jean Marie
  • 81,803
  • 1
    I have taken the liberty to change your former title "If points A,B,C,G are Given then how to locate D such that G(ABCD)=G? "into : "How to construct the centroid of a tetrahedron" – Jean Marie Apr 02 '22 at 08:22
  • 1
    The construction is very simple; Instead of dividing any median, say $AA'$, (where $A'$ is the centroid of triangle $BCD$) into 3 thirds, divide it into 4 fourths $[AA_1]\cup[A_1A_2]\cup[A_2A_3]\cup[A_3A']$ ; point $A_3$ is your point $G$. – Jean Marie Apr 02 '22 at 08:30
  • 1
    Another way to construct the centroid (similar to the way you have seen for the centroid of a quadrilateral) is to take the midpoint of the midpoints of opposite edges like $AB$ and $CD$ in your last figure. In all these cases, this is a consequence of the "associativity property" of barycenters (taking partial barycenters). Have you heard about this concept ? – Jean Marie Apr 02 '22 at 08:53
  • 1
    @JeanMarie: The question isn't about constructing the centroid; it's about constructing a fourth vertex, given the centroid (and three vertices). Moreover, the question doesn't seem to be about tetrahedra, except insofar as a planar quadrilateral is a degenerate tetrahedron. – Blue Apr 02 '22 at 10:30
  • 1
    @Blue You have read the question in a better way than me !!! I was fooled by the last figure in which I saw in 3D... I change the title I had changed previously... – Jean Marie Apr 02 '22 at 10:36
  • @Jean Marie , I don't know about these concepts of associativity property" of barycenters (taking partial barycenters) as I am in Grade 11 and I want to ask that in your above comment.
    "Instead of dividing any median, say AA′, (where A′ is the centroid of triangle BCD) into 3 thirds, divide it into 4 fourths etc.." is answer to find fourth vertex or centroid?     –  Apr 02 '22 at 12:22
  • 1
    What I said for tetrahedra is still true for quadrilaterals: if you join $A$ to the centroid $G$ of $BCD$, divide $AG$ into four segments, the global centroid $G'$ of the quadrilateral will be situated at the "hinge" between the last segment $G'G$ and the 3 others (explanation with barycentric terminology: $A$ has weight 1 vs. $G'$ has weight 3, because it gathers the three unit weights of vertices $B,C,D$ placeing the final barycenter at the 3/4 of line $AG$) – Jean Marie Apr 02 '22 at 13:18

0 Answers0