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consider 4 points in 2D in clockwise order as A,B,C,D forming a quadrilateral.

If the quadrilateral undergoes affine transformation then affine invariant will be preserved which is given by area of triangle ACD/ABC as per http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.109.5013&rep=rep1&type=pdf (see page 3 affine invariant)

My question is - Why can't it be ratio of areas across another diagonal BD i.e area of triangle ADB/BCD ?
Will both ratios be same ?
Thanks.

user1371666
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  • What happens if you relabel the vertices of the quad? – amd May 11 '18 at 09:02
  • Actually document says that this ratio is invariant to rotation ? – user1371666 May 11 '18 at 10:11
  • Actually document says that this ratio is invariant to rotation . i think acd/abc gives ratio of be/fd assuming perpendicular dropped from b to ac meets at e , similiarly perpendicular from d to ac meets at f . This is equivalent to ratio across line segment bd . see https://ibb.co/ivW0Ad . – user1371666 May 11 '18 at 10:22
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    My point is that the invariant doesn’t assume any particular labeling of the vertices, so either diagonal must work. – amd May 11 '18 at 15:57
  • sir, my doubt is that - let's say diag1 is common to both triangles then it seems the ratio of triangle areas is simply the ratio of segments of diag2 . So, is it true and provable that diagonals of a quadrilateral divide each other in same ratio ? . thanks for listening. – user1371666 May 12 '18 at 01:52

2 Answers2

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Based on your last comment I think I see what might be causing you some confusion. The claim is that the ratio of areas of the two triangles to either side of a diagonal of a quadrilateral is invariant under (nonsingular) affine transformation. This is a simple consequence of the fact that affine transformations multiply all areas by a constant factor: the ratio of the areas of any two triangles is invariant. Given this, the choice of diagonal is clearly irrelevant. However, this does not mean that the area ratio obtained by using the other diagonal is equal to the first, only that it, too, is invariant under affine transformations. Indeed, the eight possible cyclic labelings of the vertices generate up to four different area ratios.

amd
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The moduli space of (labeled) convex plane quadrangles up to affine equivalence is two-dimensional. This can be seen as follows: Given such a quadrangle $Q\subset{\mathbb R}^2$ with vertices $A$, $B$, $C$, $D$ in counterclockwise order there is a unique affine map $T$ with $T(A)=(0,1)$, $T(B)=(0,0)$, and $T(C)=(1,0)$. It follows that $$T(D)=(u,v),\qquad{\rm whereby}\quad u>0,\quad v>0,\quad u+v>1\ ,$$ is uniquely determined; hence the pair $(u,v)$ is an affine invariant of $Q$.

The two diagonals of $Q$ (resp., $T(Q)$) partition $Q$ into a pair of triangles in two different ways, generating for each pair an area ratio $\lambda$, resp., $\mu$. The numbers $\lambda$ and $\mu$ are affine invariants as well. Doing the computations for $T(Q)$ it is easily seen that $u$ and $v$ are uniquely determined by the values of $\lambda$ and $\mu$. It follows that the pair $(\lambda,\mu)$ completely characterizes $Q$ up to affinity. For a parallelogram we have $\lambda=\mu=1$.

Christian Blatter
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  • sir, that would mean that there are two invariants lambda and mu which may not be equal. I have written a program that splits quadrilateral into 2 triangle by taking diagonal from top left to bottom right .Program does not know the label of vertices.So , if we rotate the quadrilateral by 90 degree than program will report another invariant. Isn't it ? – user1371666 May 12 '18 at 01:55
  • In my answer I decidedly wanted to keep out of the labeling issue. Note that you have to draw two diagonals and obtain two parameter values $\lambda$, $\mu$. – Christian Blatter May 12 '18 at 07:50
  • actually i ran that program with quadrilateral then another time with quadrilateral rotated by 90 degree . I got different invariant . So, I thought that the invariant changed with rotation. That means there is a pair of invariant not necessarily equal .My program simply decides the order of points by sorting the slope of line from origin to the point . – user1371666 May 12 '18 at 11:13