Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

ABCD is a quadrilateral, given $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}$, then what kind of quadrilateral is ABCD? I guess it's a rectangle, but how to prove it?

If the situation becomes $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}=\overrightarrow{DA}\cdot\overrightarrow{AB}$, i can easily prove ABCD is a rectangle.

So, the question is given $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}$, can we get $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}=\overrightarrow{DA}\cdot\overrightarrow{AB}$?

thanks.

share|improve this question
1  
Why do you think these have a special name? And (for your definition of quadrilateral) it is unusual to allow the edges to cross each other, I think. –  GEdgar Sep 15 '11 at 0:58
3  
If you place B, C, and D arbitrarily, then each of the two equations between dot products defines a line that A must lie on. Putting A at the intersection between these two lines gives you a quadrilateral that satisfies the condition. So you cannot conclude anything about the angle at C (i.e., it doesn't have to be a rectangle) -- nor anything about the relative lengths of BC versus CD. –  Henning Makholm Sep 15 '11 at 1:10
3  
A concrete example would be A(2,5), B(-1,1), C(0,0), D(1,0). Doesn't look like anything that has a nice name. Neither does A(1,1), B(1,2), C(0,0), D(0,2). –  Henning Makholm Sep 15 '11 at 1:29
    
@Henning, that's great, thanks. –  Charles Bao Sep 15 '11 at 1:36
    
@Henning: Maybe you want to flesh that out into a full answer? –  mixedmath Aug 21 '12 at 22:06
add comment

1 Answer

Some informal degrees-of-freedom analysis:

  • An arbitrary quadrilateral on a plane is described by $8$ parameters: coordinates of each vertex (to simplify matter, I don't take quotient by isometries).
  • A rectangle on a plane is described by $5$ parameters: endpoints of one side and (signed) length of the other side.

We should not expect two equations to restrict the $8$-dimensional space of quadrilaterals down to $5$-dimensional space of rectangles. Three equations (also given in the post) are enough.

The above is not a rigorous proof because two equations $f=0=g$ can be made into one $f^2+g^2=0$, etc. One needs some transversality consideration to make it work. But it's easier to just quote a geometric argument given by Henning Makholm in the comments.

If you place $B$, $C$, and $D$ arbitrarily, then each of the two equations between dot products defines a line that $A$ must lie on. Putting A at the intersection between these two lines gives you a quadrilateral that satisfies the condition. So you cannot conclude anything about the angle at $C$ (i.e., it doesn't have to be a rectangle) -- nor anything about the relative lengths of $BC$ versus $CD$.

A concrete example would be $A(2,5)$, $B(-1,1)$, $C(0,0)$, $D(1,0)$. Doesn't look like anything that has a nice name. Neither does $A(1,1)$, $B(1,2)$, $C(0,0)$, $D(0,2)$.

The first of Henning's examples is below (the second isn't even convex)

example

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.