Can anybody help me on this problem please.
For any points ABCD are arranged along a line such that AC>AB and BD<BC. Draw a picture with the four points in plane. Is there more than one order possible?
Thanks a lot
Can anybody help me on this problem please.
For any points ABCD are arranged along a line such that AC>AB and BD<BC. Draw a picture with the four points in plane. Is there more than one order possible?
Thanks a lot
You need to explain the notation. What does $\displaystyle AB$ stand for?
Is it simply the length of line segment $\displaystyle \overline{AB}$?
Or is it a signed length? If we arrange $\displaystyle A-B-C$, B is between A & C, then using just length we have $\displaystyle AB<AC$.
Do you see what we need to know about the notation?
Hello, rcs!
$\displaystyle \text{Points }A,B,C,D\text{ are arranged along a line}$
. . $\displaystyle \text{such that: } AC>AB\text{ and }BD<BC.$
$\displaystyle \text{Is there more than one order possible?}$
Using emakarov's suggestion, I found 12.
The following six and their mirror-images.
$\displaystyle A - B - D - C $
$\displaystyle A - D - B - - \,C$
$\displaystyle D - A - B - - - - \,C$
$\displaystyle B - A - D - C$
$\displaystyle B - D - A - - - - \,C$
$\displaystyle D - B - A - - \,C $
Did I miss any?
This is precisely why I asked for clarification as to exactly what is involved here. I assumed that this question has to do with ordering collinear points using some form of Hilbert’s betweenness axioms. If that is correct then there is only one possible order: $\displaystyle A-B-D-C$.
There is no notion of a mirror image in the axioms.
Short of something like Ed Moise’s ruler axiom spacing makes no difference.
To be clear on what this means, given $\displaystyle P-Q-R$ we know that $\displaystyle PQ<PR$.
BUT there is no way to compare $\displaystyle PQ~\&~QR$, without knowing the full context of the question.