Thread: any more possible arrangement of AC>AB and BD<BC

1. any more possible arrangement of AC>AB and BD<BC

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

2. Have you tried drawing a line with A, B, C such that AB < AC and then trying to insert D so that BD < BC? I know it's just a restatement of the problem, but I don't see any difficulty in this problem other than trying a few cases.

3. Originally Posted by rcs
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?
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?

4. i think it is about the segment sir

5. If it is simply length of a line segment then $\displaystyle A-B-D-C$ will give $\displaystyle AB<AC~\&~BD<BC$.
Can you find another?

6. DC<DA ... is that correct sir

7. Originally Posted by rcs
DC<DA ... is that correct sir
Why do you think that works?

8. DC is shorter than DA ... am i correct sir ?

9. 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?

10. oh my God! u mean their mirror images are the opposite arrangement sir soroban?
like A-B-D-C its mirror is C-D-B-A ? am i right sir?

thanks alot

11. Originally Posted by Soroban
[size=3]Using emakarov's suggestion, I found 12.
The following six and their mirror-images.
$\displaystyle A - B - D - C$
$\displaystyle A - D - B - - \,C$
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.

12. sir plato, what do mirror images meant for?

13. Originally Posted by rcs
what do mirror images meant for?
@rcs
Unless you post the exact statement of this problem along with all definitions and relevant axioms, it is pointless to continue this thread.