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

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- Dec 17th 2010, 04:19 AMrcsany 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 - Dec 17th 2010, 04:32 AMemakarov
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.

- Dec 17th 2010, 04:33 AMPlato
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? - Dec 17th 2010, 05:04 AMrcs
i think it is about the segment sir

- Dec 17th 2010, 05:26 AMPlato
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? - Dec 17th 2010, 05:33 AMrcs
DC<DA ... is that correct sir

- Dec 17th 2010, 05:43 AMPlato
- Dec 17th 2010, 05:58 AMrcs
DC is shorter than DA ... am i correct sir ?

- Dec 17th 2010, 05:59 AMSoroban
Hello, rcs!

Quote:

$\displaystyle \text{Points }A,B,C,D\text{ are arranged along a line}$

. . $\displaystyle \text{such that: } AC>AB\text{ and }BD<BC.$

$\displaystyle \text{Is there more than one order possible?}$

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?

- Dec 17th 2010, 06:04 AMrcs
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 - Dec 17th 2010, 08:00 AMPlato
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. - Dec 17th 2010, 04:37 PMrcs
sir plato, what do mirror images meant for?

- Dec 18th 2010, 03:22 AMPlato