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

• December 17th 2010, 04:19 AM
rcs
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
• December 17th 2010, 04:32 AM
emakarov
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.
• December 17th 2010, 04:33 AM
Plato
Quote:

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 $AB$ stand for?
Is it simply the length of line segment $\overline{AB}$?
Or is it a signed length? If we arrange $A-B-C$, B is between A & C, then using just length we have $AB.

Do you see what we need to know about the notation?
• December 17th 2010, 05:04 AM
rcs
i think it is about the segment sir
• December 17th 2010, 05:26 AM
Plato
If it is simply length of a line segment then $A-B-D-C$ will give $AB.
Can you find another?
• December 17th 2010, 05:33 AM
rcs
DC<DA ... is that correct sir
• December 17th 2010, 05:43 AM
Plato
Quote:

Originally Posted by rcs
DC<DA ... is that correct sir

Why do you think that works?
• December 17th 2010, 05:58 AM
rcs
DC is shorter than DA ... am i correct sir ?
• December 17th 2010, 05:59 AM
Soroban
Hello, rcs!

Quote:

$\text{Points }A,B,C,D\text{ are arranged along a line}$
. . $\text{such that: } AC>AB\text{ and }BD

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

Using emakarov's suggestion, I found 12.
The following six and their mirror-images.

$A - B - D - C$

$A - D - B - - \,C$

$D - A - B - - - - \,C$

$B - A - D - C$

$B - D - A - - - - \,C$

$D - B - A - - \,C$

Did I miss any?

• December 17th 2010, 06:04 AM
rcs
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
• December 17th 2010, 08:00 AM
Plato
Quote:

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

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: $A-B-D-C$.
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 $P-Q-R$ we know that $PQ.
BUT there is no way to compare $PQ~\&~QR$, without knowing the full context of the question.
• December 17th 2010, 04:37 PM
rcs
sir plato, what do mirror images meant for?
• December 18th 2010, 03:22 AM
Plato
Quote:

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.