## Classical Geometry

Posted by Vago on July 9, 2010 – 12:57 pm

You have 2 points A and B. draw a line XY next to it in any direction, such that A and B do not lie on XY, but A and B should be on the same side on XY. take any point P on the line XY. let the sum of sides AP+BP be some constant K.

Now, the question is, find a point E such that the distance between E and P is equal the K. Else, give the maximum value on the line segment till which this is possible.

Note: the value for ANY point P (which can change)

July 9th, 2010 at 3:32 pm

Hi, Vago! I don’t understand the problem at all! First of all, what kind of instruments can we use? If we are given a compass then it is trivial, if not then do we use only a ruler, or we can measure some angles and distances, too (for example can we draw perpendiculars)? Secondly, do we have any restrictions for the location of the point E? Should it be on the line XY, or at least should the different points E for the different points P lie on a continuous curve? And, finally, what line segment do you have in mind? AB or PE? And what is given first the point P or the sum of the distances K?

Sorry for all the questions, but I would like to understand the problem. Thanks, in advance!

Cheers,

slavy

July 9th, 2010 at 4:16 pm

I nearly thought it wasn’t possible too. But now believe that the question is at least plausible (but It took me a couple of cigarettes before I saw that).

I take it or granted that the good ole’ compass and straightedge are allowed, but fancy that anything goes while analysing the problem.

Your job is to find where E is. We need EP = AP+BP and EP’=AP’+BP’. A,B,X,Y and E are fixed.

Having met Vago before, I don’t think it likely that he’s posted a duff question.

July 9th, 2010 at 4:19 pm

… on the other hand, I definitely have reservations about this one

July 9th, 2010 at 4:22 pm

I am not saying the question is duff – just I didn’t get that the point E is independent on the choice of P. This explains all my questions Anyway, it is after midnight here, so I will leave the problem for tomorrow…

July 9th, 2010 at 4:22 pm

Hmmm, I’ve just re-read slavy’s post, and may have jumped to a wrong conclusion. The “If we are given a compass then it is trivial” has surprised me.

The fact! that the question might be answerable has surprised me.

July 9th, 2010 at 4:24 pm

Hi slavy. Posts crossing. Sounds like you’re in my time zone (England).

I absolutely assume the E is independent of P.

July 9th, 2010 at 4:43 pm

I can see a failing case. If the line XY is close to and parallel to AB, then if P is a long way away, as P moves, AP+BP must be increasing at twice the rate that P moves. Then I can’t see how to make EP increase at more than half the required rate. But maybe that’s what the last part of the problem is about. My instinct says there is a fault somewhere (after all).

If the K is a true constant, then P would have to move on an ellipse; but then the purpose of the line XY elludes me.

July 9th, 2010 at 5:58 pm

sorry if it was kind of difficult to understand, my mistake.

I shall write it again:

You have 2 points A and B. next to it there is a line XY, such that A,B does not lie on XY. Now, let us take ANY point P on XY, and join AP and BP. find E such that EP = AP + BP

Main thing is, that P changes continuously, but E has to be 1 fixed point for any point P on XY.

So, if i take P some 10 meters away from B and 12 meters away from A, the same E point should tell you the sum of the lengths.

You are allowed only a Compass, Ungraduated scale, and a pencil. Assume these are exactly precise instruments.

Note: weirdly, part of my question has not been posted. This was a raw form of my questions, that i still had to edit.

July 9th, 2010 at 6:02 pm

Mainly a respond to slavy’s first question, sorry for the incoherence, but here E is what you have to find such that ALL points P on XY satisfy EP = AP + BP

A,B,XY is given, find E. E has no particular restrictions.

If a LINE doesnt satisfy this, then find the maximum length that can satisfy that condition. (By geometry)

Hope you understand that. And sorry for answering so late, was kind of busy.

Again, sorry for the inconvenience, not yet used to the new site

July 9th, 2010 at 6:04 pm

To Chris, after i read his last comment. AP+PB is a constant for a particular K, not for all values of P.

July 9th, 2010 at 6:37 pm

Hi Vago. I had understood the question correctly. My reference to constant K was because I was (and still am) stunned that it really is possible to determine such a point as E. The reference to this (not-really a) constant K was unnecessary. All you should have said is that [we need] EP = AP + BP independently of P [as far as that is possible].

July 10th, 2010 at 2:50 am

Good morning everybody! Chris, we are almost in the same time zone (I am in Germany) and I thought the problem is trivial with a compass before realizing that E should be independent of P. Vago, I am still puzzled (which is not that bad when you deal with puzzles) and don’t see anything. I know that when you don’t have ideas on the general picture, you start with some case studying. That’s what I assumed AB to be parallel to XY. Then, let the simetral (I think in English it is called segment bisector, but it is too long for me to write it) of AB intersects the line XY in point C. Let P and P’ be two points on XY at equal distance from C. Then AP+PB=AP’+P’B and thus PK and P’K should be equal, too. This is possible, only if K belongs to the simetral of PP’, which in this particular case coincides with the simetral of AB. Hence, there is only one choice for K (actually two up to reflection with XY or infinitely many if we decide to go to 3D and leave the plane. However, all the computations there are absolutely the same and we don’t gain anything from the extra dimension) – to lie on the simetral of AB and AC+CB=CK. This setting give us the benefit of a very nice Cartesian coordinate system, centered at the midpoint of AB and we can practically compute everything we need. So, let P be an arbitrary point on XY. We compute AP+PB and PK and after some more calculations we see that the two values are equal only for finitely many choices of P (more precisely at most 4, since we end up with finding the zeros of a polynomial of degree 4). Hence, even infinitesimal changes of the position of P will make the equality AP+PB=PK wrong… Maybe, the case of parallel lines is the worst one in this problem, but the results I got here do not motivate me to work on the general problem

July 10th, 2010 at 4:51 am

Hi slavy. I eventually realised that the trivial/compass thing was another misunderstanding on my part. But I had already made so many posts, that I decided not to mention that Each post seemed to be adding to the confusion.

I may be about to be guilty of misinterpreting your last post (Again), but you seem to have replaced E with K.

Although I hadn’t gone 3D (which sometimes can simplify a 2D problem), I too am doubtful that P can have more than discrete positions. But I haven’t tried very hard, because, like for you, every case I consider seems to fail.

However, I’m going to have a bash at it.

July 10th, 2010 at 7:13 am

I keep on finding disproofs, but they’ve all been flawed. That failure to find a disproof, makes me think it might yet be provable. Now my brain hurts. I’ve also got to see some customers, so can’t get pen to paper for hours, boo, hoo.

July 10th, 2010 at 12:16 pm

Vago, please don’t post a solution yet. I’m trying out an idea, if that fails I’ll concede defeat.

July 11th, 2010 at 12:54 am

Chris, u know me, i dont post a solution, unless, i accept somes answer, or after 10 days

July 11th, 2010 at 2:09 am

Vago, as explained above in my last post, the problem has no (nice) solution in the case of AB parallel to XY, i.e., I have proven that for any point E that is fixed apriori at most 4 different points P from the line XY (and they will not be close to each other!) satisfy the condition AP+PB=PE (yes, Chris, I replaced by mistake E with K in my post). Hence, in this case the problem is meaningless even in higher dimensions (sometimes the clever trick is to leave the plane you had your data in, although I don’t see how we can use this trick here, where only a ruler and a compass is given). Moreover, the same computations (i.e., working with Cartesian grid and computing coordinates) lead to the same conclusion for the general case (where the direction of XY could be arbitrary). So, again, I don’t see a single example, where for a continuum of points P, the point E is the same…

July 11th, 2010 at 2:52 am

Hi Vago. Thanks. I can calculate how K changes as P varies. I can then, in principle, use that to calculate the angle at the point E must be from P. But even before doing the calculus the equations look horribly tedious (involving sums of square roots of sums of squares). And when I look at them, I can’t see how to fix E.

Although I’m use full co-ordinate geometry and calculus, that’s only while I’m working it out. I assume that (if a solution exists) that I could then work out a classical geometric construction.

But I haven’t yet seen any way to satisfy the problem. I haven’t quite given up, though. I am also aware that slavy hasn’t seen a way to do it, and he should have a much better chance of doing it than I have.

July 11th, 2010 at 3:16 am

Hi again. If there is a solution, then P must be on the segment where AP amd BP are changing length in the opposite sense to each other. If P is a very long way from A and B (and hence E also), then AP + BP would be increasing almost twice as rapidly as P’s rate of change of position, so EP would only be changing at about half the required rate. i.e the solution segment must be near A and B (on the line XY).

I also note that if the line through A and B is perpendicular to the line though X and Y, the equations look likely to be much easier to deal with.

Enough of my waffle.

July 11th, 2010 at 3:30 am

… I realise that AB cannot be perpendicular to XY by the argument I made in my last post.

OK I formally give up.

July 11th, 2010 at 6:54 am

Brain seizure setting in. Now I don’t trust my last comment, para 1; because P may be nearby. So I retract my resignation. But I might tender it again.

July 11th, 2010 at 6:35 pm

I give in.