The existence of a circle as a locus of points equidistant from a given point is postulated in Euclid: Def 1.15, Post. 1.3.

1. According to Euclid I.17,I.19,III.16,III.17, the tangent to a circle at the end of a diameter is perpendicular to the diameter.
   
2. A slope of a radius-vector from the origin to (x, y) is y/x.
   
3. The product of slopes of two perpendicular lines is -1.
   
4. The slopes of the tangents to the circles centered at the origin are - x/y.
   
5. The differential equation y' = -x/y, x0 = c has a unique solution x² + y² = c².
   
6. A solution to the equation is a curve that has the same tangents as a circle centered at the origin. Therefore, the curve is such a circle. Therefore, the circle is described by the equation x² + y² = c².
   

I shall edit the proof to follow the outline given, and repost here when time permits. I suppose you and Mr. Giventhal deserve some credit for the proof? I do not think I was able to see the lack of uniqueness in the circle as the solution (many other people I approached did not notice this..) nor the fact that my real intention relied on the lemma pointed out by Mr. Giventhal.

So there's something to be said for collaboration, and perseverance!

Again I am indebted to the both of you...

BTW, I have written an article that chronicles the progression of this discussion which is undergoing review, but I suppose I should write another now...

You are welcome.

>So there's something to be said for collaboration, and
>perseverance!

Perseverance is all yours. Good will is ours.

>BTW, I have written an article that chronicles the
>progression of this discussion which is undergoing review,
>but I suppose I should write another now...

Do post a link here when it's over.

I'll need to comment on and upgrade my older page.

>

I know a German website where the m1*m2=-1 rule for perpendicular lines is proven by Euclid’s altitude rule for right triangles (h^2=p*q), and one can prove this altitude rule by the PT.

Do we have a “circulus vitiosus”?

Regards

Chris

Not necessarily. To establish that A is equivalent to B under conditions C, i.e., C⇒(A⇔B) one shows that C⇒(A⇒B) and then also C⇒(B⇒A). This does not mean there is a vicious circle but only that two statements are equivalent. This is a regular occurence in mathematics. Very often both A and B have independent proofs. It only when, unwittingly, one of the statements is used down the line in the chain of proofs to establish the other which later serves to proof the first statement that we get circular reasoning.

many thanks for your detailed answer.

Pythagoras everywhere you look…

John writ's in his proof: “Any (x,y) which is a terminal point of vectors
with length c and whose initial point is the origin will lie on this solution curve”.

However, his differential equation dy/dx =-x/y is not defined for y=0.

Chris

that's a very good point. I think it is possible to overcome it, but no doubt, unless John does that, the proof is begging.

What am I begging for? A curve with a vertical tangent at (c,0) is defined for y=0 even if the differential equation is not. Otherwise we would have a circle with a discontinuity. Perhaps I should add that the solution curve is continuous at (c,0) although not differentiable there... I don't think that would be too hard to include. Nevertheless the proof should stand without this if I used (0,c), shouldn't it?

Just pick a book on ODE and check for th existence of a solution. This is one thing where a solution is defined and another which domain it can be extended to.

Actually the latest version of the proof (see reply #2 to the post) uses the initial condition x=0 and y=c:

"Now using an initial condition (x,y) = (0,c) we have: ..."

The differential equation dy/dx = -x/y as well as the solution are both defined at this point. I suppose the question is whether or not it'should that in the case where one uses the condition (c,0) the curve has a vertical tangent there... but I didn't use that point to extract the particular solution curve, so I do not think it matters...

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0BygZeXnaKTslYjg5YjU5Y2MtYTEyMi00OGRmLTg5Y2EtYTE4YmVkYzdiZmQx&hl=en

When time permits, I should write a chronology of events that led to this and highlight the discussion between you, Wayne Bishop, and Alexander Givental (sorry I misspelled his name earlier!) This was quite an amazing chain of events, wouldn't you agree?

Again many thanks to all of you who helped me thread the needle...

Therefore, I play the “advocatus diaboli” now.

The distance from a point (x/y) to the origin (0,0) is always sqrt(x^2+y^2)
due to the PT. It doesn’t matter which curve the point is moving along.

You have proven by calculus that the distance function is valid for the
special case where the curve is a circle.

Could you do the same if y’=-y/x (a hyperbola)?

If you say “no”, how can we be sure that the distance function is still valid?

What would you answer?

Chris

true, this is stated in Alex B's original critique... but we shall continue nonetheless

"You have proven by calculus that the distance function is valid for the
special case where the curve is a circle."

Yes, and thanks to Euclid - that circle has the same tangents as those drawn perpendicular to the terminal point (x,y) on the aforementioned radius-vector. In other words, the curved traced out by all (x,y) terminal points is the same as the circle, since they have the same tangents. I simply use calculus to find the relation for the curve, and in doing so I happen to have a synonym for distance. I might even say that the curve *must* be a circle, and I'll elaborate on this below.

"Could you do the same if y’=-y/x (a hyperbola)?...."

My argument is strictly (and carefully) stated to limit radius vectors to the first quadrant, y > 0 whose initial point is the origin. But nonetheless, (and I am not sure if I'll get this one exactly right), here is my defense...

Viewing a slope field for y'=x/y (orthogonal trajectories to your differential equation) shows a hyperbolic solution because it is assumed the the radius vectors have slope -y/x for *all* values of (x,y) in *all* quadrants. This is contradictory to my diagram, but also to all quadrants.

It may be a slight oversight that you are assuming that we start with slopes that are *the opposite* of y divided by x. In each quadrant then you would wind up with different radius vector (all having opposite slopes). (Try viewing a slope field of y'=x/y.) So if one enters the 2nd quadrant slopes there are the opposite of some positive y divided by some positive x. Here you have a positive slope which again cannot support a vector drawn in that quadrant whose initial point is the origin. Similar arguments can be made for the other quadrants, but I am careful to stay in the first quadrant to satisfy the geometry of the right triangle x>0, y>0.

Regards,

Instead of "So if one enters the 2nd quadrant slopes there are the opposite of some positive y divided by some positive x."

I should have written "So if one enters the 2nd quadrant slopes there are the opposite of some positive y divided by some *negative* x."

Hopefully, in spite of this you get my "point."

You start with a line where m is positive (m=y/x). Therefore, the slope of the perpendicular line is –x/y and therefore, we have to solve the differential equation y’=-x/y.

I start with a line where m is negative (m=-y/x). Therefore, the slope of the perpendicular line is x/y and therefore, we have to solve the differential equation y’=+x/y.

Which is the right differential equation?

All the best
Chris

>Which is the right differential equation?

I think mine is. Again I ask you to look at slope fields I posted earlier. Your differential equation should be interpreted as 'all slopes where the quotient is the OPPOSITE of y divided by x.' In the second quadrant where your vector is headed the slope is still y/x - it's just that one of your coordinates is negative making the line you mentioned.

I am careful to stay in the first quadrant to avoid this confusion, but the argument should still hold in all quadrants.

>All the best
>Chris

>You start with a line where m is positive (m=y/x).
>Therefore, the slope of the perpendicular line is –x/y and
>therefore, we have to solve the differential equation
>y’=-x/y.
>

Indeed, this is true since the family of curves that are solutions to dy/dx = -x/y are orthogonal to the family of curves dy/dx = y/x. Solving the latter we have dy/y = dx/x, then ln|y| = ln|x| + C, then y = C*x which is the *line* my vector is collinear with. However, C can be positive or negative (or zero but I avoid that due to what you pointed out earlier). I just draw one where C is positive to stay in the first quadrant. Reiterating the ODE, any value of C will yield a slope y/x. (Again intentionally avoiding C = 0, but here one might use the ruler postulate instead of the PT anyway...)

>I start with a line where m is negative (m=-y/x). Therefore,
>the slope of the perpendicular line is x/y and therefore, we
>have to solve the differential equation y’=+x/y.

If you mean that you start with a line y = -C*x, then you are still in the family of curves my differential equation is orthogonal to. If you mean the differential equation defined by the family of curves where dy/dx = -y/x then you are talking about something different than lines! Here we have -ln|y|=ln|x| + C which gives the general solution y = C/x... This is precisely the reason your differential equation dy/dx = +x/y comes across across as a hyperbola.

Also from an earlier post:
>Could you do the same if y’=-y/x (a hyperbola)?

Yes, I could. I would pick an (x,y) on your hyperbola and then draw a radius vector to it from the origin. This would just make another circle, and the circle is used to find the distance. The hyperbola just happens to be a family of (x,y)'s I need to find the distance to from the origin. And thanks to Euclid and calculus, I have the relation that defines that distance.

>Which is the right differential equation?

Hopefully I have made this a little more clear, if not too elaborate.

>All the best
>Chris

Same to you, and might I add thanks for the conversation and Merry Christmas.

No, John, I’m talking about the m1*m2=-1 rule.

We have three cases for our lines:

m1=0. Our rule is not defined, no ODE.
m1 >0, m1=y/x, y’=-dx/dy , the “right” ODE.
m1<0, m1=-y/x, y’=dx/dy, the “wrong” ODE.

In other words: Our rule does not “know” where the “minus” belongs to, therefore, we get two ODEs - the principle of perpendicular lines is ambiguous.

Shall we ignore the ”wrong” ODE or do we have to explain this case?
You write: <I am careful to stay in the first quadrant…> Ok. However, your theory is wrong if we leave this quadrant…

All the best and Merry Christmas (Frohe Weihnachten).

Chris
P.S. Your proof is a challenge

>No, John, I’m talking about the m1*m2=-1 rule.

I maintain that there should be a distinction made between lines with different slopes and different ODEs. Read on...

>We have three cases for our lines:
>
>m1=0. Our rule is not defined, no ODE.

Yes we seem to agree on this, and this is not really necessary for the proof as I have stated before.

>m1 >0, m1=y/x, y’=-dx/dy , the “right” ODE.

Agreed.

>m1<0, m1=-y/x, y’=dx/dy, the “wrong” ODE.
>

Disagree. I think whether m1>0 (first quadrant) or m1<0 (second quadrant) we still have the y'=y/x trajectories and therefore the orthogonal family y'=-x/y. I think your differential equation means something different here than lines and here's why:

If you do as I do (case 2 above) and draw a vector from say (0,0) to (1,2) you have m1>0. What is the slope of this vector? y/x. What is the orthogonal slope? -x/y.

If you do as you say in case 3 and draw a vector from say (0,0) to (-1,2) you have m1<0. What is the slope of this vector? Still y/x What is the orthogonal slope? Still -x/y.

So really the ambiguity you speak of lies in whether one wants to head their vector into the first quadrant or the second. They both lead to the same "right" ODE.

The error in your "wrong" ODE is that it assumes that each slope (whether in quadrant 1 OR 2) is the OPPOSITE of y/x. So for instance in your case 3 you are saying that the slope drawn from (0,0) to (-1,2) would be m1>0 since in this case -y/x = -(-2)/1 and this is why I mention that you are not talking about lines in this case. In fact if you draw enough of these -y/x slopes at each (x,y) in both quadrants 1 and 2, you will find the family of curves y=C/x which are not lines. Additionally one finds this solution analytically as I stated in an earlier post.

In the "right" ODE the slopes y/x give us y=Cx. Here C is your m1. C<0 means 2nd quadrant, C>0 means first quadrant. It's all in there.

>Shall we ignore the ”wrong” ODE or do we have to explain
>this case?

Yes we shall since these are not lines and do not describe linear vectors drawn from the origin.

>You write: <I am careful to stay in the first quadrant…> Ok.
>However, your theory is wrong if we leave this quadrant…

Hopefully you now see that the theory is correct in either quadrant. The reason I stay in the first quadrant is twofold: 1) triangles have positive lengths and so I want to speak in terms of positive integers, and more applicable in this discussion, 2) to avoid the confusion of someone thinking that the slope of a line heading into the second quadrant is -y/x. It is not. It is still y/x.

>All the best and Merry Christmas (Frohe Weihnachten).
>
>Chris
>P.S. Your proof is a challenge

Indeed! I have been working on this since July! But I hope you can now see why I am confident that it is correct.

Regards and best wishes,

Kind regards

Chris

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=1prN98zC6aBVONNQ7qGsRYJJPyZCqtpDyfrQG5DirvhPuN7T2IGMnrtBd4AXH&hl=en&authkey=CIbZt5UL

I am quite astonished that it made that publication, and excitedly so!

Congratulations!

Chris

https://www.cut-the-knot.org/pythagoras/CalculusProofCorrectedVersion.shtml
https://www.cut-the-knot.org/pythagoras/proof89.shtml

Way to go!

https://www.cut-the-knot.org/pythagoras/CalculusProofCorrectedVersion.shtml

should instead be

https://www.cut-the-knot.org/pythagoras/CalculusProofCorrectedVersion.shtml

Again I am grateful for the inclusion of my proofs...

Merry Christmas,

Marry Christmas and a Happy New year.

The page filler will be in the April issue of the monthly.

Stuart Anderson

I am not a MAA member, so I am awaiting my copy to come in the mail. I have asked them to let me know the page in the meantime and I'll post here whenever I hear from them or whenever I receive it in the mail, whichever comes first.

I am curious, of course.