We have a triangle AB'C with angle B' = 90. AB'=a, B'C=b, AC=c

On AC from A towards C create segment AB = AB' = a.

Extend AC for length CD = AB = a. This will make AC = BD = c

For any 4 points A,B,C,D (in this order) on a line

we have Euler identity

AB * CD + AD * BC = AC * BD

which in our case is a^2 + AD * BC = c^2

Thus to prove Pythagoras we have to show that AD * BC = b^2.

Now extend AD for DC' = BC. Point C is the midpoint of AC'

(AB + BC = CD + DC' = c by construction).

Draw circle center C and diameter AC' = 2c. Draw perpendicular to

AC' at point D. It will intersect the circle at some point E.

Intersecting chords theorem AD * DC' = DE^2 = AD * BC

Triangle AB'C congruent to CDE (CD = AB' = a , CE = AC = c,

< CDE = 90)

So DE = b and AD * BC = b^2

I hope this time it is a bit more clear.

Salute,

Maj. Pestich

My intent was to get Pythagoras from Euler, but the attempt

to show that b^2 = AD * BC ended up in Proof #11.

There must be another way to show that b^2 = AD * BC in

this configuration, I hope.

Salute,

Maj. Pestich

awkward proof that encroached onto #11 , and also for standing

guard against my quite possible 'circular proofs'.

An easier and more natural proof would have been to construct

a circle centered at C and orthogonal to circumcircle of AB'C

and intersecting it in B" (where triangle AB"C is congruent to

AB'C, AB" = b, CB" = a). Then side AB" = b is tangent

and b^2 = AD * BC (tangent - secant theorem, AD - 2a = BC).

Salute,

Maj. Pestich

I think you hit it right this time. I am and shall be busy for a few days, but eventually make sure that the proof takes its well deserved place in the list.

Thank you,
Alex

It'seems to be quite difficult to come up with a completely

independent and conceptually new approach in a proof of P.T.

After all, it is not much more than simply c^2 = a^2 + b^2.

A lot of the proofs are based on an idea of an equivalent

dissection of components, but how really different are they?

It is like one starts to waltz from a corner or from the center of

the room next time. Applying combinatorics to waltz steps also

helps variety, but it is still the same rhythm in the tune.

I just thought that there was some novelty in using the E.

identity for the purpose, but after all any identity A + B = C

can be used when A is associated with a^2, etc. (Ptolemy for a

rectangle is used already).

Here is one of them: take an equilateral triangle and draw a

rectangle around it'so that one vertex is in common and two other

vertices of triangle are on the other two side of rectangle.

The 3 right triangles are such that area of one of them is

equal to the sum of other two. Now the question is:

how to tie these areas to a Pythagorean triangle? Where is this

P. triangle? I do not have the answer, maybe it is quite obvious,

but I do not see it (yet?). It would be nice to make something out

of this, there are no equilateral triangles in the over 60 proofs

here, if I'm not mistaken.

Salute,

Maj. Pestich

I confess that I also do not see how to pull the Pythagorean rabbit out of the equilateral hat. It'seems that since there are three triangles, not necessarily similar, that there are many squares to deal with here, and I do not see how to sort them out so that the squares from each triangle separately will sum up as they should. The fact that the triangle areas themselves add up ala Pythagoras would seem to be a hint, but then one must associate just one square with each triangle to take advantage of that fact. But which square, on which side? Also, there must be a single "hidden" right triangle to attach these squares to. Alas, I cannot find it.

However, you have inspired me to look again at Euclid, and I have found another proof of the theorem which seems to me very similar to Proof #7 in Alex's list. What do you think of it?

Proof:

Start with the illustration from Proof #7. Place a point E somewhere on the figure which sits on side AB, and likewise place F and G on the figure which sit on BC and CA. Consider the following 3 figures: On side AB, the figure AEBD, i.e. the triangle ABD together with the figure lying outside the triangle on side AB. On sides BC and CA, the similar figures BFCA and CGAD. These three figures satisfy Euclid's definition of "similar and similarly described figures," and therefore we can apply the propositions from Elements Book V to them. (The figures illustrating these propositions show line segments, but Euclid clearly intended these propositions to apply to any type of magnitude at all, whether length, area, or volume.)

Therefore, by similarity ABD:AEBD::BCA:BFCA::CAD:CGAD, and
AEB:AEBD::BFC:BFCA::CGA:CGAD, where the magnitudes are areas.

By Proposition V.16, and Common Notion 1, ABD:AEB::BCA:BFC::CAD:CGA.

But ABD and CAD equal BCA; therefore AEB and CGA equal BFC.

Q.E.D.

This is the "Full Euclid" way of writing the proof. The short version is that "obviously" each external figure is proportional to the internal triangle which is attached to the same side. Since the two smaller internal triangles decompose the larger, so do the two smaller external figures decompose the larger.

Alternatively, if one does not want to invoke the Book V propositions for use with areas, one can use Propositions VI.19 and VI.20 to show that the external figure and the internal triangle are each in the duplicate ratio of the side lengths, and therefore are proportional to each other. The rest of the proof is unchanged. However, I think it is cleaner to directly argue that the areas are proportional, rather than going from area ratios to length ratios and then back to area ratios, so I prefer my first version.

As a side note, both my version and the original Euclid version of Proof #7 have something in common, which dissection proofs do not share: they make it obvious why the proof requires a right angle. Of course it is because only a right triangle decomposes into 2 smaller copies which are similar to the original. The tiling proofs also show why a right angle is necessary: only right angles can fit together to form the plane tilings required for the proofs. Your proof also demonstrates why a right angle is necessary: because only for right triangles is the altitude equal to the geometric mean of the two portions of the hypoteneuse cut by the altitude.

With dissection proofs, the reason for the right angles is sometimes obscured, and it is not clear (without some external knowledge) why the same constructio could not work for some non-right triangles.

Now I will turn my thoughts back to your equilateral triangle ideal.

--Stuart Anderson

It is hard for me to see the fine distinction between your proof

and the # 7. It'should not make any difference which way

(inward or outward) the similar figures are constructed,

and constructing them inward and selecting the 3 corresponding

points approprietly in orthocenter and at the foot of altitude

from right angle makes the equality of areas easy to see.

To find the corresponding points you construct the line

of proportional division between hypotenuse and a side

(perpendiculars from this line to both segments cut the segments

in the same proportion) it happens to be hypotenuse itself, for both

sides. And this happens obviously because of right angle.

I also noticed that even though P. theorem is 'IIF' type,

it is 'almost' always proved in the direction from the right angle

towards the squares. So I have tried to go from a parabola to

prove the angle is 90, and it brings me to a configuration I had in

this thread (a la Euler), but it is again the part of getting

b out of b^2 ( you cut any Y-coordinate segment in 2 parts ,

and for the top part you have somehow to find point on

parabola that has ordinate equal to the length of the top part,

invert it upside down) gives me 'trouble', althogh I think

it again should be somewhat simple to show.

In unrelated matter, recently on Hyacinthos in post

https://groups.yahoo.com/group/Hyacinthos/message/12324

and also in post 12337 there was some material related

to the 4T problem (I think it was related). The stuff he talks

about is interesting, but too heavy for me. What do you and Alex

can make out of it?

Salute,

Maj. Pestich

The basic difference is that Proof #7 shows that the figures on the sides are proportional to the hypoteneuse segments BD and DC, which decompose the hypoteneuse. Notice that areas are made proportional to line-segment lengths. In my version I argue directly that the areas of the side figures are proportional to the areas of triangles ABD and ADC, which decompose the triangle. Notice that areas are made proportional to areas. Because of this, my proof does not depend on Proposition VI.19 at all, but only depends on propositions from book V about proportions. However, I agree that the basic IDEA of setting the figures in proportion to a decomposition is the same in my proof and Euclid's. Euclid decomposes the hypoteneuse, and I decompose the triangle.

> It'should not make any difference which way
> (inward or outward) the similar figures are constructed,
> and constructing them inward and selecting the 3 corresponding
> points approprietly in orthocenter and at the foot of altitude
> from right angle makes the equality of areas easy to see.

I agree, inward or outward cannot make a difference. However, I do not know what 3 points you are speaking of. Do you mean the E, F, G that I used to label the figures on the sides? If so, I do not see how you could construct them to lie at specific locations, since the figures on the sides are arbitrary and may not even pass through those locations. Do you mean to use the triangles ABD, ACD and ABC as the figures? In that case, of course I agree about the locations of the three points, but these are not arbitrary shapes. I was using a figure consisting of one of these triangles (on the inside) and an arbitrary shape (on the outside, but inside would work just as well).

>
> To find the corresponding points you construct the line
> of proportional division between hypotenuse and a side
> (perpendiculars from this line to both segments cut the segments
> in the same proportion) it happens to be hypotenuse itself, for both
> sides. And this happens obviously because of right angle.
>
> I also noticed that even though P. theorem is 'IIF' type,
> it is 'almost' always proved in the direction from the right angle
> towards the squares. So I have tried to go from a parabola to
> prove the angle is 90, and it brings me to a configuration I had in
> this thread (a la Euler), but it is again the part of getting
> b out of b^2 ( you cut any Y-coordinate segment in 2 parts ,
> and for the top part you have somehow to find point on
> parabola that has ordinate equal to the length of the top part,
> invert it upside down) gives me 'trouble', althogh I think
> it again should be somewhat simple to show.

It seems to me that you could construct a second cut point in the Y-coordinate segment, which would be symmetric with the first cut point, so that length b^2 is on bottom and a^2 on top with this cut point. Then construct a horizontal from the new cut point to interesect the parabola at X-coordinate = b. I think this works, but does it help with what you want to do?

> In unrelated matter, recently on Hyacinthos in post
>https://groups.yahoo.com/group/Hyacinthos/message/12324
> and also in post 12337 there was some material related
> to the 4T problem (I think it was related). The stuff he talks
> about is interesting, but too heavy for me. What do you and Alex
> can make out of it?

I have looked at these pages, and there does indeed seem to be a connection to the 4 travelers. The "equicenters" mentioned there have a physical meaning as well as a geometric one. If the three moving points are masses, the total momentum of the system is P = sum{m_i · v_i}. This can be looked at as a linear algebra problem, where a 2 by 3 matrix V of velocity components is multiplied by a 3 by 1 vector M of masses to give a 2 by 1 vector P of momentum components, i.e. P=V·M. Since there are only two equations in 3 unknowns, this system is degenerate, and there is a solution M that make P=0. But the velocity of the center of mass point is P/(total mass), which is zero, so this point never moves. Thus the equicenter is the center of mass of the moving system when the masses are chosen to make the total momentum be zero.

More direct applications of this theory to the 4 travelers problem are not obvious to me.

--Stuart Anderson

It was unusual for me that areas were made proportional to line

segments, I did not expect it from Euclid (although I

proposed it myself when I suggested that P. theorem is

everything that looks like A + B = C ).

In similar figures areas are proportional to squares of

corresponding linear dimentions. That is why I think proof #62

is the best to reflect it. As symmedian cuts the opposite side

proportionally to squares of the sides in right angle triangle

the altitude from the 90 angle is the symmedian, and symmedian

point is in the middle of this altitude. Thus the hypotenuse is

divided proportionally to the squares and that's the reason why

the sum of areas on sides is equal the area on hypotenuse.

==========

The 3 points I mentioned were vertex A=90 and point D (double

point) where the vertices of similar right triangles built on the

sides inward are. There are 3 centers of similarity of these

3 similar areas on the sides, and point D is one of them.

===========

Thanks for the idea of place change for segments on Y-coordinate of

parabola. I'm trying to prove that there is a 90 deg. angle

when 3 segments have their lengths a,b,c and a^2 + b^2 = c^2

without saying 'by P. theorem...' Not so simple.

Have to apply vector algebra and/or trigonometry (maybe something

else, I'm not that fluent in those). Alex is an expert on

parabola (wrote an article for MAA ) I know after the Big A,

and the Big A is unavailable personally. Maybe Alex will help

to squeeze 90 deg angle from parabola approach.

==================

I inched forward in the equilateral triangle problem: in my

Poncelet post (BTW, what do you think of that proof?)

I prove that area of right triangle is the product

of the segments the incircle touch point breaks hypotenuse into.

Thus the side of the P. square corresponding to this triangle is the

1/2 of chord through incenter perp. to the side of equilateral

which is the same diameter for all 3 triangles. These chords are

at 120 deg. angles . Now need a lot(?) of vector algebra

and trigo to show that if triangle is made of these chords

one of the angles will be 90.

About the 4T problem:

in msg. 12337 towards the end he says
..........................
To obtain the crashing motion, you
have just to project its worldline on the affine plane along time-axis. Then
the support line of the crashing motion will envelope an ABC in-conic
{Gamma}, projection of the apparent contour of (H) wrt time axis. Of course,
you can compute a tangential equation of {Gamma} and actually I have
succeeded to find it, leading to solutions of other new questions and so
one.. For example given 4 uniform motions, there is in general at most 4
crashing motions. If more than 4 crashing, then they are an infinity! Why?
What is fascinating in this theory is this mixing of synthetic geometry with
barycentric calculus apparently tedious for as you know where you are going,
you can use a software like Mathematica or Maple to help you.

...................................

As I said it is hard for me to understand all this, but there are

definite things in common. In my 4T posts I wrote about parabola

without any reasonable connection,

and he mentions 'the support line of crashing motion envelopes

an ABC in-conic'... His 4 lines are tangent to my parabola!

Just kidding. Can you read French?

On the entire Internet there are no site mentioning 4T problem.

How strange.

Salute,

Maj. Pestich

The gist of the matter is essentially your argument:

The Three Little Pigs built three similar houses according to their sizes (Figure 4). The two smaller attics happen to fit into the third one as in Figure 5 (in particular, this requires the attics to be right triangles.) The three living rooms are proportionately larger than the three attics and thus the two smaller living rooms add up to the third one (Figure 6).

This is also how I always thought of VI.31. I believe this is the most natural way to comprehend the theorem. I am going to amend the page with your and Givental's remarks.

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