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I understood what it means. It just means to square, cube, etc.

**Al-Allo**- Replies: 3

So, if I have three numbers such that : a/b=b/c

Then we have a/c which is a duplicate ratio of of a/b

If we have 4 numbers such that : a/b=b/c=c/d

Then we have a/d is triplicate ratio of a/b

We can continue this way with other proportions.

So, what does it mean exactly ?

bob bundy wrote:

hi Al-Allo

I'm not understanding you.

Let's say A = 8, C = 2 and E = 4. Suppose the multiplier is 7.

Then G = 56, H = 14 and K = 28.

Now suppose B = 1, D = 3 and F = 2 with multiplier 5.

Then L = 5, M = 15 and N = 10.

G > L (56 > 5) but H is not greater than M ( 14 < 15).

So I guess that was not what you meant. Please clarify.

Bob

Yes, i'm not saying that taking every number is going to work. The basic idea is that if you have a proportion of the type :

3/2 = 6/4 = 12/8

with 7 :

21 , 42 , 84

with 5 :

10 , 20, 40

21 > 10 , 42 > 20 and 84 > 40.

**Al-Allo**- Replies: 3

Hi, there's something that have been bugging me.

If we have quantities A, C, E

And if we have quantities B, D, F

And if we take the equimultiples G, H, K from A, C, E

And if we take the equimultiples L, M, N from B, D, F

And we show that whenever G is superior to L, then H is going to be superior to M. And if equal, equal. And if less, then less.

And we show that whenever K is superior to N, then H is going to be superior to M. And if equal, equal. And if less, then less.

Then, how can you be sure that whenever G is superior L, K is going be superior to N. If equal, equal. And if less, then less.

Usually we have a situation of if X is equal to B and if Y is equal to B, then X is going to be equal to Y. (This is not hard to digest, obviously)

But in the situation just described, i'm having a hard time digesting this... I have no problem understanding the statement but the conclusion is hard to digest or to be accepted by my mind.

Thank you!

bob bundy wrote:

Sorry. I could not follow what you are asking. Life for me is too short to read and follow all of Euclid's elements. Problems with geometry I may be able to do, but getting inside his head is beyond me.

Bob

It's ok. The ideas still holds and it doesn't make it false it's just the way it is done which I found weird.

pls someone.

**Al-Allo**- Replies: 3

http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV7.html

I was just wondering something . We know that if a line touches a circle at one point, then this means that this line is forming a right angle with the diameter of the circle. (“From this it is clear that the straight line drawn at right angles to the diameter of a circle from its end touches the circle.” According to corollary of proposition 16 from book 3)

http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII16.html

In our construction, we would just draw the line at right angle to the diameter and with the corollary justify that is in fact touching the circle.

So, we already know that the angle made with the line touching the circle and the diameter is right, yet Euclid goes to prove it a second time by saying : “Then, since FG touches the circle ABCD, and EA has been joined from the center E to the point of contact at A, therefore the angles at A are right. For the same reason the angles at the points B, C, and D are also right.”

There’s no problem with what he’s saying, but it seems a little repetitive to prove something which is already known. Maybe it’s me doing an error. Could someone to me what’s wrong here ? Thank you

bob bundy wrote:

EDITED VERSION.

You have a chord and similar segments but on opposite sides of the chord.

Let the chord be AB and choose points C and D, on the circumference, one for each segment.

I'm not sure what is meant by 'similar' here. If it means that the two segments are the same shape then:

Then, if segment ACB is similar to segment ADB, they must be semi-circles, and angle ACB = ADB = 90.

If it means they are segments made by the same chord then ACBD is a cyclic quadrilateral so ACB + ADB = 180.

Bob

Well, from the definition, similar just means that you have equal angles in the segments. If this is the case, we say the segments of circles are similar

**Al-Allo**- Replies: 3

So, according to this figure :http://aleph0.clarku.edu/~djoyce/java/e … III23.html

We cannot have similar segments of circles and unequal ones be built on the same side of the same straight line.

My question is : Can we build similar segments of circles but unequal ones ? (It seems to imply it)

The definition of similar segments of circles is : "Similar segments of circles are those which admit equal angles, or in which the angles equal one another."

Here is the link for it : http://aleph0.clarku.edu/~djoyce/java/e … III11.html

Thank you!

**Al-Allo**- Replies: 1

I have the following theorem : "In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base."

(Figure is in the link) http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII20.html

English isn't my first language, so I just want to make sure that I understood something correctly. We prove the theorem by putting the two angles one on the other for the circumference. I was just wondering, can I assume that the angles do not need to be one on the other and they can have different portion of the circumference, as long as the circumference are of the same lenght ? (Will the proposition still work in this way?) I guess that Euclid did the proof by putting the angles one on the other for making the demonstration less wordy. (Less long to read)

Thank you!

geometry proof-verification euclidean-geometry

**Al-Allo**- Replies: 0

We have the following statement by Euclid : "I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilinear angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilinear angle." (Book 3 proposition 16)

The figure is in this link : http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII16.html

So, I was wondering, could somebody give me an example of where it is true ? For this proposition to be true, must the given straight angle be the same for both curved angles ? Or different?

**Al-Allo**- Replies: 3

Hi,

http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII15.html

I have understood the proof in general. It is only a small detail which i'm not sure. Maybe it's because english isn't my first language. Anyway, the part of the proof which says : "Then, since BC is nearer to the center and FG more remote, EK is greater than EH." I have no problem understanding what is said here. This is supported by this definition : "And that straight line is said to be at a greater distance on which the greater perpendicular falls." (Definition 5 of book 3) Now, this is where i'm unsure. From what I understand of it, it says that if I have a perpendicular that is bigger than the other, than my straight line is said to be at a greater distance. (This is how I understand it) Now, in the proof, we do the inverse. We know that one line is at a greater distance than the other and we conclude with the definition that one perpendicular is bigger than the other. How is this correct ? Unless the definition implies that the reverse is also ok, then this works. But if the definition implies only one direction, (The one which is defined) then how is the proof valid ?

By the way, you don't need to read all of the proof. Only the things at beginning are needed.

Thank yoU!

bob bundy wrote:

hi Al-Allo

I had hoped that there would be a definition somewhere to explain the word 'cuts'. I searched all the way through books 1, 2 and 3 but Euclid seems to use the term without definition. I suppose I would reason like this:

When object A cuts object B, there is a region that they share (the overlap) and a region that they do not share (what remains after the overlap is subtracted). If A does not cut B, then there is no overlap. If a line is in the overlap, then it is in both A and B. If A does not cut B, then any line that is fully in A, cannot be in B.

Given how careful Euclid is about other proofs, I'm a bit surprised that this is omitted. But all mathematical theories have this problem: the writer may define some words, but cannot define all (for the words used in the definition would then have to be defined, and then those words, ............). He appears to take the meaning of 'cut' as obvious. Wouldn't it be great if we could ask him.

Bob

So we are assuming that there exists two cases where the circles are crossing each other and where they aren’t crossing each others to be able to do the proof ? Also, the case where they don’t cross each other, is this where definition 3 comes in ? In other words, do we prove that “ If A does not cut B” by using definition 3 ? Just want to make sure I understand the use of definition 3. Also, to be able to do the proof, I assume that it is highly important that we prove first that both lines are in each circle so that when we use definition 3, we can say that a circle that don’t cut another one then have for consequence that the line can only be located in one of the circles and not in both of them. (Since we said earlier that they had to be in the two of them) Finally, the part where “then any line that is fully in A, cannot be in B.” is something that we assume. Shouldn’t it be proved or something ? Oh, I forgot another thing, are you sure your interpretation is a good one or maybe Euclid was thinking something else ? I just feel bad when I skip something that I’m not sure of understanding. (It doesn’t quit my mind lol)

By the way, the link says that “There are logical flaws in this proof similar to those in the last two proofs.” I even found a document with comments by Heath if you want it. (The one who translated the elements)

**Al-Allo**- Replies: 2

"Then, since on the circumference of each of the circles ABDC and ACK two points A and C have been taken at random, the straight line joining the points falls within each circle, but it fell within the circle ABCD and outside ACK, which is absurd.

Therefore a circle does not touch a circle externally at more points than"

There's only one little detail which i'm not sure of. We are trying to prove that circles which touch one another will only touch at one point. Fine. I understood the first part which treats of a circle in another one. It's only the case where one circle touches another one from the outside. By using proposition 2 of book 3, we prove that the line AC will be inside both of circles since the two points are on each circumference of the two circles. Now, this is where I get lost. We say that "but it (line AC) fell within the circle ABCD and outside ACK" and we prove this by using definition 3 of book 3 (Circles are said to touch one another which meet one another but do not cut one another.) In other words, this definition says that circles which touch another do not cut one another. In our situation, we have two circles which touch one another and are not supposed to cut one another. This is where I don't understand, how does this justify this : "but it fell within the circle ABCD and outside ACK." How do we get that conclusion from the definition?

Thank you!

http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII13.html

bob bundy wrote:

hi Al-Allo

Euclid was very thorough, so I doubt that he left this out. I don't know the propositions well enough to say where, but I'm sure he has already done this, in an earlier proposition. Why not do it anyway as an exercise ?

I find I cannot do geometry unless I draw the diagrams. My brain seems to be very visual; I cannot remember names unless I see them written down.

Well done for completing book 1. Only another 12 to go ...

Bob

I'm able to prove that the ABGH and BEFG are parallelograms. BEFG is one because we constructed it. ABGH is one because we can easily prove that we have 2 pairs of parallels. We have AB that is in a straight line with BE. And we have BE that is parallel to GF. Logically, AE is parallel to GF. We extend GF in a straight line to H. This makes AE parallel to HF. Finally, this makes AB parallel to HG (If we "cut" the bigger parallels) and AH is parallel to BG because we made the construction. I need to find a way to prove that MKEB has two pairs of parallels (thus, making it also a parallelogram) and this will result in proving that the fourth figure is really a parallelogram.

By the way, is my explanation any good you think ?

Thanks! (I do hope Im not bothering you in any way)

bob bundy wrote:

hi Al-Allo

It's taken me a while to read all the props that leads up to this one. I'm not sure I fully understand your query. The lines from a parallelogram are extended so this creates several parallelograms and allows the earlier props to be applied.

Bob

What I mean is, shouldn't I prove that we have two small parallelograms ? That we are in fact dealing with such objects ? (Which I Would need to prove) By the way, I finished book 1 of the Elements. A pretty good book and interesting The only irritating part is when you have to draw many many many things.

**Al-Allo**- Replies: 5

Sorry Bob, I hope I'm not bothering you with my numerous questions.

There's again one small detail on which I'm not sure. (Proposition 44 - book 1)

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI44.html

Here's the quote :

"Then HLKF is a parallelogram, HK is its diameter, and AG and ME are parallelograms, and LB and BF are the so-called complements about HK. Therefore LB equals BF."

So, we have constructed a parallelogram HLKF and we have also constructed the diameter HK which bisects it. Now, the only part which I'm not sure is concerning the small parallelograms AG ME and the complements LB BF. We know that by propositon 43, the complements of a parallelogram equal one another. My question is : knowing that I have parallelogram that has a bisector, do we automatically know that there are two parts which are smaller parallelograms and that there are two other parts which are complements ? Can we just assume that these four smaller parts exists without proving it in proposition 44 ? Is saying "ok, here are two complements and two smaller parallelograms) enough ?

Thank you!

bob bundy wrote:

Yes. If you draw one diagonal to divide the quadrilateral into two triangles, it is not too hard to show these are congruent (side and two angles). Hence opposite sides are equal in length.

Bob

I've had a quick look at your other post but it will take a while to take in all the relevant propositions. I'll reply when I've had time to do this.

So, we do have a Rhomboid ( a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled.) ? And my question was, why does Euclid think it's enough to just show that we have opposite parallels ? I mean, a square also have its opposite sides parallels and so does a rectangle. Shouldn't he show that we have also equal opposite side and equal opposite angles ? (The side must not be equilateral and not right angled)

**Al-Allo**- Replies: 3

I have a small question regarding proposition 37 of the elements of Euclid. http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI37.html

The only problem I got with the proof is the fact that we don't seem to prove that we do have parallelogram. We have a figure with 4 side and we know that each opposite sides are parallels to each other. Is this considered enough to acknowledge that we have parallelograms ?

Thank you!

**Al-Allo**- Replies: 1

I have a small question regarding proposition 33 of the elements of Euclid. http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI33.html

We want to prove that two lines joining equal parallels at their ends are themselves equals and parallels. The only thing I don't understand is why does Euclid use the side-angle-side theorem to prove that the remaining angles are equal to one another ? Why doesn't he use the proposition 29 (A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles.) to prove that all alternate angles are equal to one another?

In my opinion, I think it's because were considering only at the begining the "finite" straight lines BA-DC, so this limits us only to two angles ABC and BCD.

Thank you

Bob did you receive my private message?

**Al-Allo**- Replies: 3

I have small question regarding this proposition : http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI26.html

To prove that one side is equal to another, Euclid assumes that one side is bigger than the other. Finally, when Euclid arrives at a contradiction, he dismisses the assumption about the inequality of sides and considers them equal. What I was wondering is, if we assume that a side is unequal to another one (A is bigger than B) and arrive at a contradiction, shouldn't we also try the inverse, B being bigger than A and assure ourselves that we arrive also at a contradiction to conclude that finally, A is equal to B ? Thank you!

**Al-Allo**- Replies: 1

I have a very short question on this proposition : http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI17.html

I understand the way the theorem was proved. Euclid proves that angles ABC and ACB is less than 180 degrees. At the end, he says the same method can be applied to prove for the other cases. I was able to do the same for the angle BAC and ACB. But how do I do it for CAB and ABC ? It doesn't seem the same method holds for this case, am I right ?

Yes, sorry it was an error. I edited. Anyway, Im going to go sleep.

**Al-Allo**- Replies: 3

I have the following integral to solve :

http://www.wolframalpha.com/input/?i=in … rom+0+to+1

I was able to integrate and I got : (y/4)+((e^(2y)-e^(-2y))/16)

I don't what to do after that. I know I can replace ((e2y-e-2y)/2) with sinh(2y)... but after that I'm lost. Help please ? Thanks

I'm using the following substitution : y=sinhθ/2