Geometry Question

[Johnny5]

Some of you have seen the formula for arc length:

 

[math] S = R \theta [/math]

 

Suppose that I have chosen my unit of distance to be the meter, and I have a rigid meterstick.

 

Suppose, i have learned this:

 

If I take a rope, and make a circle with radius one meter, that when I straighten out the rope, the length of the rope is greater than six meters, but less than seven meters.

 

I can simply see that it is greater.

 

Ok so, I then formulate the following question, which I now want answered:

 

Suppose I use a compass, and describe a circle with radius one meter. As the arc length is slowly made in the sand, it gets longer and longer. I want to know what angle to stop at, so that the arc length just traced out in the sand is one meter.

 

Call one full revolution 360 degrees.

Call one half a revolution 180 degrees.

Call one fourth a revolution a right angle of 90 degrees.

 

Now, I can easily see that a right angle is way too much, by direct measurement.

 

In other words, the length of a string, placed along the arc made in the sand, is longer than one meter, not equal to one meter.

 

So next I try a sixty degree angle.

 

Assume that I have learned that this is the measure of an angle inside an equilateral triangle.

 

I go off into the sand, and I construct an equilateral triangle. I can clearly see from the lines in the sand, that each side is one meter long. And on the far side of the equilateral triangle, there is a circular arc, because of the method I used to construct the equilateral triangle.

 

So two lines enclose a space, one is straight, the other curved.

 

The straight line is the shortest path from one point to another, hence it is painfully obvious that the curved path exceeds one meter. Thus, a sixty degree angle is too much. The arc described will still exceed one meter long.

 

Next, I consider a 45 degree angle. I figure out that this is one eigth of a full circle. Thats easy enough for me.

 

Ok so, I know the full distance around the circle is a little more than six meters, so I divide 6 by 8, and I get a number less than one, so that 45 degrees just isn't enough.

 

Thus, the angle I want is greater than 45 degrees, but less than 60 degrees.

 

What is the angle at which the circular arc will be one meter long?

 

How do I solve for it?

 

Thank you

[J.C.MacSwell]

 

What is the angle at which the circular arc will be one meter long?

 

)

 

My guess is 1 rad.

 

Edit: Can I assume you are not allowed to use 360 degrees divided by 2 pi (which would be trivial)?

[Rakdos]

A/360*C=L

 

C= circumference

A= Degree of Arc

L= Length of Arc

[ydoaPs]

Some of you have seen the formula for arc length:

 

[math] S = R \theta [/math]

 

Call one full revolution 360 degrees.

Call one half a revolution 180 degrees.

Call one fourth a revolution a right angle of 90 degrees.

 

you can't do that. that formula is in radians

 

so

 

[math]1=1\theta[/math]

 

obvious answer is [math]\frac{1}{1}=\theta{\Rightarrow}{\theta}=1radian[/math]

[Johnny5]

My guess is 1 rad.

 

Edit: Can I assume you are not allowed to use 360 degrees divided by 2 pi (which would be trivial)?

 

Oh, it is exactly one radian. There are a little more than six radians, in any full circle. mmmm

 

6.28.. something or other, the number being transcendental.

 

Which means that it isn't the root of an algebraic equation with rational coefficients.

 

The first human being to demonstrate that pi, and hence 2pi, is a transcendental number, was German mathematician 1852-Carl Louis Ferdinand von Lindemann-1939. This was in the late 1800's, i forget the exact year.

 

So yes, the answer is exactly one radian, but I want the answer in degrees.

 

Regards

 

I don't want a trivial answer. I was thinking of something that can be pictured. The steps involved in figuring out the answer.

[Johnny5]

A/360*C=L

 

C= circumference

A= Degree of Arc

L= Length of Arc

 

 

Let's see...

 

[math] S = r \theta [/math]

 

where theta is in radians.

 

S is arc length' date=' and r is the length of the radius of the circle.

 

So le me have a look at your formula here...

 

[math'] \frac{A}{360} C = \frac{A}{360} 2 \pi r = L [/math]

 

A is "degree of arc" L is arc length.

 

So here is a relationship between degrees, and radians:

 

[math] \frac{degrees}{180} = \frac{radians}{\pi} [/math]

 

So using the formula above, you can switch back and forth, i know that.

 

So of course it follows that:

 

[math] \frac{degrees}{360} = \frac{radians}{2\pi} [/math]

 

Therefore:

 

[math] \frac{2 \pi}{360} = \frac{radians}{degrees} [/math]

 

So you have L= arc length, I have S=arc length, therefore S=L

 

Therefore:

 

[math] \frac{A}{360} C = \frac{A}{360} 2 \pi r = S [/math]

 

Therefore:

 

[math] \frac{A}{360} C = A \frac{radians}{degrees} r = S [/math]

 

So with A in units of degrees, r in units of meters, S will have units of radians times meters?

[Johnny5]

you can't do that. that formula is in radians

 

so

 

[math]1=1\theta[/math]

 

obvious answer is [math]\frac{1}{1}=\theta{\Rightarrow}{\theta}=1radian[/math]

 

Yes I know the formula is in radians, it's the one I committed to memory 20 years ago.

[Johnny5]

you can't do that. that formula is in radians

 

so

 

[math]1=1\theta[/math]

 

obvious answer is [math]\frac{1}{1}=\theta{\Rightarrow}{\theta}=1radian[/math]

 

Maybe this will help. Right now' date=' I am doing a study of classical geometry, and restricting myself to only those things allowed by Euclid's postulates. What I have been doing, is busily constructing an extremely long proof sequence.

 

Recently, I posted a thread on similarity of triangles. The reason I did that, is because if I have AA, then I must also have AAA, because the sum of the interior angles of a triangle are 180 in Euclidean Geometry.

 

Now, once I have AA, and likewise AAA, then I can quite easily prove that the ratio of the circumference to the diameter of [i']any [/i] circle, is a dimensionless numerical constant, using a construction that I have found.

 

Its quite a simple construction, here let me describe it to you:

 

 

[math] \text{Theorem:} [/math]

 

Let A, B denote circles chosen at random. Without loss of generality, let them lie in a common plane, and let them be concentric (having the same point as center).

 

Ok so...

 

Using a postulate of Euclid, we can pick a point in space at random, and describe two circles with that point as center, but different radii.

 

Let C1 denote the circumference of the smaller circle, and let r1 denote the radis of the smaller circle. Let C2 denote the circumference of the larger circle, and let r2 denote the radius of the larger circle.

 

We now want to set out and prove the following statement is true:

 

[math] \frac{C_1}{D_1} = \frac{C_2}{D_2} [/math]

 

Where D1 denotes the diameter of the smaller circle, and D2 denotes the diameter of the larger circle. Recall

 

[math] D_1 = 2r_1 [/math]

 

[math] D_2 = 2r_2 [/math]

 

So let two circles have been described in a common plane, with a common center.

 

Now, from the center of the circle, produce two rays in the plane in any direction, long enough to cut both circles. (Euclids first postulate allows us to construct a finite straigth line between any two points, and his second postulate allows us to make the straight line segment longer, from either extremity, but still in a straight line.

 

So an angle has been chosen at random.

 

Thus, once the theorem is proven, it will hold for any arbitrarily chosen angle.

 

Now, a person could look at the figure and see two angles, one larger than the other, but in what follows, I am referring to the smaller one. So we have a sector of a circle we can focus upon.

 

Focus now, on the ratio of the smaller arc length, to the larger, in the sector.

 

 

Right now, these lines are curved, instead of straight, but to use AAA, we need triangles with straight lines as sides.

 

Without a diagram, this is going to be somewhat hard to communicate, but I can see the picture without even drawing it, so I will convert it into words just the same...

 

Let the center of the circles be denoted by A.

Let Ray1 cut the smaller circle at the point B, and the larger circle at point C.

Let Ray2 cut the smaller circle at point D, and the larger circle at point E.

 

Thus, we have two arcs:

 

[math] \overset{\frown}{BD} [/math]

[math] \overset{\frown}{CE} [/math]

 

The radius of the smaller circle is such that:

 

[math] r_1 = \overline{AB} = \overline{AD} [/math]

 

And the radius of the larger circle is such that:

 

[math] r_2 = \overline{AC} = \overline{AE} [/math]

 

Now comes the interesting part.

 

We are going to construct two straight lines, whose lengths are equal to the lengths of the curved arcs.

 

These two lines are going to be sides of different, but similiar triangles.

 

In other words, we are going to measure that which is curved, using that which is straight.

 

The whole point is this... we are going to use knowledge of triangles, to learn about circles. Now, I just have to explain what is actually an extremely simple construction to carry out.

 

Incidentally, once this construction has been carried out, then it can be connected to another thread, which I am still working on, which has to do with the formula for the area of a circle. In that thread, I start off with the definition of the area of a square, then move on to the area of an arbitrary circle, the pythagorean theorem gets proved, and next, I am going to cover the formula for the area of an arbitrary regular n-gon.

 

And then in the limit as n approaches infinity, I am going to obtain the following formula for the area of a circle:

 

[math] A_{circle} = \pi r^2 [/math]

 

But, in order to properly succeed in that argument, it is necessary that I have previously proven the theorem I am proving here in this thread, using the construction I am about to give.

 

At the point B, construct the tangent line to the smaller circle; the smaller circle is the circle with arc BD upon it.

 

At the point C, construct the tangent line to the larger circle; the larger circle is the circle with arc CE upon it.

 

(If you don't know how to construct the tangent, that is quite easily explainable. Knowledge of how to construct the tangent line is considered as part of the a-priori knowledge of this reasoning event going on here.)

 

So let those two lines have been drawn.

 

Now, focus upon [math] \overset{\frown}{BD} [/math]

 

Suppose that there was a finite straight line BF, whose length was exactly equal to the length of the arc [math] \overset{\frown}{BD} [/math]. Then you could describe a circle with radius [math] \overline{BF} [/math] which would cut the tangent line lying upon the smaller circle at the point F.

 

Suppose the point F was successfully found.

 

Similarly, if there was a finite straight line [math] \overline{CG} [/math] one could describe a fourth circle, such that the point G is on the tangent line currently lying upon the larger circle. Suppose the point G was found.

 

Thus, we have formed two right angles.

 

One right angle has side AB, and side BF.

 

The other right angle has side AC, and side CG.

 

We know the angles are right angles, because the tangents to a circle must be perpendicular to the radius drawn to the point of tangency. If this were not so, then the tangent would cut the circle at more than one point, and therefore not be a tangent to the circle, it would be a secant.

 

Now, since the sum of the interior angles of a triangle must equal two right angles, it is necessarily true that:

 

Angle BFA is equal to angle CGA.

 

But this could use a bit more explanation:

 

In order to use this construction to conclude that the ratio of circumference to diameter is a numerical constant, we need to be certain that:

 

The straight line AF and the straight line FG lie in one common infinite straight line.

 

Suppose we already are certain that AFG line in one infinite straight line.

 

We will then be certain that:

 

[math] \angle BAF = \angle CAG [/math]

 

We are already certain that the three points A,B,C lie on the same infinitely long straight line, because that was stipulated during the construction.

 

And we know that we constructed two right angles, so we know this too:

 

[math] \angle ABF = \angle ACG = \mathcal{R} [/math]

 

Where I have used the symbol [math] \mathcal{R} [/math], to denote a right angle.

 

Now, if we hadn't have chosen an arbitrary angle, if instead we choose the perfect angle for segment BF to equal segment AB, then triangle ABF would be isosceles.

 

And if we hadn't have chosen an arbitrary angle, if instead we choose the perfect angle for segment CG to equal segment AC then triangle ACG would be isosceles.

 

And base angles of isosceles triangles are congruent.

 

The angle at which this occurs, is called one radian.

 

But, if we choose our angle ahead of time, then the angle chosen wasn't at random... it wasn't arbitrary.

 

Since the sum of the interior angles of a triangle is [math] 2\mathcal{R} [/math], it would follow that:

 

[math] \bigtriangleup ABF [/math]

 

is similiar to

 

[math] \bigtriangleup ACG [/math]

 

And here is why...

 

Both triangles have a right angle inside of them.

 

Focus now on isosceles triangle ACG.

 

The right angle cannot be a base angle, because then the other base angle would also be a right angle, and the sides of the triangle wouldn't meet at the apex, which they clearly do. Thus, the following two angles are the base angles of triangle ACG, and they are equal as Pappus ingeniously proved several hundred years after it appeared as proposition 5 of Euclid.

 

[math] \angle GAC [/math]

[math] \angle AGC [/math]

 

In other words, each of the angles above is a base angle of triangle ACG. Therefore, they are equal in measure. Thus, the following statements would be true had we chosen a specific angle of one radian:

 

[math] \angle GAC = \angle AGC [/math]

 

[math] \angle FAB = \angle AFB [/math]

 

You know what, this argument still isn't clear enough, because we have assumed the conclusion. In other words, by saying that we chose the perfect angle, we cannot conclude we chose the perfect angle.

 

And this construction is supposed to be for the purposes of reasoning correctly.

 

Perhaps I should start over, and be clearer.

 

----------------------------------------------------------------------

 

Goal: To prove, as a theorem of some unknown set of axioms that, the ratio of circumference of any circle, to its radius, is a numerical constant.

 

Suppose that we have the ability to rotate circles and lines out of a common plane, into the third dimension, as well as translate their centers through space.

 

Then, given any two circles in the universe, we could move them from where they initially are, into a common plane, and then cause them to have the same center.

 

Now, circles and straight lines don't really exist in reality, but in geometry, we pretend they do, so that we can reason about space.

 

Now, in the theorem we are attempting to prove here, the theorem starts thusly:

 

For any circle ....

 

Now, in the construction being carried out, it is stipulated that we start off with two concentric circles lying in a common plane.

 

So someone can come along, and argue that we haven't proven the theorem is unconditionally true, that what was proved was this:

 

If two circles are coplanar, and have the same center, then the ratios of their circumferences to their respective diameters are both equal to the same numerical constant.

 

But that isn't what we want to prove.

 

Let us proceed as though we have as axioms, that we can translate the center of a circular figure through space.

 

Let us also proceed as though we have the ability to rotate a geometric figure out of the plane of construction, into the third dimension.

 

Thus, as long as there is a sequence of steps that can be carried out, which translates the center of a circle far away from a given circle, all the way to where its center coincides with the given circle, and then we can rotate it so that it lies in the same plane as the given circle, then we will be justified in saying that we proved that "for any circle"

 

And that what we proved is unconditionally true.

 

So we have found two unknown axioms, which will permit the particular construction I have in mind, to actually succeed in proving the theorem.

 

So we are going to be moving circles and straight lines around in our mind, so the question arises as to whether or not this takes place in time. In other words, are the transformation formulas functions of time or not. Historically, this is of some significance.

 

The reason why has to do with the special theory of relativity, in which moving objects shrink if they are moving in a reference frame, as compared to a measurement made in their rest frame.

 

If it can be shown that simultaneity is absolute, then SR is wrong, and we don't have to worry about this.

 

But some reasoning agents will not know a priori to this reasoning event, that SR is wrong.

 

For that reason, the motion here is considered to take place at the same moment in time. Thus, the geometric figures which we move around in our minds, cannot change size.

 

This means that the transformation formulas we are looking for are abstract. They do not represent real transformations about the center of mass of real objects, with real centers of inertia.

 

And there should be no objection to this, inasmuch as real circles don't exist anyway.

 

So what I have in mind is something very very simple.

 

 

There is a theorem here, which is to end up being true for any circle in three dimensional Euclidean space.

 

First lets state the theorem to be proven clearly:

 

[math] \text{Theorem:} [/math]

 

The ratio of circumference to diameter, of any circle, is a numerical constant. Since the constant is unique among all numbers, let us reserve a special symbol for it, which is the universally accepted symbol [math] \pi [/math].

 

So let C denote the circumference of an arbitrary circle, and let D denote its diameter. Symbolically, here is the theorem we are setting out to prove:

 

[math] \forall O[ \frac{C_O}{D_O} = \pi ] [/math]

 

There is also another way to write the theorem we are trying to prove.

 

Suppose we are given two arbitrary circles, circle A, and circle B.

 

Each circle has its own diameter, and each circle has its own circumference, and these are commonly thought of as properties of circles.

 

We can avoid being abstract, if we choose a unit of distance, so I will choose the meter.

 

So the diameter of circle A, is some amount of meters, and the circumference of circle A is also some number of meters.

 

And the same goes for circle B. The diameter of circle B is some number of meters, and the circumference of circle B is some number of meters.

 

Now we shouldn't think that the lengths in question must be an integral number of meters. For example, circle A could have a diameter of .2 meters, and circle B could have a diameter of 1.52954345 meters. We are being entirely general, and thorough in this analysis.

 

So let us denote the diameter of circle A, as follows:

 

[math] D_a [/math]

 

And let us denote the circumference of circle A, as follows:

 

[math] C_a [/math]

 

And let us denote the diameter of circle B, as follows:

 

[math] D_b [/math]

 

And let us denote the circumference of circle B, as follows:

 

[math] C_b [/math]

 

So we have two circles, chosen at random from the set of circles.

 

So, using the symbols above, the theorem we are trying to prove is:

 

[math] \frac{C_a}{D_a} = \frac{C_b}{D_b} [/math]

 

Since the circles were chosen at random, it would thus be true that for any circle A, and any circle B:

 

[math] \frac{C_a}{D_a} = \frac{C_b}{D_b} [/math]

 

But we need to have adopted translation and rotation axioms.