Math Challenge -- Pi R Round

reply to post by Ross 54





Breakdown the circle into a sequence of triangles starting with the large equilateral triangle in the middle. I would estimate a formule as below for working out area. I have

(The large equilateral triangle Height x Width/2 ) + 3 x ((one side length of large equilateral triangle X 4/7)/2) x (one side length of large equilateral triangle X 2/7) + 6 x ((one length large equilateral triangle X 4/7)/2) x (one length equilateral triangle x 1/14))

reply to post by Ross 54


1.Premise: If a right angled triangle is inscribed in a circle, the hypotenuse always runs through the centre of the circle.

2.The area of a circle is pi*diameter squared divided by four.

3. Using pythagoras. The sum of the squares of the two shorter sides of the triangle equals the diameter squared.

4.Hence, the area of the circle can be found by summing the squares of the two shorter sides of the triangle, and multiplying the answer by pi divided by four.


Area= ((a^2)+(b^2))*(pi/4) ; where a and b are the lengths of the two shorter sides of the inscribed triangle.

if you google: what is 34345434 x 23453243242 ...it will give you the answer

Brotherman
I'm not really sure but it appears you are hinting at modular arithmetic or clock mathematics. If so I am not sure how you use this to solve for area of a circle. I said before that it is possible to basically turn your circle into a polygon, I just don't remember exactly how that worked to calculate the area the more you evenly slice up the circle the closer your going to be to the area in which this method requires not a circumference nor pi. If it is clock arithmetic I'm interested in hearing how this works

so then you are using modulus 11 (if you are using 10) or you are using a different modulus in increments of 10 parts from 12 divided by Pi? or something to this effect? so essentially you are trying to gain the remainder to describe the amount of time or area inside your circle?

edit on 10-12-2013 by Brotherman because: (no reason given)

so say every 10 minutes the clock moves 5inch
10M=5in there is 60inch on the face of the clock so equally 1Minute=1inch if this is known then it would be 60*pi?(assuming one increment moves at 5in in a 10min setting which is probably absurd but just for example) Am I getting close to what you are looking for

edit on 10-12-2013 by Brotherman because: (no reason given)

edit on 10-12-2013 by Brotherman because: (no reason given)

Yes, Brotherman, you are the closest to the solution. It does involve a clock face. However the solution is simpler than the one you described.

reply to post by Ross 54


I'm still working on it ill get back to this after work

n x (d/4)^2 x theta

Where n is the number of iterations to complete a 360deg.

Or a1a2, b1b2 (mod n)

But they still both use radius/circumference when in expanded form.

reply to post by Ross 54


ok. Using the method I described in my previous post and a clock.

Draw a line from 3 o'clock to 12 o'clock. Lets call the length of the line x.
Draw a line from 12 o'clock to 9 o'clock. Its length must also equal x.

Area = ((x^2+x^2))*pi/4

Area = pi*x^2/2

reply to post by bastion


What if the hand moved from 12 to 1 = n amount of square inches you could then just add the rest of the of the sections couldn't u?

reply to post by Brotherman


Yup that's the top equation, but really they're sectors where the length of the minute hand id the radius and distance is moves a fraction of the circumference.

Can anyone offer a formula for the area of a circle that mentions neither the radius nor the circumference of the circle? Clue: This is possible, given enough time.

Well, this gives a general answer. Sorry I didn't derive it myself.

Triangle circumcurcle



Take three points on the diameter of the circle. Measure each length with a ruler. Use the "if you know all three sides" formula in that reference to get the radius

so it's area is pi * a^2 b^2 c^2 / [ (a+b+c)(b+c-a)(c+a-b)(a+b-c) ]

This seems to meet the requirement of the problem, though it doesn't use the clock hint directly.

More practically, e.g. if you were a carpenter and wanted to do this for real, you'd probably use Thales' theorem with a right-angled T-ruler (one which can measure distances off a right angle).

www.mathopenref.com...

If you can draw a right triangle in the circle and measure its sides A and B, the third side is a diameter.
You get it from Pythagoras namely sqrt(A^2+B^2) and so the area is pi/4 D^2 = (pi/4) * (A^2 + B^2).

Thank you to everyone who responded to the thread. Some ingenious answers were received.
Besides providing a challenge for people to solve, I was interested to know if anyone was already aware of, or would now think of the particular solution I had in mind.
No one did, and since there have been no more responses in nearly a full day, I will now give that particular solution.

Imagine an ordinary 12 hour clock face. Draw a chord across the circle between the points on the edge of the clock face nearest any two alternate numbers, say 5 and 7, for example. Measure the length of the chord. Multiply that figure by itself, and multiply this product by Pi. This should give the area of the circle.
When researching this problem, I found that the fact the chord I described happens to be the same length as the radius of the circle was already known. I was not able to find that anyone had made use of this fact in an alternate formula for the area of a circle, with one rather mysterious-seeming exception.

There are 12 points on the clock face, each number is 30 deg apart, any 2 numbers 2 apart are therefore 60 deg apart, measured from centre of circle. Make a triangle, two sides the same length, (radius to each number), and the angle between is 60 deg, and you have an equilateral triangle, all sides, and angles equal - so, all angles 60 deg and all sides equal to the radius (and the chord you've drawn)
so technically you're using the radius (or a value equal to the radius) without realising it.

CrastneyJPR
There are 12 points on the clock face, each number is 30 deg apart, any 2 numbers 2 apart are therefore 60 deg apart, measured from centre of circle. Make a triangle, two sides the same length, (radius to each number), and the angle between is 60 deg, and you have an equilateral triangle, all sides, and angles equal - so, all angles 60 deg and all sides equal to the radius (and the chord you've drawn)
so technically you're using the radius (or a value equal to the radius) without realising it.

Yes, of course the the chord, happens to be the same length as the radius. It not the radius, nor is it derived from the radius, as are area-of-a-circle formulas involving the diameter of the circle.
It's interesting that you should mention an equilateral triangle. The way the solution was presented originally had a larger equilateral triangle. It was drawn between the edge points nearest 12, 4 and 8 o'clock.
The largest possible square, the side being a 60 degree chord on the edge of the circle, from 5 to 7 o'clock points ,is drawn in through one side of the triangle, up to its inside edges, then across. It ends up looking like a crude drawing of a small square house with a large peaked, overhanging roof. The area of the circle is Pi times the area of the square within it.
It seems to be a novel solution for the area of a circle. The way it was presented was very unusual, too.

Ross 54

CrastneyJPR
There are 12 points on the clock face, each number is 30 deg apart, any 2 numbers 2 apart are therefore 60 deg apart, measured from centre of circle. Make a triangle, two sides the same length, (radius to each number), and the angle between is 60 deg, and you have an equilateral triangle, all sides, and angles equal - so, all angles 60 deg and all sides equal to the radius (and the chord you've drawn)
so technically you're using the radius (or a value equal to the radius) without realising it.

Yes, of course the the chord, happens to be the same length as the radius. It not the radius, nor is it derived from the radius, as are area-of-a-circle formulas involving the diameter of the circle.
It's interesting that you should mention an equilateral triangle. The way the solution was presented originally had a larger equilateral triangle. It was drawn between the edge points nearest 12, 4 and 8 o'clock.
The largest possible square, the side being a 60 degree chord on the edge of the circle, from 5 to 7 o'clock points ,is drawn in through one side of the triangle, up to its inside edges, then across. It ends up looking like a crude drawing of a small square house with a large peaked, overhanging roof. The area of the circle is Pi times the area of the square within it.
It seems to be a novel solution for the area of a circle. The way it was presented was very unusual, too.

edit on 12-12-2013 by Ross 54 because: added comma

My solutions didn't use the length of the radius at all.

Knowing 2 to find the third.

Integrate (d/2) * Cos(theta) from 0 to 2Pi d(theta) or (d/2) * Sin(theta) from 0 to 2Pi d(theta)

or just use a tape measure if you can measure your chord whats the point why not just measure the diameter, why work harder then you have to. LOL I wasnt under the impression you could use anything like that to solve the problem presented

The way the problem was originally presented was entirely graphical. No numbers, no measuring. It showed the Pi ratio, expressed both as lengths, and as areas within within the circle and square I described above.
I added the step of measuring so that a numerical value could be gotten, in line with the usual way of thinking about Pi.
The fact that the area of a circle can be obtained in this way seems to have escaped general notice, and to be outside the body of mathematical thinking. As such, it is something of an enigma.

Besides the novelty of this approach to finding the area of a circle, is the oddity of the way it was presented.
It's not known who devised the solution, or the diagram of a square inside a circle, from which it arises. No one published it or put it on the internet with his or her name attached. It simply appeared one summer's day in 1998, pressed into a field of grain, near Old Sarum, Wiltshire, in the U.K..
One may, of course, think what they please about the origin of 'crop circles'. We have the fact of an approach to the problem of the area of a circle that no one seems to have thought of before, in all the long history of mathematics. We also have the fact that it appeared by a means that some, at least, have suspected of being connected to an off-world intelligence.