Useless maths...

[RedSnapper]

You know, that maths they teach you at school but you just know you're never to need it in a real life situation so you don't retain it.

I need to be able to find the length of a side of an equilateral triangle that fits exactly inside a circle of x-diameter. Or to put it another way, I need to be able to divide the circumference of the circle into three equal parts using a pair of dividers. I have been using trial & error to get closer & closer until it's 'close enough', but I'd really like to be able to just calculate the correct measurement and get on with it.

'DAK' how to do this seems rather redundant - CSWKHTDTPHM* seems more appropriate as I'm sure plenty will....

TIA

* Could Someone Who Knows How To Do This Please Help Me

[modellingman]

Draw an equilateral triangle inside a circle. Let radius be r, call the vertices (points) of the triangle a, b and c and call the centre of the circle o. you want the length from a to b. Draw a line from o to the circumference of the circle so that it crosses line ab at its midway point. Call this midway point x.

o, a and x form a right-angled triangle whose hypotenuse (longest side) is the radius r. The angle at point o is pi/3 radians (or 60 degrees in protractor units) so if you recall the SOHCAHTOA rule, the distance from a to x is r*sin(pi/3).

The length from a to b is twice this length so 2*r*sin(pi/3).

There is a simple trick involving Pythagoras' Theorem which uses half an equilateral triangle with sides equal to 1 and leads to the deduction that sin(pi/3) equals sqrt(3)/2, so the value you require is r*sqrt(3) or 1.732*r. In practice the diameter of a circle is usually easier to measure so use 0.866*diameter

[GoSeigen]

The center of the triangle is also the center of the circle. Therefore the radius of the circle is the distance from the center of the triangle to a point of the triangle. The center of the triangle is at its center of gravity, one third of the distance from base to apex.

Imagine your triangle has sides of length y. The distance from base to apex is:

(√3/2)y

and the distance from center to apex is

(2/3)(√3/2)y = (1/√3)y

This is half your diameter x, i.e.

x/2 =  (1/√3)y

and rearranging to get the length of the triangle edges y in terms of the diameter x:

y =  (√3/2)x = 0.866x

GS

[kempiejon]

You know, that maths they teach you at school but you just know you're never to need it in a real life situation so you don't retain it.

I need to be able to find the length of a side of an equilateral triangle that fits exactly inside a circle of x-diameter. Or to put it another way, I need to be able to divide the circumference of the circle into three equal parts using a pair of dividers. I have been using trial & error to get closer & closer until it's 'close enough', but I'd really like to be able to just calculate the correct measurement and get on with it.

'DAK' how to do this seems rather redundant - CSWKHTDTPHM* seems more appropriate as I'm sure plenty will....

TIA

* Could Someone Who Knows How To Do This Please Help Me

I seem to remember that if you use a pair of compasses to draw a circle then place the point any where on the circumference, mark with the pencil the two points on the circumference, move the point to either cross and mark again, complete all the way round that give you a hexagon inside the circle. Does every other point make your triangle?

I remember right, here youtube https://www.youtube.com/watch?v=C6FiPa-aQ-Y

[quelquod]

Nearly but not quite. The circumference of the circle is 2 x pi x r where r is the radius and pi is a little more than 3 so there is a little more than 6 radii distances around the circumference.
(pi doesn't have an exactly describable value, it's transcendental, but it is roughly 22/7 - a little more than 3).

[kempiejon]

Nearly but not quite. The circumference of the circle is 2 x pi x r where r is the radius and pi is a little more than 3 so there is a little more than 6 radii distances around the circumference.
(pi doesn't have an exactly describable value, it's transcendental, but it is roughly 22/7 - a little more than 3).

Sorry quelquod I must have described my method poorly, lets try again. We know that pi is not 3 so the radius doesn't fit into the circumference exactly. Crossing the radius on the circumference as I described and found in the linked video is not dividing the circumference by the radius, it's drawing 6 lines (chords) of length r to form a hexagon within the circle and joining alternate points to form a triangle. Did that make sense, I could do this better with a bit of paper in real life.

[woodenman]

https://docs.google.com/drawings/d/16UH ... sp=sharing

Try the pic at this link. That should work, even though I've not done proper trig in 15 years!

[jfgw]

There are some rather complicated answers to such a simple problem.

If you know where the centre of the circle is, set your dividers to the radius and divide the circumference into six. This is basic, primary school "pretty patterns" stuff.

If you don't know where the centre is, you can find it. Take a 90 degree square and position it so that the 90 degree corner is touching the circumference. Mark the two places where the square crosses the circumference and draw a straight line through these points. This line should cut the circle exactly in half. Repeat the process to draw another line at a different angle. Where the lines intersect is the centre of the circle.

Alternatively, measure the diameter and divide by 2 to get the radius.

If you want to set your dividers to divide the circumference into three, multiply the diameter of the circle by 0.866024. (i.e., half of the square root of 3).

Julian F. G. W.

[RedSnapper]

In practice the diameter of a circle is usually easier to measure so use 0.866*diameter

Many thanks to all that replied. This is exactly what I was after.

[quelquod]

Sorry quelquod I must have described my method poorly, lets try again. We know that pi is not 3 so the radius doesn't fit into the circumference exactly. Crossing the radius on the circumference as I described and found in the linked video is not dividing the circumference by the radius, it's drawing 6 lines (chords) of length r to form a hexagon within the circle and joining alternate points to form a triangle. Did that make sense, I could do this better with a bit of paper in real life.

I must be getting old (well, I am of course!). I should have looked at the video, still haven't.

However, 6 chords of length r don't form a circumscribed regular hexagon. The outer ends of the 'first' and 'last' chords don't coincide. (The relationship for the length of a side of such a hexagon follows the maths above for the isosceles triangle.)

[kempiejon]

Thanks for coming back quelquod as I feel this is the heart of the matter.

However, 6 chords of length r don't form a circumscribed regular hexagon.

I'm sure they do.
You hold the contrary view. Can you help me see it your way perhaps? I was sure if pi were 3 circles would be hexagons.
here's what I think. There are 6 and bit radii in a circumference. Yes.
If we divide the circumference into 6, join each point then the joining lines will be r.
Set the dividers to r and mark each point on the circumference there is no remainder.
To draw an equilateral triangle join alternate points.

[quelquod]

Thanks for coming back quelquod as I feel this is the heart of the matter.

However, 6 chords of length r don't form a circumscribed regular hexagon.

I'm sure they do.
You hold the contrary view. Can you help me see it your way perhaps? I was sure if pi were 3 circles would be hexagons.
here's what I think. There are 6 and bit radii in a circumference. Yes.
If we divide the circumference into 6, join each point then the joining lines will be r.
Set the dividers to r and mark each point on the circumference there is no remainder.
To draw an equilateral triangle join alternate points.

I was so surprised by your comment that I hauled out an ancient pair of compasses and did exactly what you suggested on a piece of paper. The end points don't coincide (by much more than experimental error).

As I mentioned earlier, you can follow the same sort of geometric analysis as earlier for the triangle to prove the point.

[quelquod]

I was so surprised by your comment that I hauled out an ancient pair of compasses and did exactly what you suggested on a piece of paper. The end points don't coincide (by much more than experimental error).

And I did it again more carefully and indeed they do coincide.

My apologies, I feel a need for a large whisky!

Actually after a bit of research, I found that for a regular polygon with n sides where r is the radius of the circumscribing circle, the side length is given by 2 x r x sin(pi/n). For a hexagon this is r.

[kempiejon]

quelquod, have the whisky, thanks for the exchanges
I'm glad I had the courage of my convictions, you seemed pretty resolute. Something else helped me maintain my assertion, it was going to be another idea of an example for you, I can visualise a regular hexagon consisting of 6 equilateral triangles side of length r, meeting at the centre of a circle.

[GoSeigen]

There are some rather complicated answers to such a simple problem.

[...]

multiply the diameter of the circle by 0.866024. (i.e., half of the square root of 3).

Julian F. G. W.

Well, if you pluck this magic number out of the air, of course it's simple. but how exactly did you derive that number? Other Fools showed their working so that readers could correct any mistakes...


GS

[Bubblesofearth]

Using the Sine rule also works and is pretty simple.

Label 2 points of the equilateral triangle a and b and the centre of the circle, o.

Draw triangle aob.

Internal angles of an equilateral triangle are 60 degrees so angles abo and bao are half that, i.e. 30 degrees, and aob is 120 degrees.

By the Sine rule;

ab/Sin 120 degrees = ao/Sin 30 degrees

ab = sq rt 3 x x/2

It's useful to be aware of the Sine and Cosine rules as they enable you to find angles and sides of triangles that may be less amenable to conversion into right angled triangles.

BofE

[jfgw]

Well, if you pluck this magic number out of the air, of course it's simple. but how exactly did you derive that number? Other Fools showed their working so that readers could correct any mistakes...
GS

This isn't an exam where you have to show workings. If the answer is correct, there is probably not a mistake. (It is possible to get the correct answer by incorrect means but that is not relevant here.)

Julian F. G. W.