If I plot the number of sides in a regular polygon (x) versus the 2 interior angles* of the triangles that make it up (y), the two lines intersect at EXACTLY 6 sides. Why?
In other words, why do equilateral triangles fit perfectly into a hexagon? Why isn’t a hexagon made up of triangles with angles of, say, 64-58-58 degrees?
This is the same reason that six circles fit perfectly around a circle of the same size. It is the minimum energy configuration of natural systems and allows regular close packing.
I am probably looking right at the obvious answer, but I cannot see it. What gives?
*only 2, because 2 are the same
If you cut a pie into six equal slices, the angle at the point of each slice is 60 degrees, because there are 360 degrees around a point - so dividing those 360 degrees into six equal parts yields 60 degree angles.
Remember that a ‘degree’ is an arbitrary number. We could decide that there were 270 degrees in a circle; in which case, each slice would be 45. 360 sounds a funny number to choose, but it works pretty well. One degree is small enough for most uses and has the useful feature of being divisible by 24 integers (whole numbers) - 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180 and 360.
But more fundamentally, why are the triangles equilateral?
You are saying, if you cut a pie into 6 equal pieces, each piece is one sixth of the pie. The absolute angle doesn’t matter, a circle could be made out of 720 degrees, for example - I presume historically 360 was agreed on since it was conveniently 3x4x5x6.
Correct, and an isosceles triangle where one angle is one-sixth of a full circle (whether you denote that 60°, or pi/3, or whatever) is already equilateral. I don’t know if you could prove that – it might be an axiom of Euclidean geometry.
The units of measurement are indeed arbitrary - we could be using radians and the math still works.
Yes - dividing a pie into six results in six time one-sixth of a pie.
The cuts meet the rim of the pie dish at equal intervals along its perimeter.
Joining those points together with straight lines yields a hexagon.
Now, why are the other angles 60 degrees?
OK, if you walk along the perimeter of a hexagon in a clockwise direction, you have to make six turns to the right.
You end up facing the same way you did at the start, so you have turned 360 degrees - six turns of 60 degrees each.
Those aren’t the angles we’re looking for, but their straight-line complement is - because that’s the angle on the inside of each vertex of the polygon - 180 degrees(a straight line) minus 60 degrees(the amount you have to turn at the corner).
So the internal angles of the hexagon are 120 degrees. The line from the centre point must divide them exactly in half if everything is regular, so the other two angles of the slices have to be 60 degrees.
NB - that’s not the only way you can slice up a hexagon into pieces - you could join lines from the centre point to the centre of each side, dividing it into 6 kites - each of which has internal angles of 60, 90, 120, 90 degrees
And as to why circles nest together in the same pattern…
Put two identical coins together - the distance between their centres is twice their individual radius - no matter which bit of the perimeter you touch against which (it’s all the same by virtue of a circle’s perimeter being equidistant from its centre. A circle can only touch another circle in such way that the distance between their two centres is twice their individual radius.
Add a third coin so it touches the other two - it can only touch the other coins in the same way as above, so the distance between the centre of any coin and any other coin must be the same - and there are three of them - so that’s an equilateral triangle (which we know can be fitted together to make a hexagon)
Of course you can prove it using Euclidean geometry, but you’ll end up using an axiom that’s just as arbitrary. The infamous Fifth Postulate is needed for the proof. Many throughout history have felt that the Fifth Postulate ought to be a theorem, and tried to prove it, but they always found that the proof required some other replacement postulate. One option for a replacement postulate is that the sum of the angles of a triangle is exactly equal to a straight angle: In other words, given that you already have the other axioms, that statement is equivalent to the traditional statement of the Parallel Postulate.
Look at it this way: as Mangetout showed, the interior angle of the triangle must be 60 degrees because there are six triangles. Also, in every triangle, the sum of the three angles is 180 degrees (this is quite easy to prove). So, if a triangle has one angle that is 60 degrees and the other two are the same, they must both also be 60 degrees for the three of them to add up to 180.
Take a convex polygon (a regular hexagon is, of course, convex). Pick a point in its interior (any point will do). Draw segments from the point to each vertex of the polygon. You will now have divided the polygon into exactly as many triangles as there are sides of the polygon. For a polygon of n sides, that is n triangles.
The sum of the interior angles of a triangle is 180º (this is true by application of the 5th Postulate of Euclid, as Chronos noted). Thus, you now have in our n triangles a total of *n**180º worth of interior angles. Those interior angles do one of two things: they either have a vertex at the point we picked, or they have vertices at the vertices of the n-gon we are using. The vertices at the interior point sum to 360º, because they make a full turn around that point. That’s two triangles worth of degrees. The remaining vertices all sum up to the total of the interior angles of the n-gon. Thus, the sum of the angles of a polygon with n sides is (n - 2) * 180º.
Applying this to our regular hexagon, we see that the sum of the interior angles is 720º. Since it’s a regular hexagon, each interior angle has equal measure, so each interior angle is 120º. Each of those angles is split equally by the radii of the hexagon (the segments we drew to the vertices from the center), so each of the base angles of the triangle is 60º. Thus, the triangles are equiangular (60º each angle), and by application of the corollary to the Base Angle Theorem, equilateral.
You can, of course, work the problem backwards. Take two equilateral triangles, which share a side. Pick one of the shared vertices, and call that the “center” of our soon-to-be hexagon. Pick the other shared vertex, and establish that the two interior angles that vertex is the vertex of together form one of the interior angles of our hexagon. Since that combined angle is 120º, and since we are putting nothing but more equilateral triangles together, all the vertices will be 120º. Proving that this will result in a complete regular hexagon is relatively trivial.
Speaking of six triangles forming a circle, sort of - actually, they don’t since the bases are straight lines and not curved. So, if you drew a circle that touched the points of the outer wall you’d have six dome shaped shapes, one at the base of each triangle. Does the area of that shape have a direct relationship to the area of the circle? I’m assuming it must. If you know the area of the circle you should be able to determine the area of the domes, right? Regardless of the number of slices/triangles, assuming you know that number. I’m also guessing that there is a more direct way of determining the area of the dome. Does that shape have a name?
If you’re asking, “Why is there a regular polygon that forms equilateral triangles when the center is joined to the vertexes? and why is it a six-sided polygon in particular?” then I think there’s no interesting, deep, answer. Only “because that’s just the way it is in flat 2-D space.”
Kind of like asking “Why is the ratio of a circumference to a diameter 3.14etc.?” That’s just the way (flat, 2-D) geometry is.
Dome? If you are talking about the circular segments that remain when you remove the hexagon/polygon, the area of the circle is πr[sup]2[/sup] and the area of the polygon is nr[sup]2[/sup]/2 sin(2π/n), so, subtracting, what remains is nr[sup]2[/sup]/2(2π/n - sin(2π/n)) .
Or, if you want the area of a single circular segment, it is ½r² (θ - sin θ), where θ is the central angle, e.g., 2π/n. Either way, by scaling considerations its area varies in direct proportion to the area of the circle.
Ahaa. Thanks everyone. Got it!
So it’s simply because a triangle has 3 sides. If you evenly divide a straight line by 3, then a triangle formed by one angle will be equilateral because it has 3 sides.
And a hexagon is just 3 triangles back-to-back.
So, a polygon can only be made out of equilateral triangles where the number of polygon sides is twice the number of sides of a triangle.
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This is false even for regular polygons, as exemplified by a single equilateral triangle (by itself, or divided into four smaller equilateral triangles, what have you).
But I think you understand why an isosceles triangle is equilateral exactly when its angles are 60 degrees.