#regular polygons

[noelements-setempty]

Since everyone likes regular shapes so much:

Favourite regular polygon

Equilateral triangle (3)

Square (4)

Pentagon (5)

Hexagon (6)

Heptagon (7)

Octagon (8)

Nonagon (9)

Decagon (10)

Dodecagon (12)

Heptadecagon (17)

257-gon

Something else not listed

Voting ended onApr 8, 2024

[paulspuzzles-blog]

Solution #9

Click “Keep reading” to see the solution!

So we need to know the perpendicular height of this triangle to find out its area (yes, yes, we could use ½*a*b*sin(C) = ½a²sin(60°).  I’ll be using that at the end, funnily enough.  But humour me a minute).  Well, let’s drop a perpendicular and see what we get:

We’ve now split this equilateral triangle into two right-angled triangles.  These right-angled triangles have a hypotenuse of a, and one of their short sides is a/2 (half the length of a side of the triangle).  With a bit of Pythagorean magic, we get:

x² + (a/2)² = a² x² = a² - a²/4 x² = 3a²/4 x² = (a√3)/2

So the perpendicular height of an equilateral triangle side length a is a√3/2.  This is a very important relationship in trigonometry, as it lets us get exact values for sin(30°) = cos(60°) = ½, sin(60°) = cos(30°) = √3/2, tan(30°) = 1/√3, and tan(60°) = √3.

Anyway, area of a triangle is ½*base*perpendicular height, so:

Area = ½*a*a√3/2 = a²√3/4.

Nice.  And yes, this checks out with the ½a²sin(60°) formula from earlier, since sin(60°)  = √3/2.

Moving on... well, a regular hexagon is, after all, just six equilateral triangles grouped together:

So its area is simply 6*a²√3/4 = 3a²√3/2.

The octagon goes in a slightly different direction.  We can split it up into nine parts, as follows:

By Pythagoras again, we can see that the lengths of the short sides of the isosceles right-angled triangles (let’s just call them iRATs) in the corners is a/√2.  This gives us:

1 square, area a²

4 iRATs, each with area ½(a/√2)² = a²/4

4 rectangles, each with area a*a/√2 = a²/√2, or a²√2/2 because it’s easier to multiply top and bottom through by √2 and get your surd on the top.

Let’s put that all together:

a² + 4a²/4 + 4a²√2/2 = a² + a² + 2a²√2 = 2a² + 2a²√2 = 2a²(1 + √2)

There we go.  You might even have noticed a shortcut here: those four iRATs can be combined together geometrically to make another square side length a, area a².  We showed this this algebraically here, but one of the great things about maths is how many ways there are to skin a cat.

Now then, on to that dodecagon.  As I said at the beginning, I’ll be using the area = ½*a*b*sin(C) formula here.  We can split this thing up into twelve triangles meeting at a centre point.  Since there are 360° in a full circle, dividing that by twelve gives us an angle of 30° for the pointy bit of this triangle:

So the area for one of these triangles is ½*r²*sin(30°), and we showed earlier that sin(30°) is ½.  So the area of a triangle is ¼r².  And since we have twelve triangles, the total area of this dodecagon is 3r².

That’s a rather pleasing result, I think. If nothing else, it sets a nice lower bound for π of 3 (granted, you can do a lot better than that).

I probably should have expected this of course, but Wikipedia has a lovely table of regular areas for polygons in terms of different properties of them here: https://en.wikipedia.org/wiki/Regular_polygon#Circumradius .  But it’s still fun to work this stuff out for yourself.

[paulspuzzles-blog]

Puzzle #9

Oh wow, sorry how in depth those last couple of puzzles got.  Let’s scale things back with a few small puzzles about some nice, friendly regular polygons.

Find a formula for the area of an equilateral triangle in terms of the side length, a:

Hence find a formula for the area of a regular hexagon in terms of the side length, a:

And now a formula for the area of a regular octagon in terms of its side length, a:

Finally, and slightly different, find the area of a regular dodecagon in terms of its outer radius, r - that is, the distance from the centre to one of the vertices:

Now, admittedly there’s probably a general formula out there for finding the area of any regular polygon in terms of these things.  And you know, feel free to do that.  Might spoil the puzzle aspect, though.  Mostly I picked these because I like the specific methods I used for their solutions.

Solution tomorrow :).

[mathhombre]

Polygon Rings

@mathrecpics had 2 posts recently that really made me curious. One was a ring of 7 heptadecagons and the other was a ring of 14 heptagons. In this sketch you can choose the number of sides for the regular polygon and how many sides of the polygon the copies are offset. There is also a tool for making more polygon rings: two vertices, number of sides, and the offset. You can use the sliders for the two numbers as well.

Edit: oops! Forgot the link to GeoGebraTube. Thanks, @Farrarscott!

[mathhombre]

Regular Fit                                                        

Sam Shah posted this intriguing picture of a 42-gon, heptagon and an equilateral triangle fitting together at a point, out of which he was making a geometry lesson. That made me wonder about how to make a sketch that let you find out if, given two regular polygons, there was a third to fit. Turns out, this is a pretty interesting application for rational functions, too.

On GeoGebraTube. Thanks, Sam!

What would you guess is the least sides that has no 3 polygon fit?

[spring-of-mathematics]

Platonic Soild

Proof That There Are Only Five Regular Solids:

For a regular solid, denote:

the number of vertices as V.

the number of edges as E,

the number of faces as F,

the number of edges on a face as s, and

the number of faces at each vertex as c.

We have established that V = 2E/c and   F = 2E/s. From Euler’s Formula we have 0 < 2  = V-E+F = 2E/c-E+2E/s  = E(2/c-1+2/s) = E(2s/cs-cs/cs+2c/cs)  = E(2s-cs+2c)/cs. So the numerator satisfies 2s-cs+2c > 0, or cs-2s-2c < 0, or cs-2s-2c+4 < 4, or (c-2)(s-2) < 4.  Since c and s are both greater than or equal to 3, then (c-2)(s-2) is a positive integer less than 4. Namely, (c-2)(s-2) is in {1,2,3}.  This gives five cases:

(s-2)(c-2) = 1 and s = 3, c = 3.  Then we have the tetrahedron.

(s-2)(c-2) = 2 and s = 4, c = 3. Then we have the cube.

(s-2)(c-2) = 2 and s = 3, c = 4.  Then we have the octahedron.

(s-2)(c-2) = 3 and s = 5, c = 3.  Then we have the dodecahedron.

(s-2)(c-2) = 3 and s = 3, c = 5.  Then we have the icosahedron.

In three-dimensional space, a Platonic solid is a regular, convex polyhedron. It is constructed by congruent regular polygonal faces with the same number of faces meeting at each vertex. Five solids meet those criteria, and each is named after its number of faces

The tetrahedron, with 4 triangular faces:

The cube, with 6 square faces:

The octahedron, with 8 triangular faces:

The dodecahedron, with 12 pentagonal faces.

The icosahedron, with 20 triangular faces:

[assignment-expert]

Regular polygons can be formed by applying to each other on the sides of an isosceles triangle. Those isosceles triangles that make up the regular polygon have an apex angle φ_n, equal to 360 / n and a common vertex O. They are obtained from one another by a rotation around the point (in either direction) by the angle φ_n, and fold it at any angle. Therefore, a regular polygon has rotational symmetry with a rotation angle equal φ_n. The center of a regular polygon the center of rotational symmetry of the polygon.

www.assignmentexpert.com

[mathhombre]

Experimenting with GeoGebra animation using one slider to control two others. Using the dividend command Div[ , ] and remainder command Mod[ , ].

But it also makes pretty patterns! On GeoGebraTube; try it yourself.