Thirteen Dodecagonal Rings of Dodecagons

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Thirteen Dodecagonal Rings of Dodecagons

Pentagon of Decagons #3

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Pentagon of Decagons #3

Pentagon of Decagons #2

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Pentagon of Decagons #2

Pentagon of Decagons #1

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Pentagon of Decagons #1

A Polygon with “n” Sides — How Many Diagonals Does It Have?

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Please do not thing of the figure above as a decagon, but as a polygon with n sides — an n-gon, in other words. How many diagonals does it have? Well, first, there are n vertices for diagonals to come from, and three vertices they cannot go to — themselves, and the two immediately on either side of them, since the segments to those vertices are sides, not diagonals. That’s n vertices firing diagonals at n-3 other vertices, or (n)(n-3). However, that counts each diagonal exactly twice (once from each side), so the actual number of diagonals is half that: d=(n)(n-3)/2.

Now we can look at the polygon above, and use it to check this formula, by “remembering” that it is a decagon. With n = 10, d = (n)(n-3)/2 = (10)(7)/2 = 35.

Are there really 35 diagonals in the decagon above? Well, I made those of the same length into color-groups, to make them easier to count. There are five green ones, ten yellow ones, ten red ones, and ten pink ones, which does indeed total 35.

Suppose you know a polygon has, say, 104 diagonals. Can this formula be used to find the number of sides? Yes! Substituting 104 for d leads to this:  104 = (n)(n-3)/2, which then becomes 208 = (n)(n-3) = n² – 3n. To set this up for the quadratic formula, I’m rearranging it to n² – 3n – 208 = 0. The quadratic formula then states that n = (3 ± sqrt(9 – (4)(1)(-208)))/2 = (3 ± sqrt(9 + 832))/2 = (3 ± sqrt(841))/2 = (3 ± 29)/2 = (32 or -26)/2 = 16 or -13, and only one of these answers, 16, can be the number of sides of a polygon. Voila!

Variations On Octagonal Prisms

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Variations On Octagonal Prisms

UnnrealamedUnnreadsfaslamedUnnam888edUnfgdnreadsfaslamedrealCo888nvexyhgtehullrealCo888nvehgdgcxyhgtehullrealCo88jfhg8nvehgdgcxyhgtehullConvex hul88888lConvex h8888ullConhgdfvex hul88888lCon888vex hullCongvex hull

I made these with Stella 4d, a program you can find at http://www.software3d.com/Stella.php.

Variations On Decagonal Prisms

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Variations On Decagonal Prisms

Faceted Dualyhbuzzdgtof Convex hullFaceted Dualyhdgtof Convex hullStellahgdjjfjted Conhghjhgfgghvgdfyex hullStellahgdted Conhgghvgdfyex hullStellahgdted Conhghjhgfgghvgdfyex hullStellahgdted Conhjhgfgghvgdfyex hullStellahgdted Convgdfyex hullConvex hullConvex hufsdfjllStellated Convex hullStellated Convgdfyex hullConvex hull

I made these with Stella 4d, a program you can find at http://www.software3d.com/Stella.php.

A Faceted Version of a Truncation of the Icosahedron

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A Faceted Version of a Truncation of the Icosahedron

I made this with Stella 4d, a program you can find at http://www.software3d.com/Stella.php.

Seven Variants of the Truncated Dodecahedron

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Several Variants of the Truncated Dodecahedron

The smaller pics may be enlarged with a click.

UnbvxcgnamedDual of Convedgex hullConvex hullUnnamed DualDual of Conhdgsavex hullanother compoundStefgasdllated Dual of Convex hull

I used Stella 4d to assemble these polyhedra. You may try this program for free at http://www.software3d.com/Stella.php.

Given Equations for Two Circles, How Does One Find the Coordinates of Their Point(s) of Intersection, If Any?

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Given Equations for Two Circles, How Does One Find the Coordinates of Their Point(s) of Intersection, If Any?

Before attempting to create a method for solving this problem with any given circles, I’m going to attempt solving it for this arbitrary pair of circles. The small circle has radius 5, is centered at (-8,3), and has the equation (x+8)² + (y-3)² = 25. The larger circle has a radius of √85, is centered at (4,-3), and has the equation (x-4)² + (y+3)² = 85.

The first thing I will do is add 60 to each side of the equation of the smaller circle, changing that equation to (x+8)² + (y-3)² + 60 = 85. There are now two equations with 85 on one side, so I will set their other sides equal to each other:  (x+8)² + (y-3)² + 60 = (x-4)² + (y+3)². Next, I square each term in parentheses, turning this equation into x² + 16x +64 + y² -6y +9 + 60 = x² – 8x +16 + y² + 6y +9. After x², y², and 9 are each cancelled from each side, the equation looks like this:  16x +64 – 6y +60 = -8x + 16 + 6y. This simplifies further to -12y + 124= -24x +16, then -12y = -24x -108, then, after dividing both sides -12, that yields the simplified equation for the straight line y = 2x + 9. Any solutions which exist, therefore, must be found on the line y = 2x + 9.

(x+8)² + (y-3)² = 25 is the equation of the smaller circle, and substituting 2x + 9 for y in this equation turns it into (x+8)² + (2x + 9 – 3)² = 25, which is equivalent to (x+8)² + (2x + 6)² = 25. By squaring the terms in parentheses found in this last equation, this turns into x² + 16x +64 + 4x² + 24x + 36 = 25, which simplifies further to 5x² +40x + 75 = 0, and then x² +8x + 15 = 0.

Applying the quadratic formula (because I don’t enjoy factoring trimonials) to this last equation, x = {-8 ± sqrt[64 – (4)(1)15)]}/2, which means that x has two values in this case:  x = (-8 + 2)/2 = -6/2 = -3, and x = (-8 – 2)/2 = -10/2 = -5. When x = -3, then y = 2(-3) +9 = -6 + 9 = 3, so one solution is the ordered pair (-3,3). By constrast, when x = -5, then y = 2(-5) +9 = -10 + 9 = -1, so the other solution is the ordered pair (-5,-1).

For other circle-pairs, the method is the same, but appropriate substitutions will need to be made in the numbers with which I started, likely leading, of course, to different answers. For some pairs of circles, the quadratic formula will not yield two real ordered pairs as solutions. If all values of x and y are nonreal, this simply means that the two circles do not touch at all. If, on the other hand, only ordered pair can be found as a solution, this means the circles are tangent to each other — touching only at a single point.