1. For an equilateral triangle also, it is D=0.

2. An equivalent formulation is to say, take for each side a unit vector that is perpendicular to the side and points to the inside of the polygon. D=0 if these vectors all add up to zero.

For triangles, it can only be equilateral. For quadrilaterals, it can only be a parallelogram. I do not have a general result at this time.

3. I think that's a pretty good answer--necessary and sufficient, and gets all the cases we've noted so far, e.g. even number of sides and regular, or more generally, opposite sides are parallel. Also, it shows D=0 for any regular polygon, since the unit normals must add to 0 for the same reason that the nth roots of unity add to 0.

4. I am just realizing this point, why do we use the word "triangle" for three but "quadrilateral" for four? Why not "trilateral" or "quadrangle"? They are both words for other things But it would be consistent to use always "angle" or "lateral," but we don't.

5. In "projective geometry" a quadrangle is four points, no three collinear, and a quadrilateral is four lines, no three concurrent. (no parallel lines in projective geometry, so any four lines, no three concurrent, have four intersection points, actually six). So, it really depends on what you emphasize.

I don't know why in ordinary Euclidean geometry we choose to emphasize the vertices of a triangle but the sides of a quadrilateral.

6. And, it's not pentangle/pentalateral nor hexangle/hexalateral neither!
Originally Posted by tdvance
I think that's a pretty good answer--necessary and sufficient, and gets all the cases we've noted so far, e.g. even number of sides and regular, or more generally, opposite sides are parallel.
I'm not sure it's sufficient. What if three of the normals are parallel? Say, by adding a square to one side of a regular septagon? Of course, that's no longer convex.

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Ok no 1 is posting questions/answers so I'll post 1.

98 to the power of 7 = 8.68 × 1013
(Hope non of you are useing calculators, took me 10 mins to figure that out)

and my question, 934 to the power of 11

8. Originally Posted by connor240287
(Hope non of you are useing calculators,
we all are!

9. In which direction is the bus travelling?

left or right?

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It's travelling right, just a gess cos I'm looking at the wheels and the circles inside are more to the right.

12. Sorry for not responding. Connor's answer is incorrect.

Clue: Imagine the bus in in the US, and not in the UK (or commonwealth countries).

13. well, assuming the "cameraman" is not in a position to get run over, it's going to the right.

14. It's going left, where's the door?

15. correct.

16. What if the door's on the back?

Clue: Imagine the bus in in the US, and not in the UK (or commonwealth countries).
Well, it could be a Canadian bus

18. Ok, here we go.

What is the next number in this sequence?

1 4 6 15 17 19 ___

19. Well, it could be 92, but I don't think this is the answer you are looking for.

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The bus could be backing up to the right :P

Unless it's a finite sequence, we probably need another term. I'd guess 21 but don't have a complete pattern yet.

21. Originally Posted by mr obvious
The bus could be backing up to the right :P

Unless it's a finite sequence, we probably need another term. I'd guess 21 but don't have a complete pattern yet.
Definitely a finite sequence. Just almost certainly not the sequence you expect. There's a maximum of 10 terms.

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Hmm. I'd guess the next term is 26. Problem is, I have no idea how to get that. Then 27, 30, 31.

23. My guess was based on simply polynomial fitting, but I didn't think that was the approach. The finite series confirms it

24. Nope. All terms of the series are <=20.

25. ## One more term in the series...?

Ok.

1, 4, 6, 15, 17, 19, 16, __, __, __.

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1, 4, 6, 15, 17, 19, 16, 14, 5, 3

27. 1, 4, 6, 15, 17, 19, 16, 5, 2, 1
?

28. 1, 4, 6, 15, 17, 19, 16, 11, 9, 5
Originally Posted by yalius
there's a maximum of 10 terms.
Originally Posted by yalius
all terms of the series are <=20.
The board

29. Originally Posted by hhEb09'1
1, 4, 6, 15, 17, 19, 16, 11, 9, 5
The board
Absolutely!!! Got it.

30. Cool!

Can you construct three dice by changing the number of dots on their faces so that, if you were to roll a pair of them, the probability that the first would beat the second, and the probability that the second would beat the third, and the probability that the third would beat the first, are all greater than half?

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