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**jacks****Member**- Registered: 2012-11-21
- Posts: 113

Find the number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly 2 sides common with the polygon

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

number the sides: 0 to 9: (0 1 2 3 4 5 6 7 8 9).

Okay... now you can do the adjacent sides first with the point going out to meet across somewhere, and then add that set to the ones where the two sides are not adjacent.

Now letter the vertices and number the sides as follows:

0Z1A2B3C4D5E6F7G8H9Y

Okay, now we can talk of the first set where two adjacent polygon sides are used like this list of them:

90A

90B

90C

90D

90E

90F

90G

then rotate thse around to get more of them...

for example later you rotate to

45F

45G

45H

45Y

45Z

45A

45B

This is the first set, partially shown, but only 2 tenths of the set is shown above.

So seven times ten is 70, so the first adjacent sided set has 70 four-sided figures.

See next post for the other set...

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Now i will try to list the two sides that are not adjacent...

03

04

05

06

07

14

15

16

17

18

25

26

27

28

29

36

37

38

39

XX30 is already listed above (alphabetical numerical order neglect repeats)

47

48

49

XX neglect repeats

58

59

69

Okay just count 'em up and add to 70 for answer...

Zave a byeutizfull daze.

**igloo** **myrtilles** **fourmis**

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