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# If convex polygon C has 7 sides, then the number of distinct

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If convex polygon C has 7 sides, then the number of distinct [#permalink]

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01 Sep 2006, 13:27
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If convex polygon C has 7 sides, then the number of distinct quadrilaterals than can be formed by drawing one or two straight lines between non-adjacent vertices of C is...
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03 Sep 2006, 08:14
Giving it a try...

n = 7

Now we can form a polygon in 4 ways
1) 3 sides of the polygon are common to the quadrilateral.
we can choose x consecutive sides from an n sided polygon in n ways.
so we can choose 3 consecutive sides from an 7 sided polygon in 7 ways.

Now given any three sides of a polygon there is only 1 way to form a quadrilateral ( by joining th two points not connected)
so total # of quad = 7*1

2) 2 consecutive sides of a polygon are taken as sides of a quadrilateral.

We can select 2 consecutive side of a polygon in 7 ways.

Given two consecutive sides we can have two possible quadrilaterals
As total # of vertex = 7;
3 are linked by the 2 consecutive sides=> remaining = 7-3; 4 other vertex;
2 of these 4 are already connected by the sides of the polygon; => 4-2 = 2 vertex can be connected by diagonals
so total # of quad = = 7*2 = 14

3) 1 side of the polygon is shared by the quad; # of ways in which a side can be selected = 7
Total vertex = 7
vetex on the common side = 2; so remaining = 7-2 =5
2 other vertex are directly linked by the previous 2 to form the polygon; so remaining = 5-2 = 3
we can choose 2 points from these 3 in 3C2 ways = 6
so total # of quad = 7*6 = 42/2 = 21

4)the polygon and quad do not have any side common;
we can choose vertex 1 in 7 ways;
not of the remaing 6 two are connected by sides of polygon;
so remaing vertex = 4
now among these 4 we CANNOT choose 3 non consecutive vertex; so this cmbination does not seem possbile

so total # of quad = 21+14+7 = 42

anyone else? ans?
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06 Sep 2006, 13:36
Interestingly used the same method:-

Quad has 4 sides. Can share 1, 2, or 3 sides with the polygon.

Sharing one side:
7 (sides) x 3C2 = 7 x 3 = 21

Sharing 2 sides:
7 (set of 2 adjacent sides) x 2C1 = 7 x 2 = 14

Sharing 3 sides:
7 (set of 3 adjacent sides) x 1 = 7

Kev?
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13 Sep 2006, 05:59
so kavincan, you are saying it's 42?

it does not make sense to me because if quadrilateral is to share only one side with the polygon than we'd need to draw 3 lines. But we can only draw 1 or 2 lines?

I must be missing something.....
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13 Sep 2006, 08:01
EconGirl wrote:
so kavincan, you are saying it's 42?

it does not make sense to me because if quadrilateral is to share only one side with the polygon than we'd need to draw 3 lines. But we can only draw 1 or 2 lines?

I must be missing something.....

I'm not saying it's 42, but the work so far is on the right track. When they say one side, they mean two non-adjacent sides. But there is a small mistake in their reasoning. Can you find it? Alternatively, an elegant solution can be found using the first reply as a base.
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13 Sep 2006, 08:12
EconGirl wrote:
so kavincan, you are saying it's 42?

it does not make sense to me because if quadrilateral is to share only one side with the polygon than we'd need to draw 3 lines. But we can only draw 1 or 2 lines?

I must be missing something.....

The work so far is on the right track, but the answer is not 42. When they say one side, they should be thinking about two non-adjacent sides. But there is a mistake in their reasoning. Can you find and fix it? Also, can you use the first reply to come up with an elegant solution?
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14 Sep 2006, 04:53
Ah, at last I see what you mean though I may not be able to solve it. but let me try:

Sharing 2 sides:
7 (set of 2 adjacent sides) x 2C1 = 7 x 2 = 14

Sharing 3 sides:
7 (set of 3 adjacent sides) x 1 = 7

Total: 14+7+7 = 28?
14 Sep 2006, 04:53
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