[Guest]

Using N toothpicks of equal size placed end to end a regular polygon is created. The polygon is then changed by moving exactly two of the toothpicks. The shape that results is still a polygon but it has 31/3% less interior diagonals.

What is the value of N?

[Guest]

Make that 3 1/3% less interior. Sorry!

[robo]

I think...

N=15

[dark_magician_92]

though i am just kinda guessing, i think

180

[BobbyGo]

N = 22

5x6 rectangle = 30 units squared

Invert one corner = 29 units squared

29/30 = 96.6...% = 96 2/3 %

100% - 96 2/3 % = 3 1/3%

Edited November 4, 2011 by BobbyGo

[Guest]

N = 15

3 1/3% is equivalent to 1/30; therefore, for each diagonal removed, there must have been 30 others. (i.e. #diagOriginally = 30*#diagRemoved)

Since the resulting shape must remain a polygon, the two toothpicks moved must share a vertex. Since the equation for number of diagonals in a regular polygon with n sides is n*(n-2)/2, we can find the possible regular polygons.

z=number diagonals to be removed

n=number of edges in regular polygon

n(n-3)/2=number of original diagonals in regular n-edged polygon

30z = n(n-3)/2

60z = n^2 - 3n

For z=1, no integer solution

For z=2, no integer solution

For z=3, n=15

For z=4, no integer solution

I could go on, but we need to define when an interior diagonal is eliminated. To me, this would be the case when the vertex goes from convex to concave on the regular polygon thus causing one diagonal to fall outside the polygon. Also, if two diagonals fall on the same line, then only one would count thereby removing a diagonal.

After playing around with angles and looking at the shapes created, it became clear to me that the minimum polygon to test would have to be a pentagon (5-edged regular polygon) as a square would simply collapse and render itself no longer a polygon when a vertex is indented. Let's standardize all the possible regular polygons by choosing any vertex to indent and labeling it as vertex 3, vertices 4 and 5 follow clockwise while vertices 2 and 1 are labeled in that order anti-clockwise from vertex 3.

When vertex 3 is indented, edges 2,3 and 3,4 are rotated about vertices 2 and 4 respectively. These edges are rotated by an amount equal to twice the exterior angle (supplementary to the interior angle) of the regular polygon. Edge 2,3 rotates clockwise and 3,4 anti-clockwise. This action causes vertex 3 to bisect the diagonals 1,4 and 2,5. Now, since the diagonal 1,4 is equivalent to the diagonal 1,3 plus the edge 3,4, only one of those diagonals counts (-1 diagonal). Also, the same is true about diagonal 2,5 being equivalent to diagonal 3,5 plus edge 2,3; so it can be removed as well (-1 diagonal). Finally, diagonal 2,4 lies outside the irregular polygon which precludes it from being counted as an interior diagonal (-1 diagonal). We have thus removed a total of three interior diagonals from the polygon by moving two edges to form an irregular polygon. Since the edges cannot change length, we are limited to only those three diagonals being removed from the polygon as any additional rotation of the edges would leave a gap making it ineligible for being classified a polygon. This is true for any regular polygon with at least 5 edges.

Therefore, with the minimum and maximum numbers of interior diagonals able to be removed being 3, there must have been thirty times three (30*3) originally to meet the conditions of the OP (that would be 90 interior diagonals). The only regular polygon that meets the OP conditions is a pentadecagon (15-edged regular polygon).

[Guest]

dirty dozen

12

[voider]

My interpretation of "interior diagonals": It has n sides where n is even.

Under this interpretation, solve n / 2 = (1 + 1/30) * (n / 2 - 1)

yields n = 62

I know I know my paint skills are pretty good

Edited November 6, 2011 by voider

[Guest]

Can't get it :

By moving 2 sticks of a regular polygon you can make only one vertex cave in. That means that only ONE INTERNAL DIAGONAL will be reduced. So we would have started with an n sided regular polygon where the number of diagonals were 30 to start with .

Count of diagonals for n sided regular polygon:

n sided polygon has n vertices.

Each vertex can make a diagonal with another vertex except for itself and the two adjoining vertices. So each vertex can make n-3 diagonals.

n points can make = n(n-3) diagonals

But in the above figure a diagonal between two vertices have been counted twice.

So the distinct diagonals for n sided polygon is n(n-3)/2

getting back to solving for n :

n(n-3) / 2 = 30

n² - 3n -60 = 0

but the above equation has no integer roots

Edited November 7, 2011 by nak