BdMO National 2013: Primary 6

[BdMO]

A polygon is an area bounded by three or more edges in a plane. For example a triangle has three edges, a quadrilateral has four edges, a pentagon has five edges and in this way they are given names according to their number of edges. A regular polygon is a polygon that has edges of equal lengths and all of it's angles are equal as well.

Your father is going to set tiles on your room’s floor. He asked your choice of tiles(polygon). But there are some conditions.

(a) You can choose only one type of regular polygon.

(b) When tiles will be set, there must be no gap among them.

(c) No overlapping is allowed.

(d) You should not be concerned about your room size, rather the above three conditions.

Which type of the regular polygon can you use to draw such a figure that satisfies the above conditions? Write down the logic of not using other regular polygons rather than the one/s you chose. Here is a sample picture for you.

[Raiyan Jamil]

Here only a rectangular regular polygon can be used . Because all the angles of the tile is 90 degree . And a rectangular regular polygon has 4 sides as a square and one angle has ((2n-4)*90) / n or 360 /4 or 90 degree . Here only this type of polygon has sides of 90 degree . Therefore it is the only possible polygon .

[dshasan]

A regular hexagon also meets the condition.

[SA'AD]

The condition will be met, when the corners will perfectly fit with each other.
so, the sum of the corner's angle must be 360.
the angle of a vertex of an 'n-gon' is 180(n-2)/n.

so, 360/{180(n-2)/n} must be a natural number.
let, 360/{180(n-2)/n}=k , where k is a natural number.
This implies, 2n/(n-2)=k
the values of n which satisfies that 2n/(n-2) is a natural number are {3,4,6}

How we understood from 2n/(n-2)=k,(k,n are natural numbers) that n=3,4,6 :

2n/(n-2)=k
Case 1: When n is odd
If in 2n/(n-2), n is odd, n-2 will be the previous odd number.So, n is not divisible by n-2. So in order for 2*n/(n-2) to be a natural number, n-2 must be equal to 1.
So, n-2=1,So n=3.
Case 2:When n is even
Let, n=2m, where m can be both even or odd.
So, 2n/(n-2)=2*2m/(2m-2)
=2m/(m-1)
Now, in 2m/(m-1), m is not divisible by m-1, cuz they are consecutive numbers,
Except, i) when m-1=1, So,m=2 and thus n=4.
ii) when m-1=2, So, m=3 and thus n=6.

Hence, equilateral triangle,square and hexagons will do..