BdMO National 2013: Primary 6

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

BdMO National 2013: Primary 6

Unread post by BdMO » Fri Jan 10, 2014 1:18 am

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.
Attachments
Primary 6.JPG
Primary 6.JPG (12.93 KiB) Viewed 2449 times

User avatar
Raiyan Jamil
Posts: 138
Joined: Fri Mar 29, 2013 3:49 pm

Re: BdMO National 2013: Primary 6

Unread post by Raiyan Jamil » Sat Mar 15, 2014 9:43 pm

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 .
A smile is the best way to get through a tough situation, even if it's a fake smile.

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm
Location: Dhaka,Bangladesh

Re: BdMO National 2013: Primary 6

Unread post by dshasan » Thu Jan 21, 2016 3:00 pm

A regular hexagon also meets the condition.
Last edited by SANZEED on Sun Jan 24, 2016 10:58 pm, edited 1 time in total.
Reason: Rephrasing the statement.

User avatar
SA'AD
Posts: 5
Joined: Wed Jan 16, 2019 10:17 pm

Re: BdMO National 2013: Primary 6

Unread post by SA'AD » Wed Jan 16, 2019 11:08 pm

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..

Post Reply