Geometry

I'm not a fan of geometry so I haven't done this,but some might like this weeks poser from my newspaper:-
The Queen drew a regular polygon and saw that it's perimeter was equal to it's area.The King drew the largest circle he could inside the polygon. what was it's radius?

@venda

Hi Venda

I shall rewrite my deleted answer with a little more information.

I think the 'regular polygon' has to be a square. The area of a regular polygon is found using the function tan(180/n), where n is number of sides. This will be a rational number not equal to zero only if n = 4, so get tangent of 45 (1). If the area is an irrational number, then there is no rational answer for each side.

The square that has area equal to perimeter has side 4 (both equal 16)

So the inscribed circle will have radius 2.

@blood-on-the-tracks said
@venda

Hi Venda

I shall rewrite my deleted answer with a little more information.

I think the 'regular polygon' has to be a square. The area of a regular polygon is found using the function tan(180/n), where n is number of sides. This will be a rational number not equal to zero only if n = 4, so get tangent of 45 (1). If the area is an irrational number, ...[text shortened]... as area equal to perimeter has side 4 (both equal 16)

So the inscribed circle will have radius 2.

Correct Mr tracks
However the answer does say that whatever polygon you think of,the circle inside will always have a radius of 2 inferring the polygon is nor necessarily a square

@venda said
Correct Mr tracks
However the answer does say that whatever polygon you think of,the circle inside will always have a radius of 2 inferring the polygon is nor necessarily a square

It works for an equilateral triangle as well:

Let the side length be S

P = 3S
A = 1/2* ( 3/16 ) ^( ½ ) * S^2

P = A → S * ( ( 3/16 ) ^( ½ ) * S - 3 ) = 0

S= 0, 4*( 3 )^( ½ )

let the radius of the inscribed circle be "r"

Then;

r / ( S / 2) = (1/2 ) / ( ( 3/4)^( ½ )

r =2

@joe-shmo said
It works for an equilateral triangle as well:

Let the side length be S

P = 3S
A = 1/2* ( 3/16 ) ^( ½ ) * S^2

P = A → S * ( ( 3/16 ) ^( ½ ) * S - 3 ) = 0

S= 0, 4*( 3 )^( ½ )

let the radius of the inscribed circle be "r"

Then;

r / ( S / 2) = (1/2 ) / ( ( 3/4)^( ½ )

r =2

Actually...the general proof is pretty simple:

For any regular n-gon with side length S

P = n*S

The Area is can be divided into n triangles of base S and a vertex terminating at the centroid :

A = n * 1/2* S* h

The height of each triangle is the perpendicular bisector of the side length S, as such it is exactly the radius of the inscribed circle "r"

A = n * 1/2 * S * r

And we have constraints that:

P = A

Thus,

n*S = n * 1/2 * S * r

r = 2 ( the radius of the inscribed circle )

for all regular n-gons where their perimeters is equal to the areas.

Nice problem!

@joe-shmo said
Actually...the general proof is pretty simple:

For any regular n-gon with side length S

P = n*S

The Area is can be divided into n triangles of base S and a vertex terminating at the centroid :

A = n * 1/2* S* h

The height of each triangle is the perpendicular bisector of the side length S, as such it is exactly the radius of the inscribed circle "r"

A = n * ...[text shortened]... ed circle )

for all regular n-gons where their perimeters is equal to the areas.

Nice problem!

Pleased you both got something out of it.
I do look at the puzzles every week, but often they're just rubbish