# Thread: Polygons at a Point

1. a scalene and isosceles triangle are also "regular" - don't tell me they aren't cause last yr in maths comp, a scalene was considered "regular" and i put it down as irregular and i got it wrong.

2. Originally Posted by confused
a scalene and isosceles triangle are also "regular" - don't tell me they aren't cause last yr in maths comp, a scalene was considered "regular" and i put it down as irregular and i got it wrong.
It is not only me telling you that a regular 3-gon is an equilateral triangle,
see MathWorld.

Also, your pentagon is not regular.

RonL

3. then i can only conclude that the board of studies are bull****

4. Originally Posted by confused
then i can only conclude that the board of studies are bull****
It is not unknown.

RonL

5. i can't even think why skool is "compulsory".....so sad

6. Hello, Chuck_3000!

Certain sets of regular polygons fill space around a point without gaps or overlapping.
e.g. imagine 3 hexagons joined together and the common vertice is the point they surround.
That is classified a [6,6,6] point fill set (3 six sided shapes)

Although the polygons can be arranged in a different order, we count these as the same.

We call a set of regular polygons filling space around a point a point-fill set.

a) Find a point-fill set of 3 polygons containing a 24-gon
b) Explain why it is not possible to have a point fill set containing a triangle and a pentagon
c) Find all point-fill sets that contain at least one square.

The interior angle of a regular $n$-gon is: . $\frac{n-2}{n}\cdot180^o$

Armed with this formula, you can make a list:

$\text{Triangle: }60^o\quad\text{Square: }90^o\quad\text{Pentagon: }108^o\quad\text{Hexagon: }120^o$

$\text{Octagon: }135^o\quad\text{Decagon: }144^o\quad\text{12-gon: }150^o\quad \text{15-gon: }156^o$

$\text{18-gon: }160^o\quad\text{20-gon: }162^o\quad\text{24-gon: }165^o$

$\text{30-gon: }168^o\quad\text{36-gon: }170^o$

(a) A 24-gon takes up $165^o$ of the circle about the point,
. . leaving $360^o - 165^o \:=\:195^o$ to be filled.
This can be accomplished with a triangle $(60^o)$ and an octagon $(135^o)$.

(c) Of course, four squares comprise a point-fill set.

With two squares, there are $180^o$ to be filled.
This can be accomplished with:
. . three triangles: $3 \times 60^o$
. . a hexagon and a triangle: $120^o + 60^o$

With one square, there are $270^o$ to be filled.
This can be accomplished with:
. . two triangles and a 12-gon: $60^o + 60^o + 150^o$
. . a hexagon and a 12-gon: $120^o + 150^o$
. . two octagons: $135^o + 135^o$
. . a pentagon and a 20-gon: $108^o + 162^o$

I hope I didn't miss any . . .

7. The question asked in an Australian take-home competition. In case you wanted just to check, it is the:
2006 Maths Challenge Stage
Mathematics Challenge for Young Australians
TERM 2

JUNIOR STUDENT PROBLEMS

An Activity of the Australian Mathematical Olympiad Committee
A Subcommittee of the Australian Mathematics Trust in association with the Australian Academy of Science and the University of Canberra

...Just google Australian Mathematics Trust Maths Challenge or something like that

This person happened to be trying to complete one of the hardest questions in the book - well, it's hard until you figure out the key point to it, then it's easy. I'm just not satisfied this thread is still up there, and she/he's been given the answers. Unfortunately/Fortunately, I don't think she/he'll be able to write a 1 to 2 page explanation on Egyptian algebra and other techniques =) Especially not since they're probably about 13. There's another easier way...

This competition is due on Wednesday ((GMT+10:00 Sydney))...so it won't make much of a difference if it isn't deleted or not, just as long as no-one else posts answers anymore.

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