# Polygons at a Point

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• May 1st 2006, 02:00 AM
Chuck_3000
Polygons at a Point
Hey people, i have another problem i'm having trouble with!

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 polygones 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.

Any help would be fantastic
• May 2nd 2006, 05:12 AM
CaptainBlack
Quote:

Originally Posted by Chuck_3000
Hey people, i have another problem i'm having trouble with!

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 polygones 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.

Any help would be fantastic

The internal angle at any vertex of a regular $n$-gon is:

$
$

If $m$ regular $n$-gons with $n_1,\ n_2,\ \dots\ n_m$ sides constitute a point-fill set then
the sum of the internal angles at their vertices sum to $2\pi$ radians, that is:

$
2\ \pi=\pi(1-2/n_1)+ \dots+\pi(1-2/n_m)
$

or:

$
m-2=2(1/n_1+\dots+1/n_m)\ \ \ \dots(1)
$
.

Now lets look at part (a).

$m=3$, and $n_1=24$, so plugging these into equation $(1)$ and simplifying:

$
\frac{11}{24}=\frac{1}{n_2}+\frac{1}{n_3}\ \ \ \dots(2)
$

then trial and error will allow you to find $n_2$ and $n_3$ to complete the point fill set.

RonL
• May 2nd 2006, 07:04 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack

$
\frac{11}{24}=\frac{1}{n_2}+\frac{1}{n_3}\ \ \ \dots(2)
$

A solution to $(2)$ can also be found by applying the greedy algorithm for
Egyptian fractions, and techniques for Egyptian fractions can be used
to find other solutions if they exist.

RonL
• May 3rd 2006, 12:36 AM
Chuck_3000
thanks a lot

I follow what you've done in a), but one question - where you've said m=3 and n1 = 24, what does the m=3 refer to?

And i have no idea what Egyptian fractions are, or how to use them for b) :eek:
• May 3rd 2006, 03:26 AM
CaptainBlack
Quote:

Originally Posted by Chuck_3000
thanks a lot

I follow what you've done in a), but one question - where you've said m=3 and n1 = 24, what does the m=3 refer to?

And i have no idea what Egyptian fractions are, or how to use them for b) :eek:

The stuff at the beginning is about a general point fill set, with polygons
of $n_1,\ n_2 \dots\ n_m$ sides, so we see here that $m$ is the number of
polygons in the point-fill set. In part (a) we are asked for a point-fill set with
three polygons so we use $m=3$ for this case.

You don't really need to know about Egyptian fractions to solve the problem,
in fact I did it by trial and error. If you do want to know about Egyptian
fractions though there is another thread discussing them here

RonL
• May 8th 2006, 11:28 PM
Chuck_3000
does anyone know how one would go about question c) ?
• May 22nd 2006, 03:58 AM
lozer
need help
hi, am stuck with the same question. does anyone know how to do part b and c. thanks
• May 22nd 2006, 06:24 AM
CaptainBlack
Quote:

Originally Posted by lozer
hi, am stuck with the same question. does anyone know how to do part b and c. thanks

Part b) can be done by looking at the angle left after you put a triangle and
pentagon together at a vertex. Then try to fill this angle with a combination
of other polygons. You will quickly find that there is no combination of
polygons that will perfectly fill this angle.

RonL
• May 27th 2006, 05:49 PM
confused
Part b) can be done by looking at the angle left after you put a triangle and
pentagon together at a vertex. Then try to fill this angle with a combination
of other polygons. You will quickly find that there is no combination of
polygons that will perfectly fill this angle.

This is not true as if u use 3 triangles and a pentagon, u can get a point-fill set of (3,3,3,5).
• May 27th 2006, 07:50 PM
confused
wat i don't get is a)
• May 27th 2006, 09:05 PM
CaptainBlack
Quote:

Originally Posted by confused
Part b) can be done by looking at the angle left after you put a triangle and
pentagon together at a vertex. Then try to fill this angle with a combination
of other polygons. You will quickly find that there is no combination of
polygons that will perfectly fill this angle.

This is not true as if u use 3 triangles and a pentagon, u can get a point-fill set of (3,3,3,5).

The internal angle of a vertex of a 3-gon is $\pi/3$, the corresponding angle
for a 5-gon is $\frac{3}{5}\pi$ so the sum of these angles for three 3-gons and a 5-gon is $\frac{8}{5}\pi$.

But if [3,3,3,5] where a point fill set the angle sum would be $2\pi$

RonL
• May 27th 2006, 09:56 PM
confused
Quote:

Originally Posted by CaptainBlack
The internal angle of a vertex of a 3-gon is $\pi/3$, the corresponding angle
for a 5-gon is $\frac{3}{5}\pi$ so the sum of these angles for three 3-gons and a 5-gon is $\frac{8}{5}\pi$.

But if (3,3,3,5) where a point fill set the angle sum would be $2\pi$

RonL

who cares about that technical stuff?? u try drawing it on paper - a pentagon, and three triangles. it worxs
• May 27th 2006, 10:05 PM
CaptainBlack
Quote:

Originally Posted by confused
who cares about that technical stuff?? u try drawing it on paper - a pentagon, and three triangles. it worxs

For three triangles, since they have to be equilateral, the internal angles a
vertex are $\pi/3$ or 60 degrees, three times sixty is 180 degrees.
so for a regular pentagon to complete a point fill set with three triangles the
internal angle at a vertex would have to be 180 degrees, which it is not. So
[3,3,3,5] is not a point fill set (not even nearly!).

So check what you have done, it may be wrong, then check the
wording of the problem you may have misunderstood it.

RonL
• May 27th 2006, 10:46 PM
confused
Quote:

Originally Posted by CaptainBlack
For three triangles, since they have to be equilateral, the internal angles a
vertex are $\pi/3$ or 60 degrees, three times sixty is 180 degrees.
so for a regular pentagon to complete a point fill set with three triangles the
internal angle at a vertex would have to be 180 degrees, which it is not. So
[3,3,3,5] is not a point fill set (not even nearly!).

So check what you have done, it may be wrong, then check the
wording of the problem you may have misunderstood it.

RonL

plz look at my attachment. i'm not the best of drawers on computer so excuse my graphics but you can get the general idea.
p.s. who said anything about the triangles being equilateral??
• May 27th 2006, 10:51 PM
CaptainBlack
Quote:

Originally Posted by confused
plz look at my attachment. i'm not the best of drawers on computer so excuse my graphics but you can get the general idea.
p.s. who said anything about the triangles being equilateral??

The problem asks for regular polygons. Regular means all sides and angles
are equal, so a regular 3-gon is a equilateral triangle. Also the pentagon
has to be a regular pentagon.

RonL
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