Drawing a six pointed star

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Peter Rooney - 30 Mar 2006 14:05 GMT

Good afternoon, all!

I'm trying to draw a six pointed star. I created an Isosceles triangle, drew
lines from each point to the centre of the opposite side, linked and copied
the whole thing, flipped the copy over and tried to position it using the
intersections of the bisecting lines in each triangle.

However, whilst it looks "nearly" OK, it doesn't look accurate enough for
what I have in mind. Where some of the lines should overlap each other,
they're out my a mm or so.

I held Shift down when I created the first traingle, so it would be
equilateral, but this doesn't seem to have helped.

Can anyone suggest how I could improve on this process?
Also, if you have any ideas for a five pointed star, i'd be be eager to
listen!

Thanks in advance

Pete

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Ed Bennett - 30 Mar 2006 15:49 GMT

> Can anyone suggest how I could improve on this process?

Create an equilateral triangly by using the Shift key on the isosceles
triangle AutoShape.

Then zoom in so it fills your screen.

Use the shift key when drawing your construction lines to constrain them to
15° increments. Ensure you position the crosshair precisely at the point,
and then you will be certain within the limits of accuracy of your screen
that the line passes through the centre. (Unless you're 15° off and you
divide the line into ¼ and ¾.)

Then group, copy, flip, and align. You can then ungroup and delete your
construction lines. Some internal lines would remain visible.

An alternative method would be to draw a hexagon (using the Shift key to
make it regular), and then draw an equilateral triangle side the same length
as the side of the hexagon. Copy and paste to create six, flip three of
them, and place them. Again edge-tracing would be necessasry to remove some
internal lines.

> Also, if you have any ideas for a five pointed star, i'd be be eager
> to listen!

Try the 5-point star in Autoshapes > Stars and Callouts?

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Ed Bennett - MVP Microsoft Publisher

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Peter Rooney - 30 Mar 2006 16:12 GMT

Ed,

Perhaps I should try trusting Publisher rather than my eyes and my screen.
Even the [Shift]ed five pointed star looks squashed when I stand on my head
in the office.
I'll give your other suggestion a go in the morning.
Regards and thanks

pete

> > Can anyone suggest how I could improve on this process?
>
[quoted text clipped - 22 lines]
>
> Try the 5-point star in Autoshapes > Stars and Callouts?

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Mary Sauer - 30 Mar 2006 17:53 GMT

Hey Ed, I used a hexagon, using centimeters, Height 6.60cm Width 7.6cm
Created a triangle, 3.3cm high, 3.8cm wide. Copied the triangle, rotated the
copy 60 degrees, pasted the triangle again, rotated this one 120 degrees. Copied
the two rotated triangles, flipped and placed on the other side.

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Mary Sauer MSFT MVP
http://office.microsoft.com/
http://msauer.mvps.org/
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>> Can anyone suggest how I could improve on this process?
>
[quoted text clipped - 22 lines]
>
> Try the 5-point star in Autoshapes > Stars and Callouts?

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Ed Bennett - 30 Mar 2006 18:33 GMT

> Hey Ed, I used a hexagon, using centimeters, Height 6.60cm Width 7.6cm
> Created a triangle, 3.3cm high, 3.8cm wide. Copied the triangle,
> rotated the copy 60 degrees, pasted the triangle again, rotated this
> one 120 degrees. Copied the two rotated triangles, flipped and placed
> on the other side.

Yes, but using equilateral triangles, rotation is not necessary - rotating
through 60° gives the same triangle as flipping vertically.

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Ed Bennett - MVP Microsoft Publisher

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Peter Rooney - 31 Mar 2006 10:51 GMT

I'm still not getting this.

If I draw an equilateral triangle 15 wide by 12.99 high (using sines and
cosines here!), then draw a line from each point of the triangle to the
middle of the opposite side, all three lines intersect at the same point.
Looking at the line that goes up through the centre of the triangle, it's at
right angles to the base. The other two lines AREN'T at right angles to their
respective sides.

If I then lock the whole thing together and rotate it through 120/240/360
degrees, whichever side of the triangle is at the bottom has a line running
up from it at 90 degrees, but the other sides never do - the line always
appears to run at an angle to the side.

Do we have some wierd parallel universe thing going on here? If all the
sides of the triangle are equal and all the angles are equal and all the
bisecting lines meet at the same point, why is it only the bottom side of the
triange that has a line running from it at 90 degrees, and why does this
change when you rotate the object?

Either I've forgotten something about basic trigonometry, my screen
resolution's WAY out or the ants have taken over my PC.

Any thoughts welcome...Help!

Pete

> > Hey Ed, I used a hexagon, using centimeters, Height 6.60cm Width 7.6cm
> > Created a triangle, 3.3cm high, 3.8cm wide. Copied the triangle,
[quoted text clipped - 4 lines]
> Yes, but using equilateral triangles, rotation is not necessary - rotating
> through 60° gives the same triangle as flipping vertically.

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Ed Bennett - 31 Mar 2006 11:16 GMT

> Do we have some wierd parallel universe thing going on here? If all
> the sides of the triangle are equal and all the angles are equal and
> all the bisecting lines meet at the same point, why is it only the
> bottom side of the triange that has a line running from it at 90
> degrees, and why does this change when you rotate the object?

It's probably a quirk of the monitor.

The bottom line runs along a line of pixels. As does the perpendicular
bisector. Hence it will look "obviously" at right-angles.

The other two lines don't run along the pixel axes, so will not look as
"straight", and anglegs might not look as "right".

Insert a square using the rectangle tool (it needn't actually be a square,
but use the Shift key if you want one).

Rotate that through 30° and 60° and use a corner to test against the angles
that should be right angles.

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Ed Bennett - MVP Microsoft Publisher

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Peter Rooney - 31 Mar 2006 11:41 GMT

Ed,

It looks fine when printed - thank you very much!

Pete

> > Do we have some wierd parallel universe thing going on here? If all
> > the sides of the triangle are equal and all the angles are equal and
[quoted text clipped - 15 lines]
> Rotate that through 30° and 60° and use a corner to test against the angles
> that should be right angles.

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