Re: Enigma 1513 - Two polygons

Ilan Mayer wrote:
On Nov 12, 9:37 pm, Chappy <petergregorychap...@xxxxxxxxxxx> wrote:
Enigma 1513 - Two polygons
New Scientist magazine, 27 September 2008.
By Richard England.

I drew a regular polygon that had X sides
and a second regular polygon that had Y
sides. Just as my second polygon had (Y-X)
more sides than my first polygon, so each
internal angle of the second polygon was
(Y-X) degrees greater than each internal
angle of my first polygon. (X+Y) was a
perfect square.

How many sides did each of my polygons have?

Ciao,
Chappy.

SPOILER

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For n sides the angles are 180-360/n

Integer values:
n angle
3 60
4 90
5 108
6 120
8 135
9 140
[...]
I'm not sure it's obvious that the angles have to be integers.
Anyway, you can make it obvious (and cut down the number of
possibilities) by solving
(180-360/Y) - (180-360/X) = Y-X
(180XY - 360X) - (180XY - 360Y) = XY(Y-X)
360(Y-X) = XY(Y-X)
0 = (XY-360)(Y-X) .
since X and Y are not equal, this means that XY = 360.
We are also given that X+Y is a square, which can only be
0, 1, or 4 mod 16, leaving only the possibilities
X Y X+Y
1 360 361
9 40 49
which turn out indeed to be squares. Of course, we can't have
a monogon (with a -180-degree interior angle!) so the enneagon
and tetracontagon form the unique answer.

Dan Hoey
haoyuep@xxxxxxx
.

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