All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 A Common Vertex (Posted on 2004-07-12)
Three regular polygons, all with unit sides, share a common vertex and are all coplanar. Each polygon has a different number of sides, and each polygon shares a side with the other two; there are no gaps or overlaps. Find the number of sides for each polygon. There are multiple answers.

 See The Solution Submitted by Brian Smith Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 I got six | Comment 16 of 19 |

If we let i, j, and k be the number of sides of the polygons, then

180*(i-2)/i + 180*(j-2)/j + 180*(k-2)/k = 360   or

k = 2*i*j/(i*j-2*(i+j))

Using the following Perl program:

for(\$i=3;\$i<1000;\$i++) {
for(\$j=\$i+1;\$j<1000;\$j++) {
\$m = \$i*\$j-2*(\$i+\$j);
if (\$m>0) {
\$k = 2*\$i*\$j/\$m;
\$m = int(\$k);
if (\$k==\$m) {
if (\$k>\$j) {
print \$i,\$j,\$k;
}
}
}
}
}

I get the six answers found by Charlie.

 Posted by Jerry on 2004-07-17 17:08:47

 Search: Search body:
Forums (0)