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.

(In reply to re(3): solution- Missed Solution! by SilverKnight)

Actually it was a bug in the program.  The portion of the program which checks for any one or two angles above 120 degrees (number of sides > 6) checked the number of degrees in the angle to be integral, instead of properly checking the number of sides in the polygon to be integral.

The wrong code:

    FOR j = 7 TO i - 1
      t = j * 180 - 360
      a1 = t / j
      a2 = remain - a1
      IF a2 < 180 THEN
        IF ABS(a2 - INT(a2 + .5)) < 1E-12 THEN
          GOSUB reportToHex
          PRINT "1"; STR$(j); "-gon; ";
          PRINT "1"; STR$(360 / (180 - a2)); "-gon; "
        END IF
      END IF
    NEXT

The right code:

    FOR j = 7 TO i - 1
      t = j * 180 - 360
      a1 = t / j
      a2 = remain - a1
      s2 = 360 / (180 - a2)
      IF a2 < 180 THEN
        IF ABS(s2 - INT(s2 + .5)) < 1E-12 THEN
          GOSUB reportToHex
          PRINT "1"; STR$(j); "-gon; ";
          PRINT "1"; STR$(INT(s2 + .5)); "-gon; "
        END IF
      END IF
   NEXT

The correct output:

 3 hexagons;
 2 pentagons; 1 10-gon;
 1 square; 2 8-gons;
 1 square;  1 hexagon; 1 12-gon;
 1 square;  1 pentagon; 1 20-gon;
 4 squares;
 1 triangle; 2 12-gons;
 1 triangle; 1 7-gon; 1 42-gon;
 1 triangle; 1 8-gon; 1 24-gon;
 1 triangle; 1 9-gon; 1 18-gon;
 1 triangle; 1 10-gon; 1 15-gon;
 1 triangle;  2 squares;  1 hexagon;
 2 triangles;  2 hexagons;
 2 triangles;  1 square; 1 12-gon;
 3 triangles;  2 squares;
 4 triangles;  1 hexagon;
 6 triangles;

Posted by Charlie on 2004-07-13 15:37:32