Let T be a triangle with sides 7, 8 and 9 (it is a solid triangle consisting of both interior and boundary points). Find the perimeter of a planar figure formed by all points with distance at most 1 from the triangle T.
Along the edges of course, the distal end of a perpendicular to the sides sweeps out lengths of 7, 8 and 9.
When that perpendicular's base reaches a vertex it starts to sweep out an arc whose central angle is the supplement of the angle of the triangle at that vertex. As the ends of the rectangles are parallel, the radii of successive circular sectors are parallel, so that the three arcs can slide together to make up one complete circle, with circumference 2*pi.
The total perimeter asked for is 7+8+9+2*pi = 24 + 2*pi.
Edited on January 21, 2025, 9:42 am
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Posted by Charlie
on 2025-01-21 09:38:16 |
solution
