Conference rusure::math

614.0. "concentric equilateral triangles?" by ECCLES::DOUG (doug dickinson, ex-patriot) Thu Nov 20 1986 07:06

               |        /\
               |       /  \
               |      /    \
               |     /  /\  \
               |    /a /  \ b\
               |   /--/    \--\
               |  /  /______\  \
               | /     |c       \
        _______|/______|_________\ ______________________
               |
               |
               |
               |

    (i can hear the ooh's and aah's for this fantastic diagram, but gosh it
    was nothing (i can hear you agreeing with that too)) 

    now suppose that we have two equilateral triangles here, one inside the
    other.  if the larger triangle's lower left-hand corner is at (0,0),
    what is the formula for finding the co-ordinate of the lower left-hand
    corner of the inner triangle such that a=b=c, where a, b, and c are the
    lenghts of lines drawn from the edge of the outer to the edge of the
    inner triangle. 

    this is one of those things that i'm sure i knew in the 10th grade,
    felt i'd never need again and forgot, and now find myself doing
    graphics. 


    dd 
    

    I take it the lines marked with a and b are perpendicular to the sides
    of the triangles, not horizontal as shown?
    
    Let the side of the outer triangle be s.  The highest vertex is at
    (s/2, s/2 sqrt(3)).  The rightmost vertex is at (s, 0).  The midpoint
    of the segment between those two vertices is (3s/4, s/4 sqrt(3)).  A
    line from the origin to that midpoint is y = x/sqrt(3).
    
    The leftmost vertex of the inner triangle is on this line.  Its y
    coordinate is c, so c = x/sqrt(3), or x = c sqrt(3), so the vertex is
    at (c sqrt(3), c). 
    
    
    				-- edp