Correct solutions were received from Pat Shea, Daniel Marquette, Scott Hoffmeyer, Jeremy Light. Correct and complete solutions were also received from Dan Menard, Juan Carlos Marivela, Al Zimmermann, Bill Webb, Lou Cairoli, Paul Botham, Jens Voß, Ahmed Aboadham, Nancy Schwarzkopf, Ronald Welch, Alan O'Donnell.  

Let the bottom row contain n circles.  The total number of circles within the triangle is 1 + 2 + ^... + n = n(n + 1) / 2.  To find the radius of one circle, note that the light blue triangle in the corner, see the figure on the left, is a 30-60-90 triangle, therefore the base of the large equilateral triangle has length 2rÖ3 + 2(n - 1)r.  Since the base also has length 1, you get r = 1/(2Ö3 + 2(n - 1)).  The total area covered by the circles is given by A_n = = .39....  A straightforward calculation gives lim _{n® ¥} A_n = p/8 = .41....  Comparing this to the area of the equilateral triangle, Ö3/4, shows that the circles will never fill up the triangle.  

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