CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

Round 5: Problem 2

with 3 comments

A pile of identical spheres (each of radius 1 meter) are placed in a pile so that they form a triangular pyramid that culminates with one lone sphere at the top of the pile. If the pile of spheres is higher than 2009 centimeters, what is the minimum number of spheres that can be assembled to form the pyramid?

Written by Administrator

May 10, 2009 at 14:42

3 Responses

Subscribe to comments with RSS.

  1. Height of the tetrahedron formed by 4 spheres
    = Increase in height by one layer
    = sqrt(24)/3.
    let h = number of layers.

    2 + (h-1) * sqrt(24)/3 > 20.09
    h-1 > 11.0778
    h > 12.0778
    h = 13

    We can find number of spheres in layer k by
    summing 1 to k, so
    there are 455 spheres.


    May 13, 2009 at 18:23

  2. You forgot to subtract the top and bottom halves of the top and bottom squares. You should be using 19.09 rather than 20.09.

    Franklin Lee

    May 13, 2009 at 22:04

  3. Sorry, no, I’m wrong. You should be using 18.09, which seems to be what you used.

    I’m convinced.

    Franklin Lee

    May 13, 2009 at 22:24

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: