Thelema Posted May 16, 2009 It's an interesting rejoinder/extension to this Q: How many coins of 1cm diameter can you fit into a 100cm x 100cm square? : Reason might dictate between 10600-10700 pieces. But first of all, you need to know how high 3 coins stacked together in a triangle is. r+r+sqrt3r. Share this post Link to post Share on other sites
ThunderIdeal Posted May 19, 2009 ok now translate the following "sqrt3r" and i'll give this a crack Share this post Link to post Share on other sites
Alchemica Posted May 19, 2009 ok now translate the following "sqrt3r" and i'll give this a crack The square root of three times the radius. So, height = 2r + sqrt(3r) = 1 + sqrt(1.5) ~ 2.2247 cm Share this post Link to post Share on other sites
ThunderIdeal Posted May 19, 2009 (edited) thanks for translating but you've made a mistake somehow since the answer must be i think 1.8660254037844386467637231707529cm, in a height of 100cm this fits 53.589838486224541294510731698852 53 times 199 gives 10547 coins but i'm gonna guess cigarette style isn't the solution remainder of .589 of a triangles height means you can fit an extra row and it may as well be 100 not 99 10647? Edited May 19, 2009 by ThunderIdeal Share this post Link to post Share on other sites
ThunderIdeal Posted May 19, 2009 haha ive made a huge oversight Share this post Link to post Share on other sites
ThunderIdeal Posted May 19, 2009 100 - 1.8660254037844386467637231707529 (first double-row of 199) = 98.133974596215561353236276829248 98.133974596215561353236276829248 (remaining height) divided by 0.8660254037844386467637231707529 (height needed for each following single row) = 113.3153532995459013728114585395 remainder is useless, 113 / 2 is 56.5 so i'll take 57 rows of 100 and 56 rows of 99, add the 199 and i get 11443!!! Share this post Link to post Share on other sites
alkatrope Posted May 19, 2009 Has anyone really been far even as decided to use even go want to do look more like? Share this post Link to post Share on other sites
Thelema Posted May 20, 2009 OK, but I get 11444. The first double row contains 200 coins, not 199. Compare this to the best cigarette packing, which is 57 rows of 100 and 57 rows of 99 = 11343. Is 11444/11343 a better improvement than 106/105? No, as it turns out. But the total number is much greater than the scale-wise prediction of 10600. Now, can you devise a formula for similar solutions to the problem for sides of squares in increasing multiples of 10? Share this post Link to post Share on other sites
ThunderIdeal Posted May 20, 2009 damn. remainder was almost useless, but turned out to be worth one extra coin. can i devise a formula lol (side - 1) / 0.866 = A A + 1 = number of possible cigarette rows = B lol no i can't, anybody? Share this post Link to post Share on other sites
