Sunday Times Teaser 3066 – Leaning Tower of Pesos
by Howard Williams
Published Sunday June 27 2021 (link)
I have six old coins worth an odd number of pesos, comprising a mixture of one- and two-peso coins. Both denominations are of the same diameter and made of the same metal, but the two-peso coins are twice the thickness of the one-peso coins.
After making a vertical stack of the coins I then slid each of the top five coins as far as possible to the right, to make the pile lean as much as possible in that direction, without toppling. I managed to get the rightmost edge of the top coin a distance of one and a quarter times its diameter further right than the rightmost edge of the bottom coin.
Starting from the top, what is the value of each of the six coins?
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GeoffR permalink12345678910111213141516171819# Reference: Brian's diagram for weights/offsetsfrom itertools import productd1 = 1.0 # radius of top coin and initial offset# d2, d3, d4, d5 are offsets of 2nd, 3rd, 4th, 5th coinsfor W in product(range(1,3), repeat=6):W1, W2, W3, W4, W5, W6 = Wd2 = d1 + W2 / (W1 + W2)d3 = d2 + W3 / (W1 + W2 + W3)d4 = d3 + W4 / (W1 + W2 + W3 + W4)d5 = d4 + W5 / (W1 + W2 + W3 + W4 + W5)# The cumulative top coin offset is 2.5 (5/2 of the radius)if abs(d5 - 2.5) < 0.000001:# Six old coins are worth an odd number of pesosif (W1 + W2 + W3 + W4 + W5 + W6) % 2 == 1:print(f"Coin Values (from top)= {W1},{W2},{W3},{W4},{W5},{W6}")