Howard Williams
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?
Sunday Times Teasers Discussion Group 
Fairly quick to solve – not many options available.
The calculation of the location of accumulative gravity centres top down was not quick for me.
I considered the case with three 1-peso and three 2-peso coins. I assumed that 1-peso coins are at top and bottom and that, for the sake of calculations, the top coin is sideways displaced by exactly 5/4 times the diameter. There are four options for the location of the third 1-peso coin.
I found that two locations are unstable, one location gives marginal stability / instability and the fourth one permits stability. The top -down sequence for stability could be anticipated without calculations.
I believe that Howard Williams had the stable option in mind.
Correction:
reading the text again I think that Howard Williams had the “marginal” option in mind. The “stable” option permits exceedidng the 5/4 upper coin displacement a little. But “I managed to get the rightmost edge ….one and a quarter times its diameter further right…”, no more.
So the wording of the puzzle invites us to check options by calculations rather than to identify intuitively the most stable coin sequence.
Hi Peter,
I may be wrong here but my solution suggests that the 5/4 x coin diameter displacement is the limit above which there are no stable solutions. In this arrangement for each coin, the centre of gravity of the coins above it are at its extreme right hand edge.
Thanks, Brian
I found that the gravity centre of the five upper coins of the “stable” option of coin sequences is exactly 1/24 th diameter to the left of the right edge of the bottom 1-peso coin when the top 1-peso coin is displaced to the right by exactly 5/4 times the diameter and when the four middle coins are kept as far left (!) as possible. In this limiting case the equilibrium is marginally stable /unstable.
Therefore one has a very small freedom to shift the four middle coins to the right until the gravity centre of all upper coins reaches the stability limit at the right edge of the bottom coin.
One could even shift the top coin by a small amount to the right without loosing stability. This seems to be forbidden by the wording of the Teaser.
The option “permitting stability” would therefore have to be excluded and a less comfortable sequence of coins has to be sought for. This turns out to give marginal stability, with fixed positions of all coins.
The best coin sequence to obtain stability may be guessed as the one where the three heavier coins are directly above the bottom coin, giving a good counter-of weight for the two upper 1-peso coins.
HI Peter,
What bothers me is your sentence “One could even shift the top coin by a small amount to the right without loosing stability.”
You seem to be saying here that there is a stable arrangement of the coins in which the top coin is further to the right of the bottom coin than 5 / 4 of a coin diameter. I don’t think this is possible.
Hi Brian,
I obtained the marginal limit of the “stabllity permitting sequence” stepwise by imposing marginal stability for the top coin. Initially assuming the top coin 5/4 times more right than the bottom coin.
Next one can place the third coin from top such as to obtain marginal stability for the top two coins (without changeing their relative positions).
Next one can place the fourth coin from top such as to obtain marginal stability for the upper three compound coins.
One can continue this procedure down to the fifth coin from top.
In all these steps the middle four coins are moved to the left (!) as far as compatible with individual and compound marginal stability. The gravity centre of the upper five coins is then calculated to be left of the right edge of the bottom coin by 1/24 times the coin diameter.
So there is some freedom to move each of the middle coins to the right relative to the next coin above, maintaining individual and compound stability. In the simplest case one can postulate the same tiny right relative shift relative to the next coin above, say mu.
One can then calculate the maximum possible mu for overall marginal stability.
Now, if we reduce mu to a smaller value mu’ < mumax, then we obtain some reserve for stability. We can then move also the top coin somewhat to the right from the “5/4 position” on the expense of reducing mu’ to, say, mu” < mu’.
The mu’s could be made the same including the top coin’s right shift from the 5/4 diameter position. Maybe I should perform the calculation to find the maximum possible common mu” and to confirm or disprove if the position of the top coin can be more right than 5/4 times diameter. Sofar, my arguments are only qualitative.
Sorry for my lenghty exlanation attempts.
Sorry, I should have said that the mu is the right shift of a coin relative to the gravity centre of the compound of coins above it rather than relative to the coin above it.
My apologies Peter, I completely misunderstood what you were trying to explain. I now see what you are referring to and there are in fact three arrangements that exceed the displacement that is the solution (top down, R = coin radius):
displacement coins
529.R/210 (1, 1, 1, 2, 2, 2)
811.R/315 (1, 2, 2, 2, 2, 2)
31.R/12 (1, 1, 2, 2, 2, 1)
I see a modified harmonic sequence…
Based on classical mechanics (statics), and testing of several odd-pesos-combinations, I found one where the coins’ off-sets (relative to the cion below) in clean unit fractions (of diameter) with one-digit dominators turned out to meet the text’s requirements exactly.
Some combinations are close, so Howard Williams must have had good measure instruments at hand. Unless there is some hidden patterns around that can be argued analytically, as Truner perhaps hints to above.. I am not there yet.
Does anyone else agree that the wording implies more than one of each coin, and as the total is odd that’s three of each coin?
Hi Truner,
Yes, the two ‘the’s in the sentence “but the two-peso coins are twice the thickness of the one-peso coins.” implies this and it reduces the number of combinations that need to be considered. Without the two ‘the’s it would be harder work manually.
I think you are probably right and Brian is right about the wording. It did take me a bit more work as I checked a fuller number of coin sets, it just meant 32 arrangements of coins instead of 20. Still doable and does not lead to multiple solutions so I assume the implied restriction to 3 of each coin is just to make the puzzle a bit less work.
I have modelled the coin-stable in GeoGebra.
A ‘close-to-solution’ is obtainable by inspection.
Hi Erling,
Your wonderful model shows that the most promising top to bottom coin sequence 1 1 2 2 2 1 permits to shift the top coin by more than (5/4) x diameter relative to the bottom coin to maintain stability throughout.
I suspected this but did not yet demonstrate it by calculation.
Surely, the centre of gravity of the coins above any level must lie on the edge of the coin at that level (Neutral Equilibrium but still equilibrium). One ends up with a stack in multiple states of equilibrium, which conforms with “all coins are shifted the maximum to the right”. The weight of the bottom coin doesn’t enter the analysis except to make-up the odd number calculation. Perhaps being a Civil Engineer helps in visualising this !!
But what is the maximum offset for an odd value of pesos coins !!!
I make it 31D/24 for coin diameter D.
Agree.
In my notes it appears as ‘1.2916..’ just before the ‘1.25’ – the former followed by “Maks?” the latter by “Bingo!”
I did enjoy this one very much but it took two sides of A4. I had to draw diagrams to help my calculations. First I drew the final situation, labelling the overhang of each coin relative to the one below as x1,x2,x3,x4,x5. The total of these needs to be 5r/2 for coin radius r. Corresponding masses of the overhanging coins are m1,m2,m3,m4,m5, which will be 1 or 2 ‘units’ to be determined.
Then step by step from the top moving each coin to its overhang and equating mk terms either side of the pivot, where k is the distance of the centre of the coin base from the pivot, gives a formula for each x in terms of r and some m values.
I did wonder whether the question was hinting there should be at least two of each m and hence must be 3 of each for odd total, or whether cases with just one of a certain m should be included. I decided to allow the wider possibility for completeness, which meant I considered 32 mass sequences instead of 20.
Then it was just a case of tabulating lists of m, calculating the x values in terms of r and totalling them. Only one possible list of m gives the required total of all x values. This is the answer.
We might just check the known formulae for harmonic series in numerators and denominators…
Funny, I remember a similar question in my (Scottish) half-Higher in Statics back in the late 1960s.
(The other half-Higher was Dynamics)