Irrationality in the balance

Question

Sensibilly -T ™ scales were out of stock, so Gramazon filled the order with a lousy Lopside® from Irreturnable Irrationals International, Inc.73205080...  How irredeemably irritating.  Sigh, might as well unpack it anyway and just start weighing things.

Parts
• One factory-calibrated lopsided beam balance
• One unit weight,          weight =  1   unit
• One disposable weighing pan,    weight =  p   units ( p is an irrational amount)
• Another disposable weighing pan, weight =  p   units
• Another disposable weighing pan, weight =  p   units
   ⋮     (infinitely many identical pans)

Specifications
• The scale balances when nothing hangs from either beam’s hook.
• The scale also balances with the unit weight on the left side and one pan on the right.
• Any given weight that’s a positive integer W units heavy may be balanced by hanging one or more pans, placing W in one of those pans, and optionally hanging the unit weight as well.
• Each pan is good for a single balancing only and is then wastefully discarded.
• The unit weight may be used again and again.

Action plan
Day  1.   Balance W =  1  unit   (then discard all used pans).
Day  2.   Balance W =  2  units (and again discard the pans)
Day  3.   Balance W =  3  units (and discard pans)
     ⋮
Day 30. Balance W =  30  units (discard pans one last time)

Among positive irrational numbers, what value for p allows the fewest pans total to be used up during these 30 action-packed days?

Notes

All items may hang on either side, but W always needs a pan on the same side to hang in.

Although the balance’s left beam is depicted above as being longer than its right beam, that disparity would be reversed if p were between 0 and 1, which is allowed.

Answer 1

With a weight $L$ on its left arm and $R$ on its right, the balance balances exactly when $pL=R$.

Suppose we hang $L$ pans on the left and $R$ on the right. Take $e$ to be 0 if we don't use the unit weight and 1 if we do. There are four options for where we locate our weight $W$ and our unit weight.

(L,L) : $p(Lp+W+e)=Rp$ or $L p^2 + (W+e-R) p = 0$
(L,R) : $p(Lp+w)=Rp+e$ or $L p^2 + (W-R) p - e = 0$
(R,L) : $p(Lp+e)=Rp+W$ or $L p^2 + (e-R) p - W = 0$
(R,R) : $p(Lp)=Rp+W+e$ or $L p^2 - R p + W+e = 0$

We can ignore case (L,L) because it cannot be true if $p$ is irrational. For every positive integer $W$ at least one of the others must be true, for suitably chosen $L,R$.

So, clearly $p$ satisfies some quadratic equation $Ap^2+Bp+C=0$. In this case it satisfies another quadratic equation precisely when that equation is a multiple of $Ap^2+Bp+C=0$.

When can case (L,R) work? Well, that constant coefficient is always either 0 or -1, and in fact it can't be 0 if $p$ is irrational. So if case (L,R) works then we have $L=A,W-R=B,-1=C$. The number of pans used will be $L+R=A+W-B$. If this case always works then we must have $B\leq1$ and we'd better take $A=B=1$, so we need $W$ pans every time. This requires that $p=\frac{\sqrt5-1}2\simeq.618$. But of course there might be solutions where we use this only for some $W$, and in that case we might be able to use a larger $B$.

When can case (R,L) work? Well, then we will need to take $C=-1$ again since $W$ always has to be a multiple of $C$. Then we do best to take $A=B=1$ giving the same $p$ as we needed to make (L,R) work. In this case we have $L=W,R=W+e$ so we'd better always take $e=0$ (i.e., not use the unit weight) and then we take $2W$ pans. This is always worse than case (L,R) so we can ignore it.

When can case (R,R) work? Here we can do a bit better because we get to choose whether the constant coefficient is $W$ or $W+1$. So we can make our constant coefficient be 2, and then we can scale the coefficients by about $W/2$ instead of $W$. But there is a snag. Our quadratic needs to have real roots! This means that $B^2\geq4AC$ which if $C=2$ requires $B^2\geq8A\geq8$ so $|B|\geq3$. So the best we can do is $A=1,B=-3,C=2$. But this won't do because this quadratic equation has rational roots. Since we end up needing about $(|A|+|B|)W/2$ pans, this is going to end up worse than case (L,R) again.

Conclusion: we can never do better than case (L,R) which requires exactly $W$ pans to achieve balance with weight $W$. So we will need, in total,

$1+2+\cdots+30=31\times15=465$ pans.

And we should take

$p=\frac{\sqrt5-1}2\simeq0.618$.

Perhaps "Gramazon" is so-called

in honour of the golden ratio.

Cautionary note: This is exactly the sort of thing in which I easily make mistakes, so I will be unsurprised if any part of the above is wrong. The overall approach is clearly right, though.