Is tossing out ice that forms in the buckets overnight helping with that efficiency, or are we actually throwing out valuable sucrose that could be made into syrup.
There's an easy way to test: take a sample of the discarded ice, melt it, measure the volume $V$, evaporate it until only solids remain, weigh the solids, and compute the percent dissolved solid, $X=m/V$.
This will give you a more straightforward (and more accurate) measure of exactly how much you're throwing away than theoretical calculations, and it's syrup-tapping season, so it should be very convenient to conduct the experiment.
I suspect that there is some sugar that is adhering to the ice (either on the surface or trapped in interstitial pockets that form during freezing due to solute precipitation), and this sugar will be lost in the process. However, whether it is cost-competitive with the additional energy expenditure that would otherwise be required to evaporate the ice that is discarded will crucially depend on the total dissolved solids content $X$, and the market price of maple syrup.
If you get a number after evaporating a sample of the ice, feel free to post it; it would be easy to calculate whether it is a cost-positive or cost-negative process.
The answer is here.
In the article "The structure of the energy bands and optical absorption in osmium" (Sov. Phys. JETP 63, 115 (1986)), Nemoshkalenko et al. report measurements of the complex refractive index of osmium (which can be extracted from reflectivity and related to complex conductivity). Unlike cubic metals such as gold, osmium has a hexagonal crystal structure, which means that its optical properties are not isotropic. Light with electric field in the plane of the hexagon ($\mathbf{E}\perp \mathbf{c}$) has different reflectivity than light with electric field perpendicular to the hexagon ($\mathbf{E}\parallel\mathbf{c}$), where $\mathbf{c}$ is the lattice vector normal to the hexagons.
Check out Figure 1(b), which is the measured reflectivity (the visible range corresponds to ~1.75-3 eV). As @JohnRennie pointed out, there is a reflectivity dip in the red (what the authors call absorption band B), especially for $\mathbf{E}\parallel \mathbf{c}$, which leads to a bluer color.
The authors explain this behavior by computing the band structure of osmuim. They find that the theory predicts the absorption band B to occur due to a couple of electronic transitions (their bands $7\to 8$ and $8\to 9$), which they describe as $d\leftrightarrow p$ type transitions.
As is usually the case with metals, the color is described by interband absorption. Essentially, you have a crystal with a mess of energy bands, the details of which are resulting from the type of lattice, the various electron orbitals in each atom, the couplings between them, and other interactions like spin-orbit coupling. For low energies, the conductivity is typically dominated by free-electron (Drude-like) behavior. When the photon energy matches the energy difference between an occupied and an unoccupied band, you get interband absorption. This is, for example, why copper and gold have their colors, but platinum and silver appear colorless (Pt and Ag don't have interband transitions in the visible range or lower). For osmium, apparently a band with $d$-orbital character is full of electrons, and with photons in the 1-1.5 eV range (with $\mathbf{E}\parallel \mathbf{c}$ to make the matrix elements work out) you can promote those electrons to another band with $p$-orbital character. What's a little interesting about osmium is that there are a number of lower-energy (infrared) transitions too, which distinguishes it from Pt, Ag, Au, Cu, etc.
Best Answer
Some more misconceptions:
The ChemGuide website quoted above might be a useful reference for "UK-based exam purposes" as stated there, but it certainly does not help in solving the question. The arguments given above that followed the comments on ChemGuide are inaccurate. A simple quantum chemistry calculation of gold in its ground state will give you that the electron in the s orbital (A1G) is the most energetic in this atom. Hence, ionization will most easily be accomplished by removal of this electron, and not of d electrons, and this is easily proved by another computation for ionized gold, which will show you that the 5d orbitals will remain filled while the 6s orbital is no longer occupied. Actually, it is known that if the most external d shell is filled, the energies of these orbitals will be effectively lowered, and there is a very high probability that the ionized electron will not come from it, but from more energetic s or p orbitals. (I have just done a few of these calculations in order to make sure this point is right)
In order to analyze why some metals are more inert than others, various effects come into play. Relativistic effects, such as the contraction of s orbitals, for example, are a major factor in making gold less reactive than silver, and in lowering the oxidation potential of gold. So, in addition to looking at chemical potentials when discussing the inertness of metals in different environments, it is better not to reduce the arguments to simple electron configuration trends which usually work quite well for main group elements, since, though they might generate insights for the understanding of the behavior of metals, these insights might be either right or wrong.