Another week, another
batch of riddles. The first one, involving the vitamins just took me a few minutes and I'm pretty confident in my answer. As you might expect, your chances of having your favorite flavor is pretty dang good.
The problem involving the grasshopper is much harder and I've made very limited progress on it. I began by working on the more general case involving a lawn of arbitrary area. This allowed me to see that the optimal solution is not a circle, as was hinted. Suppose, for example, the area of the lawn were exactly pi*(15 cm)^2. Then we surely wouldn't want a circular lawn because any jump would necessarily take you out of the circle, since the radius of the grasshopper's jump exceeds the circle's diameter.
I then speculated that the lawn might be a sort of narrow slit-- tapered at the ends, bulging slightly in the middle. Since that's a difficult shape to work with, I turned my efforts to finding the optimal dimensions of a rectangular lawn of infinitesimal area. This was much easier, requiring only a little calculus, but did require some attention as to whether the width of the rectangle is less than 2R (R being the radius of the grasshopper's jump) or greater than that amount, since these two cases break the rectangle up into regions where successful jumps by the grasshopper are given by 0, 1, or 2 arcs of its circle.
Anyway, I'm not sure how clear any of that is, but I found that the optimal width of a rectangle of infinitesimal area is 3R. Incidentally, this lends a bit of credibility to my thin slit theory: if we were to add a small amount of area, it would likely be best to put it somewhere in the middle where the grasshopper has two ways to jump to this area and, should the grasshopper first fall in that new area, will have two exit points.
So my next step will be to find the optimal dimensions of a cross-shaped lawn of infinitesimal area. That's about as far as I intend to go with the riddle this week unless I have some grand insights. It seems that it's a calculus of variations problem and a pretty pathological one at that. It probably would be best solved through programming. I can sort of picture a methodology in which individual pixels are included/excluded from the lawn based on maximizing how many other pixels it allows a jump from or to. I just don't care to program it myself.
Edit: I've finished analyzing the cross-shaped lawn. Assuming the lawn remains divided into three regions each of width R (it gets tricky to analyze it otherwise), one half of its area is distributed at the thin ends and the other half is distributed in the fat middle. I'll leave the short step of determining its dimensions to anyone else.
This suggests a starting point for the 1 m^2 lawn would be to assume it has two wings each of dimension 83.3 cm by 30 cm, plus a fat middle whose overall dimensions are 166.7 cm by 30 cm. This is almost assuredly less than ideal, since it rests on the assumption that the arcsine of x is approximately equal to x, which holds for sufficiently thin regions. If anyone is interested in continuing this analysis through some program, I'd draw up the region as a few thousand pixels, then move the pixel that contributes the least probability from its current spot to a spot that maximizes the gain in probability.
By the way, it still seems to me that we're trying to maximize the number of pixels that are exactly 30 cm away from any pixel in question. For our cross-shaped region, I believe the optimal placement is somewhere not adjacent to the cross. This leads me to wonder if the lawn is necessarily contiguous and, if not, the solution might be some very unusual shape. Maybe it's a bullseye pattern? Hmmm...
Edit 2: Well, I've done something wrong somewhere. I realized I never actually calculated the probability of the grasshopper landing in the lawn for either the rectangular lawn or the cross-shaped lawn. For the rectangular lawn, the probability was A/(3*pi*R^2). For the cross-shaped lawn, I found the probability to be A/(4*pi*R^2). This makes no sense because it indicates the probability
decreases by including a fatter middle. I even confirmed that the derivative is strictly positive with respect to my one free parameter, consistent with an increasing probability. I'll need to check the consistency of my formulas.
The diagram above shows the view from above (looking down the z-axis) of the sphere (black and blue colors) with two cylindrical holes drilled through it (in gray), according to the problem Warp stated. Note that the resulting shape has symmetry, such that we need only take the volume in the first octant and multiply it by 8.
We use cylindrical coordinates. Focusing first on just the part of the sphere on the plane z=0 and in the first quadrant (indicated by the red lines in the diagram), the angle θ goes from 0 to pi/2 and the radius r goes from the inside which is sin θ (the cylindrical hole) to the outside which is 1 (sphere's equator). Now the height of the sphere above the plane z=0 is given by sqrt(1-r2). So the volume of the whole thing is:
=
=
=
=
=
So the volume of the resulting sphere with cylindrical holes in it is 16/9. Note that this is a rational number, without any pi in it!
