Pickover Problems 1 through 11

Solution to prob.7

Since the creature doesn't ask for the expected ratio or an estimate of the ratio, it evidently expects us to calculate the exact average ratio. A quicky experiment shows that this value isn't invariant on how the bones are matched, so I conjecture that solution involves coming up with an algorithm for matching bones that can be executed in two days or less. Supporting evidence: why would the alien specify this time limit rather than (say) an hour to run a simulation?

Brute force doesn't work. A motivated sorter could compare bone fragments for a match faster than a Floridian precinct worker can evaluate chads and dimples (about 5 seconds per ballot)

Solution for 5a)[5b) solved already]:

line 1 from center to circumference line 2 from circ to opp point in circ(normal to line 1) line 3 as an extension of line 1 starting from center to circ

part 2 seems impossible.

SPOILER

-2, 0, 2

2 2 2 3 (-2) +0 +2 = 8 = 2

3 3 3 3 (-2) +0 +2 = 0 = 0

Found by inspecting solutions of type -x, 0, x.

Integrating (1-x)/x from 0 to 1/2 indicates that the average ratio approaches infinity as the number of bones approaches infinity. Some empirical trials indicate that the average ratio is usually about 20 to 1. To get the exact average for this set of 20,000 half-bones, arrange them in order of length, then match longest to shortest; this works perfectly if all bones were the same length before being broken, and is a good heuristic if they were roughly the same length.

Yes. Consider the finite set S of all planes containing two or more of the points. Choose a plane P not parallel to any member of S, a line L perpendicular to P, and establish Cartesian coordinates with L as the X-axis. For any given X-value, the set P' of all points with that X-value is a plane parallel to P, thus is not in S, thus contains at most one of the points. So all points have different X-values. The plane P* of all points with an X-value equal to the median of the points' X-values will have exactly one thousand of the points on each side.

This is equivalent to problem #1.

[ quoting Cliff ]

Let the lines extend outside the circle.

Superimpose two of the lines.

Cliff Pickover weote

results three different square numbers? How > did you solve this? ***** SPOILERS

No solution is possible with rational numbers. What we would need is an arithmetic sequence of squares differing by 100.

Let’s look at rational square sequences. I believe that they must be of the form [n(a^2-b^2-2ab)]^2, [n(a^2+b^2)]^2, and [n(a^2-b^2+2ab)]^2 for any integers a and b and any rational n (including fractions). [See Note 1].The difference between these squares is n^2*4ab(a^2-b^2).

Setting this equal to 100 and doing some basic math yields sqrt[ab(a+b)(a-b)]=5/n where 5/n is rational and a and b are integers. If there are solutions to this equation, they can be expressed in lowest terms where a and b are mutually prime. If so, then neither a nor b can divide (a+b) or (a-b), meaning that no rational square root is possible.

[Note 1. Take any Pythagorean triangle. take the difference between the legs , the hypotenuse, and the sum of the legs. Multiply each by n. The squares form an arithmetic sequence.] ***** results three different cube numbers? ***** The number -101 will do.

Bill Ryan

A couple of thoughts on this: does average always imply arithmetic mean in this forum? If by average ratio the alien meant product((/(r(i),i=1,n)/))**(1.0/n), where r(i) is the ith ratio, then you get something like 1/(2*exp(1)). If it really does indicate arithmetic mean could you change your simulation so that perhaps the contribution of the 100 smallest fragments to the mean are evaluated directly and the others are estimated by an integral? How close do you get to the actual mean by this

Perhaps I haven't read all the responses carefully enough, but this problem seems simpler to analyze if you consider that all three sums must be quadratic residues mod 3. Problem 4 is also easy if you think about it mod 8.