HACK TO THE FUTURE
21 September 2008


Okay now it can be told. The reason you haven't been able to get to the Mindsport website for the last couple of months is because it had been hacked into. But, wow, is that something or is that something! Looks like MS has finally arrived after a quarter century of trying to no avail. Now we're in the company of NASA, the Bank of England, CERN's Large Hadron Collider superblog and the CIA -- not to mention the White House itself -- who all pride themselves no end each sundown for hosting hackers and often, I understand, show off who has more.

Anyway the way all of us found out the www.mindsport.org presence on the Net was tainted was that those who tried to browse their way in were told by the sublime Google gods who rule over our ineffectual little online lives "proceeding to this website can be dangerous to your pancreas" or "abandon all hope ye who enter for here be demons" or "geddouddahere scumbag" or whatever but the message metaphors were never mixed. As a result, the mailcall dwindled by a warp factor of 10 and, what's more, a lot of people missed a lot of puzzles. Therefore, by comparing normal response loads with what was actually throughput, I've figured out which ones were relegated to cyber oblivion. Two particular weeks were dumped on the hardest. Thus, I'm going to repeat the gist of those e4s for the benefit of our beloved loser-outers. (People who had access to print editions, please bear with us, pretending you have a choice in the matter.)



1. (e4) "A boy presses the side of a blue pencil to the side of a yellow one, holding both pencils vertically. One inch of the pressed side of the blue pencil, measuring from its lower end, is smeared with paint. The yellow pencil is held steady while the boy slides the blue pencil down 1 inch, continuing to press it against the yellow one. He returns the blue pencil to its former position, then slides it down 1 inch again. He continues until he has lowered the blue pencil five times and then raised it five times -- making 10 moves in all. Suppose that during this time the paint neither dries nor diminishes in quantity. How many inches of each pencil will be smeared with paint after the tenth move?"


2. (Endgame) "A lady usually buys large bundles of asparagus, each 30 centimetres in circumference. One day the man had no large bundles in stock so he handed her instead two small ones, each 15 centimetres in circumference. "That is the same quantity," she said, "and therefore the price should be the same." But the man insisted that the two bundles together contained more than the large one, and charges her a little bit extra. Who was correct, the lady or the grocer?
(Submitted by Avinash Goel, Mumbai, India)



 




3. (e4) As you move from place to place during your life, including visits to foreign countries, visualise the corresponding path of your antipodal point -- that's the point traced exactly through the centre of the Earth to the opposite side of the globe. What percentage of the time, would you say, is your antipodal trace on water and what on land? Most people allow for the fact that about two-thirds of the globe is covered by water and so they answer that this antipodal point would be on water anywhere from 65 to 90 percent of the time. But here's something startling: for most Americans the answer's closer to 100%! That's because there are only a few square kilometres of the continental United States that map onto land. Most of western Europe and Israel also map into water, except for parts of Spain. So what are the chances for people in India to exit on the other side of midnight?



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To which one near-correct answer had been received from our resident veteran Shamik Banerjee. However there was a numerical error in his calculation. Here's the correct explanation and the exact calculation: The aim is to clear the width of the wall. Call it AB at the top. So let's assume, for the moment, that the stone is thrown with a velocity V ft/sec from point A at an angle of . . . (Lots of lurid hair-raising stuff here like "cos(á)^2 = ½", "Hence Ux = Vx = 35.8887*0.7071 = 25.377 ft/sec" and "From (1), we can write, y = x - 32.2(x/35.8887)^2" till we get to . . . -- MS) Now, if we extend our parabola backwards to O, then at y = - 25 we find x = - 17.41 ft.








So the stone must be thrown 17.41 ft from the wall at an angle of inverse of [tan(Uy/Ux) = 1.87084] or the initial angle of projection should be 61.97 deg to the horizontal. The primary requirement of the problem is that at A the velocity must make an angle of 45 degrees to the horizontal and this velocity at A must be just adequate to have a horizontal range of 40 ft. The rest follows automatically. Now do you think I can legitimately claim the Fields Medal you had mentioned in your postscript when the solution was put up?

Ajit Athle, ajitathle@gmail.com

In-The-Mean-Time-Dept:

(Hundreds of years ago the problem was: "A stone is thrown from 5 ft above the ground so that it clears a wall 30 ft tall and 40 ft wide. Assuming that the mass of the stone and air resistance are negligible and that one may select the initial angle of projection and the distance from the wall suitably, what is the minimum effort (i.e. minimum initial velocity) with which the stone must be thrown so as to just clear the wall? What will be the maximum height attained by the stone and how far does it land from the point of projection in the horizontal direction?" -- MS)








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Endgames
If you've seen police action serials like CSI, Law & Order and junk you know that when the alleged perps are taken into cop custody for the old third degree there's often this huge one-way mirror on one wall which lets the other cops watch the fun and games from the outside while all that the poor sweating sap sees from the inside is a moving mugshot of his own making. Meaning, if you were in his or her position and wanted to know if it was, in fact, a real mirror or one of those sneaky one-way tricks, what's the easiest way to find out?


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