Thanks for all the fun entries! The past few weeks have been entertaining, to say the least.

I saved one of my favorites, and one of the more tricky problems, for last. It’s not the simplest to set up mathematically, but can be solved intuitively.

Last week’s problem:

Hole in the Sphere

A cylindrical hole 6-inches long has been drilled straight through the center of a solid sphere. What is the volume remaining in the sphere?

I picked at random this week, and Don nailed it with his simple explanation.

The problem doesn’t state the width of the cylindrical hole. Therefore, if the problem has a unique solution, the answer must be independent of the hole’s width.

Therefore, we can safely assume a limiting case, that the width of the hole is zero. In this case the ‘remaining’ volume is simply the entire volume of a sphere three inches in radius: that is to say, 4/3 * pi * 3^3, or simplifying, 36*pi cubic inches.

I ran these contests as a way to launch the book here at our blog; when I started we just had some pre-release copies sitting around. It’s out now, and available nation-wide.

For previous contest entries too entertaining to tuck away in my inbox, see the Hall of Fame.

Last chance!

You guys rock–it makes me sad that I don’t have enough spare copies to give to the great entries I’ve been getting. I will be back momentarily with this week’s puzzle. First:

Four bugs sit 10 inches apart in a square. They move toward each other, forming a logarithmic spiral.

How far do they move until they hit the center?

Does this address any confusion? Guha and Mikey, if nothing else, demonstrate that upside-down sunglasses really do look like bug-eyes. A relativity game indeed.

They answer correctly, and are this week’s winners!

Several people did solve the puzzle with calculus–and it looks something like this (thanks B. Li):

Suppose the beetles always move with velocity v towards the next beetle. The radial component of v will be v*cos(pi/4)=v/sqrt(2) for the entire path.

The radial distance from each beetle’s starting point to the centre of the square is 10/sqrt(2). Distance travelled radially = (time taken along entire path)* (radial speed). So each beetle will take a time of (10/sqrt(2))/(v/sqrt(2)) = 10/v to reach the centre of the square.

Since the velocity of each beetle is v and distance=velocity*time, each beetle will travel a distance of v*10/v=10 inches before reaching the centre of the square.

However, Gardner made sure to say that it could be done without calculus. The simplest expression I got was this (thanks, D. Graham):

The motion of each target bug is such that from the chaser’s perspective their distance doesn’t change (other than through the motion of the chaser), only the angle at which they must walk, due to the continuous perpendicularity of the paths. Therefore, each bug will travel 10 inches before meeting the next bug (less the length of a bug!

I got two good poems this week, and they deserve special mention. They’ll go in the Hall of Fame page later today.

THE FINAL PUZZLE, after the jump.

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October 1, 2008

For this week’s contest, click here!

Hello, Leiter Reports! I’m very glad that I have a couple philosophy degrees, because the conditional logic that you guys threw my way … well, wasn’t too complicated per se, I just don’t see much of it during a regular work day! That wasn’t supposed to rhyme.

Anywho, I’m gonna pick THREE winners this week rather than one; because the response was so overwhelming, and I want to raise the chances for everyone. So I’ll announce the winners of Hexaflexagons, Probability Paradoxes, and the Tower of Hanoi.

If you don’t win, don’t fret! Along with the winners, I’ll post the next contest. Two more weeks to go.

There are three simple ways to do it, and you guys nailed them all.

Ask about the other guy: “If I were to ask a member of the other tribe the if this (pointing) road leads to the village, what would he say?”

The soul-searching method: “If I were to ask you if the road over there leads to the village, what would you say?”

The Biconditional (hey, I remember these): Warren C. Haggstrom of Ann Arbor suggested the following phrasing to Gardner (it’s in the book) to take the ambiguity out of questions within questions: “Of the two statements - ‘You are a liar’ and ‘This road leads to a village,’ is one, and only one of them true?”

I like Haggstrom’s language; it’s simpler than the “if and only if” conditional.

Check this out. In a 1957 letter to Scientific American, a couple of writers (one a Cambridge cosmologist) pointed out that you could know that two words, say, ‘pish’ and ‘tush’ mean yes and no, but not know which means which. Even so, given the liar/truth-tellers a crafty logician could ask:

“If I asked you whether the road I am pointing to is the road to the village, would you say pish?” If the response is “Pish,” then that’s the road to take. If “Tush,” take the other road.

Though not categorically correct, I also would have accepted: “Did you know that they are serving free beer in the village?”

This week’s winners:

M. M. Smith with a nice little poem, Gathering for Gardner-style. He changed the logician to a young girl for clarity.

The lass questioned the teen on her plight
But feared he might lie by birthright
So she quizzed the youth’s ken:
“Would the other tribe’s men
Say to the left, or the right?”

The lass then took the other road.

Oh, man! A. Andrews managed to tie this one to our economic crisis:

There once was man from a U(niversity),
got stuck at a fork, what to do!?!
A truth-teller he sought, but discern he could not,
whether what he was saying was true.

He said to his new friend “Dear sir,
If an opponent you were (that is, a member of the other tribe),
Would you tell me to stay on branch B, or A?
Oh, and please do try not to slur.” (For the man from the University was old, and his hearing had declined).

“My friend,” said the native, “I know
that my opponent will tell you to go,
down this road right here, letter B, but I fear,
that with you this way I must go. (For the man’s sight had also declined, and this was apparent to the native).

Now the university man was quite bright,
and he knew well that the native might
be telling the truth, but with out any proof,
he could not be sure he was right, (at least, he could not be sure whether the native were a truth-teller).

But this quandary could not overcome
the mind of the man who was from
the University, for the answer, you see,
was clearer for him than for some.

He started down branch letter A
But why, you ask, did he stray,
from the path advised by the native disguised?
Since the native’s true tribe was still vague.

The answer will now be secured:
The question the prof asked ensured
that the native’s report, or his opponent’s retort
would certainly be a false word.

So along down branch A he went
Till every last dime he had spent
It was gone very fast, for no bill was passed,
that would, an economic crisis, prevent.

Winner #3, picked at random, was B. Bix. with an “Ask the other guy” method. There were some other memorable answers, some insightful, others, well, wrong, but good for a laugh. I’ll post them in the Hall of Fame page up top later today.

Thanks, Leiter readers, I knew I could count on you. Enter again!

And the next contest, after the jump.

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September 24, 2008

[Update] This week is over! The winner, and the current contest, will be announced here.

Today, we’ll start on the puzzles from Hexaflexagons, Probability Paradoxes, and the Tower of Hanoi. I’ll run three; that’s three more weeks of puzzle goodness.

Incidentally, don’t miss Don Albers’ lengthy interview with Gardner, updated weekly.

Last week: the final puzzle from Origami, Eleusis, and the Soma Cube. Gardner’s puzzle last week saw a bank teller inadvertently doubling a man’s withdrawal, having switched dollar and cent values. Doubling, that is, if one takes into account that pesky five-cent newspaper.

I’ve chosen a winner at random this week:

And the book goes to J. Snyder who, incidentally, correctly identified the equation as:

…a linear diophontine equation which is c + d/100 - 5/100 = 2*(d + c/100), where c=cents and d=dollars

The answer: $31.63

This week’s puzzle (a logic one, for all the philosopher-types out there) after the jump:

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Welcome back!

Go to the current contest here >>

The entries are really coming in. Thanks to everyone for entering!

Also, check out the interview with Gardner that I’ll be posting over the next few weeks.

Last week’s question asked how many planes, re-fueling in midair, are required for one circumnavigation, assuming that a full tank=1/2 the trip, and no lost time for re-fueling. Plus, they all have to return safely.

I had to pick this entry by Andrew McFarland because the title cracks me up every time. Putting the word “Plan” in quotation marks, the faded old paper, gas can icon, and old planes — it all amuses me so. Plus, he got the right answer, and depicted it very clearly.

Click to enlarge

This week’s puzzle after the jump!

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