Futility Closet An idler's miscellany of compendious amusements

Because the moon’s shadow is smaller than Earth’s, a total solar eclipse is visible only from within a relatively narrow latitude on Earth’s surface, while a lunar eclipse is visible from anywhere on the night side of the planet. For the same reason, a total solar eclipse lasts for only a few minutes at any given location, while a lunar eclipse lasts for several hours.

1. Qb1 and every Black move loses.

Sea level gives us a convenient equilibrium water level. Since Everest rises higher, the water would flow toward Chimborazo.

1. Qh5 Ke4 2. Rxc4#. (If the knight moves then the queen can mate immediately from e5.)

Follow a sector on the inner wheel through a complete revolution: That sector will find exactly 50 matches. The same is true for each of the 100 sectors on the wheel; altogether, as the wheel turns through 100 positions, there will be 100 × 50 = 5,000 matches. This means that the average number of matches per position is 50, so a position must exist with at least 50 matches.

The area of the deck is the area of the large circle minus the small circle. That’s πA² – πB², or π(A² – B²).

We don’t know A or B, but we do know the length of the green line. That’s 42. So the Pythagorean theorem can tell us that B² + 21² = A², and hence that 21² = A² – B². Happily, we can plug this directly into the equation above and get our answer: The area of the deck is 21²π, or 441π square feet — regardless of the size of the carousel!

Remarkably, this is true — so long as the green chord stays locked in position between them, the two circles can assume any size and the difference in their areas will remain constant. A similar principle holds in three dimensions.

In The Universal Book of Mathematics, David Darling writes, “An even more amazing fact is that if you slide a chord of fixed length around any convex shape C so that the chord’s midpoint traces out another figure D, the area between C and D doesn’t depend on what shape you started with.”

(Thanks, Dave.)

No. After n hops, Alice will have traveled at most 1 + 2 + 4 + … + 2^n-1 = 2ⁿ – 1 feet from the starting point. But her very next hop will carry her 2ⁿ feet. So she must always overshoot the goal.

Here are two ways:

1. “Dog-ear” the page, folding one corner across to the opposite side. The area below the dog-eared portion measures 8.5 x 2.5. Reopen the paper and fold this 8.5 x 2.5 portion up into the page. The area that remains uncovered measures 8.5 x 6; it can be folded in half to produce a height of 3 inches.
2. Fold the paper in half to make a doubled sheet measuring 5.5 x 8.5. Now “dog-ear” this doubled page, folding one corner across to the opposite side. The area below the dog-eared portion measures 5.5 x 3.

Consider what happens if you leave your opponent 5 pennies. At that point, no matter how many he removes, you can play so as to leave the final penny to him. Now, if leaving him 5 pennies is a worthy target, then by the same principle so is leaving him 9 pennies: No matter how he plays, he must now give you the opportunity to leave him with 5. And going one step further, aiming for 13 pennies will ensure that you can reach 9, then 5, and victory. So, on your first turn, take 2 pennies.

It sounds as if there’s not enough information to solve this, but there is if you remember that when a right triangle is inscribed in a circle, its hypotenuse is the diameter. The pool is 100 feet wide.

1. Qa8 and the queen will take the pawn, giving mate.

Yes. Let x, y, and z be the number of green, brown, and gray chameleons, and consider the remainders left when each of the quantities x-y, x-z, and y-z is divided by 3. These remainders won’t change as the chameleons vary their colors. This means that if two populations are to reach 0 simultaneously, they must differ now by a multiple of 3. In this case two do: The populations of green and brown chameleons differ by 6, so it’s possible for both to disappear without any leftovers:

13, 19, 23
15, 18, 22
17, 17, 21

… and then green and brown chameleons can disappear in pairs until all have turned gray.

From Dick Hess, All-Star Mathlete Puzzles, 2009.

In the U.K., October is one hour longer than any other month because of the end of Daylight Saving Time.

See Fugue State.

Imagine the process in reverse. If Monkey 1 reaches Banana G and then descends again according to the same rules, it will return to the foot of Ladder 1; its path is uniquely determined. Further, because there’s no occasion for loops or dead ends, each monkey will eventually reach the top of some ladder. If every monkey reaches a goal, and no two monkeys reach the same goal, then every monkey gets a banana. The solution generalizes to any number of ladders.

1. After the sneaky 1. Qa8 Black can choose among six different mates.

It’s impossible. Each of the 13 reels must go through an even number of transfers in order to finish with its original tape. (Likewise, the empty reel must go through an even number of transfers to remain empty.) But each of the 13 tapes must be transferred an odd number of times to end up reversed. So the total number of transfers must be both even and odd, an impossibility.

Each can be decomposed into two 3-4-5 triangles … so they’re equal.

1. Nc4 and White will mate by 2. d5, 2. Qg8, or 2. Qc8.

“Potatoes.”

1. Qh6 and the queen or knight will mate.

One in four. There are three intersections, so there are eight possible ways the lace can cross itself. Of these, only two (over/under/over and under/over/under) will produce a knot.

1. Qa6 and White will mate by 2. Qh6, 2. Nd3, or 2. Qa1.

Consider first the endpoints, A and K: The sum of the distances to these two points will be the same for any point between them. Similarly, any point between B and J will have the same total distance to points A, K, B, and J. Proceeding in this way and taking the points in pairs, we find that every point between E and G produces the same total distance to the outermost 10 labeled points. That leaves only F. So the answer is that point F has the minimum total distance to the labeled points.

Strangely, this means that we can shift the points around without affecting this result. If points A through E crowded F closely on the left, and point K were 90 light-years away on the right, point F would still yield the smallest total sum, so long as the points appeared in the order shown above.