Kepler’s second law holds that a line segment connecting an orbiting planet to its sun sweeps out equal areas in equal periods of time: In the diagram above, if the time intervals t are equal, then so are the areas A.
If gravity were turned off, would this still be true?
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Yes. The planet would pass into space in a straight line, and each time interval would define a triangle with height h and base t. Since a triangle’s area is half the product of its height and base, they’re all equal.
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April 12, 2014 | Categories: Puzzles, Science & Math
By Sam Loyd. White to mate in two moves.
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Typically for Loyd, the answer is the least likely move on the board:
1. Nd2! (threatening 2. Rxe2#)
If 1…Kxf2 then 2.Qg3#
If 1…Kxd2 then 2.Qc3#
If 1…Rxe3 then 2.Qxe3#
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April 11, 2014 | Categories: Puzzles
Prove that log10 5 is irrational.
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Suppose it’s rational. Then log10 5 = a/b, where a and b are positive integers. Then 10a/b = 5 and 10a = 5b. But these two values can never be equal — any positive integral power of 10 will end in 0, and any such power of 5 will end in 5. So log10 5 cannot be rational.
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April 10, 2014 | Categories: Puzzles
By George Edward Carpenter. White to mate in two moves.
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1. Qh2! threatens 2. Nf5#. If 1. … Kf4 then 2. Ne4#, and if 1. … Kd4 then the queen mates from h8.
From The Chess Players’ Chronicle, 1874.
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April 4, 2014 | Categories: Puzzles
Imagine a 1000 x 1000 chessboard on which a white king and 499 black rooks are placed at random such that no rook threatens the king. And suppose the king goes bonkers and wants to kill himself. Can he reach a threatened square in a finite number of moves if Black is trying actively to avoid this?
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Yes. Imagine that the king is on the lower left square and that on each move he takes a step along the main diagonal toward the upper right. After his first move and Black’s response, there must be no rooks on the lowest three ranks or the leftmost three files, or the king could immediately reach a threatened square. After his 998th move, the king arrives at a similar situation in the upper right: At this point there must be no rooks on the uppermost three ranks or the rightmost three files, for the same reason. This means that when the king begins his journey, all the rooks must be northeast of him, and when he finishes, all must be southwest. During the trip, then, each of Black’s 499 rooks must change both its rank and its file, in order to keep out of his way. This requires 499 x 2 = 998 moves. So, at the latest, Black will run out of time just a half-move too soon: White will squeeze Black’s maneuvering room down to zero on his own 998th move, just before Black evacuates the last rook.
From Miodrag Petkovic, Mathematics and Chess, 1997.
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March 26, 2014 | Categories: Puzzles
By Hyacinth R. Agnel. White to mate in two moves.
March 21, 2014 | Categories: Puzzles
By William Anthony Shinkman. White to mate in two moves.
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If White could pass his turn, Black would have to promote the pawn or play 1. … Kd1, inviting mate in every variation:
2. Kd1 Qxd2#
2. d1=Q(or R) Bh4#
2. d1=B (or N) Bd2#
To reach these lines, White needs to play a waiting move now. But this is tricky to find: If he marks time by moving the bishop, then Black can mark time by advancing the g-pawn. And if White moves the king to b3, b2, b1, or a1, then Black can promote the pawn with check. The only safe move is 1. Ka3!, and now Black must submit to one of the lines above.
From the St. John Globe, 1905.
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March 14, 2014 | Categories: Puzzles
The names of two U.S. state capitals end with the same eight letters. What are they?
March 9, 2014 | Categories: Puzzles, Trivia
By Sámuel Gold. White to mate in two moves.
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With 1. Qf3 White threatens 2. Qh5#. If Black captures the queen then 2. d4 is mate. His only other defenses require moving the bishop (Bxb3/Be6/Bf7), which permits an immediate 2. Qxe4#.
From The Chess Players’ Chronicle, 1872.
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March 7, 2014 | Categories: Puzzles
Walter Penney of Greenbelt, Md., offered this poser in the August 1969 issue of Word Ways: The Journal of Recreational Linguistics. Below are five groups of English words. Each group appears also in a foreign language. What are the languages?
- aloud, angel, hark, inner, lover, room, taken, wig
- alas, atlas, into, manner, pore, tie, vain, valve
- ail, ballot, enter, four, lent, lit, mire, taller
- banjo, chosen, hippo, pure, same, share, tempo, tendon
- ago, cur, dare, fur, limes, mane, probe, undo
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- Dutch: aloud — very old, angel — sting, hark — rake, inner — collector, lover — foliage, room — cream, taken — jobs, wig — wedge
- Finnish: alas — down, atlas — satin, into — enthusiasm, manner — mainland, pore — bubble, tie — road, vain — only, valve — awake
- French: ail — garlic, ballot — bumpkin, enter — to graft, four — oven, lent — slow, lit — bed, mire — gun sight, teller — to send out roots
- Japanese: banjo — sovereign, chosen — challenge, hippo — style of calligraphy, pure — pulley, same — shark, share — witticism, tempo — supplement, tendon — bowl of rice and fish
- Latin: ago — I act, cur — why, dare — to give, fur — thief, limes — boundary, mane — early in the morning, probe — thoroughly, undo — wave
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March 6, 2014 | Categories: Language, Puzzles
In projective geometry, every family of parallel straight lines intersects at an infinitely distant point. Chess problem composers in the former Yugoslavia have adapted this idea for the chessboard, adding four special squares “at infinity.”
Now a queen on a bare board, for example, can zoom off to the west (or east) and reach a square “at infinity” from which she attacks every rank on the board simultaneously from both directions. She might also zoom to the north (or south) to reach a different square at infinity; from this one she attacks every file simultaneously, again from both directions. Finally she can zoom to the northwest or southeast and attack all the diagonals parallel to a8-h1, or zoom to the northeast or southwest and attack all the diagonals parallel to a1-h8. These four “infinity squares,” plus the regular board, make up the field of play.
N. Petrovic created the problem below, published in Matematika Na Shahmatnoi Doske. White is to play and mate in at least two moves. Can you find the solution?
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After 1. Kg1 Black has no safe move:
- If he moves the king to e4 or f3, then the queen zooms westward to infinity and mates, attacking simultaneously from left and right on every rank.
- If he moves the king onto the g-file then the queen zooms southwestward to infinity, attacking simultaneously along all the southwest-northeast diagonals from both directions.
- If he moves the d-pawn then again the queen can mate from the southwest/northeast.
- If he moves the b6 knight (anywhere), then the queen zooms northwest and attacks along the northwest-southeast diagonals.
- If he moves the c7 knight then the queen zooms north and attacks along the files.
Given all this firepower, White’s main task is really to find a square where his king is safely out of the way. But that’s tricky. After 1. Kh1?, for example, Black has 1. … Nbd5! and there’s no mate.
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March 5, 2014 | Categories: Puzzles
A Russian problem from the 1999 Mathematical Olympiad:
A father wants to take his two sons to visit their grandmother, who lives 33 kilometers away. His motorcycle will cover 25 kilometers per hour if he rides alone, but the speed drops to 20 kph if he carries one passenger, and he cannot carry two. Each brother walks at 5 kph. Can the three of them reach grandmother’s house in 3 hours?
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Yes. The father sets out with his first son, leaving the second boy to walk alone. Father and son drive 24 kilometers, which takes 6/5 hours. Then the father puts down the first son and rides back to meet the second son 9 kilometers from home, which takes 3/5 hours. Finally he and the second son ride 6/5 hours more and reach grandmother’s house just as the first son is arriving there on foot. Each son spends 6/5 hours riding 24 kilometers and 9/5 hours walking 9 kilometers, so the three of them reach grandmother’s in exactly 3 hours.
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March 3, 2014 | Categories: Puzzles
By Ottmar Weiß. White to mate in two moves.
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The surprising 1. Qh3! leaves Black with no options. If he moves his knight then White can take the rook with mate, and if he moves the rook then White’s knight will mate.
From Teplitz-Schönauer Anzeiger, 1921.
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February 28, 2014 | Categories: Puzzles
From Sam Loyd:
Two children, who were all tangled up in their reckoning of the days of the week, paused on their way to school to straighten matters out.
“When the day after tomorrow is yesterday,” said Priscilla, “then ‘today’ will be as far from Sunday as that day was which was ‘today’ when the day before yesterday was tomorrow!”
On which day of the week did this puzzling prattle occur?
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“The two children were so befogged over the calendar that they had started on their way to school on Sunday morning!”
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February 24, 2014 | Categories: Puzzles
Six boys are accused of stealing apples. Exactly two are guilty. Which two? When the boys are questioned, Harry names Charlie and George, James names Donald and Tom, Donald names Tom and Charlie, George names Harry and Charlie, and Charlie names Donald and James. Tom can’t be found. Four of the boys who were questioned named one guilty boy correctly and one incorrectly, and the fifth lied outright. Who stole the apples?
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In presenting this problem in Ingenuity in Mathematics, Ross Honsberger gave a complex analysis, but his friend Scott Vanstone showed a simple way to visualize it, building on an observation by John Annulis:
Each boy is represented by a point, and each pair of boys named in an accusation are joined by a line. Now we know that each line has a thief at one end and an innocent at the other, except for one line, which has an innocent at both ends (due to the outright liar). Since four boys named one guilty boy correctly, the two points that represent the thieves are attached to a total of four lines. And because no boy made two accurate accusations, the thieves themselves are not joined by a line. The only two points that satisfy all these requirements indicate Charlie and James.
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February 23, 2014 | Categories: Puzzles
A group of four missionaries are on one side of a river, and four cannibals are on the other side. The two groups would like to exchange places, but there’s only one rowboat, and it holds only three people, and only one missionary and one cannibal know how to row, and the cannibals will overpower the missionaries as soon as they outnumber them, either on land or in the boat. Can the crossing be accomplished?
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Yes. First three cannibals cross; then the two rowers return, bearing an extra missionary; then the rowing missionary crosses with a cannibal and returns with two missionaries; and finally the rowing cannibal crosses:
Mmmm Cccc
MmmmCcc c
mmcc MmCc
Mmmccc mC
ccc MmmmC
Cccc Mmmm
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February 22, 2014 | Categories: Puzzles
By Sam Loyd. White to mate in two moves.
February 21, 2014 | Categories: Puzzles
A brainteaser by Chris Maslanka:
A packet of sugar retails for 90 cents. Each packet includes a voucher, and nine vouchers can be redeemed for a free packet. What is the value of the contents of one packet? (Ignore the cost of the packaging.)
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80 cents. It’s tempting to say that the answer is 81 cents, since paying for nine 90-cent packets will give us 10 packets total, meaning we’ve paid ($0.90 × 9) / 10 = $0.81 for each packet. But the free packet itself includes a voucher, which is worth something.
Maslanka suggests this thought experiment: Borrow a packet from a friend, then buy eight additional packets for $0.90 × 8 = $7.20. Consume the nine packets, redeem the nine vouchers for a free packet, and use this to repay your friend. You’ve consumed nine packets of sugar for $7.20, so each packet is worth 80 cents.
From Maslanka’s “A Bouquet of Brainteasers,” in Tribute to a Mathemagician, 2005.
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February 19, 2014 | Categories: Puzzles
A variation on yesterday’s puzzle:
Suppose there are six bottles of pills, and more than one of them may contain defective pills that weigh 6 grams instead of 5. How can we identify the bad bottles with a single weighing?
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Take 1 pill from the first bottle, 2 from the second, 4 from the third, 8 from the fourth, 16 from the fifth, and 32 from the sixth. This gives us 63 pills that ought to weigh 315 grams. As before, the bad pills will add some excess weight, but now that weight can tell us which combination of bottles is bad. Suppose the 63 pills weigh 336 grams, or 21 grams too much. That extra 21 grams can only have come from the fifth bottle (16 grams), the third (4 grams), and the first (1 gram). The same is true for any overage: Because any integer is uniquely expressible as the sum of powers of 2, the excess weight will always identify uniquely any combination of bad bottles.
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February 16, 2014 | Categories: Puzzles
An efficiency-minded pharmacist has just received a shipment of 10 bottles of pills when the manufacturer calls to say that there’s been an error — nine of the bottles contain pills that weigh 5 grams apiece, which is correct, but the pills in the remaining bottle weigh 6 grams apiece. The pharmacist could find the bad batch by simply weighing one pill from each bottle, but he hits on a way to accomplish this with a single weighing. What does he do?
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He numbers the bottles from 1 to 10, then takes one pill from bottle one, two pills from bottle two, and so on. This gives him a pile of 55 pills that ought to weigh 275 grams. Its difference from this weight can tell him which bottle is bad: If the pile is 1 gram overweight then the bad pills are in bottle one, if 2 grams then they’re in bottle two, and so forth.
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February 15, 2014 | Categories: Puzzles
This week’s puzzle has a twist: Imagine that the board has been rolled into a cylinder so that the a- and h-files are joined and pieces can move across the boundary. How can White mate in two moves?
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1. Ra5! The rook tours the fifth rank and returns to its original square, forcing 1. … Kc1 2. Ra1#. The normal-seeming 1. Rxa6, which would ordinarily lead to the same mate, here fails because Black can take the rook.
By Kuznecov and Plaskin, from Miodrag Petković, Mathematics and Chess, 1997.
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February 14, 2014 | Categories: Puzzles
The first 10 letters of the alphabet, ABCDEFGHIJ, form a cipher that conceals the name of a number less than 100. What is the number?
February 10, 2014 | Categories: Puzzles
- How does a deaf man indicate to a hardware clerk that he wants to buy a saw?
- How can you aim your car north on a straight road, drive for a hundred yards, and find yourself a hundred yards south of where you started?
- What runs fore to aft on one side of a ship and aft to fore on the other?
- A very fast train travels from City A to City B in an hour and a quarter. But the return trip, made under identical conditions, requires 75 minutes. Why?
- Does Canada have a 4th of July?
- Exhausted, you go to bed at 8 p.m., but you don’t want to miss an appointment at 10 a.m. the next day, so you set your alarm clock for 9. How many hours do you sleep?
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- He says, “I want to buy a saw.”
- Drive backward.
- Its name.
- An hour and a quarter is 75 minutes.
- Yes.
- One.
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February 9, 2014 | Categories: Puzzles
