Projective Chess

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.”

projective chess 1

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?

projective chess 2

	Select Click for Answer
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.

Click for Answer

March 5, 2014March 16, 2014 | Puzzles

Reunion

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?

	Select Click for Answer
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.

Click for Answer

March 3, 2014March 9, 2014 | Puzzles

Black and White

By Ottmar Weiß. White to mate in two moves.

	Select Click for Answer
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.

February 28, 2014March 9, 2014 | Puzzles

Calendar Trouble

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?

	Select Click for Answer
“The two children were so befogged over the calendar that they had started on their way to school on Sunday morning!”

February 24, 2014March 2, 2014 | Puzzles

Crime Story

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?

	Select Click for Answer
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.

Click for Answer

February 23, 2014March 2, 2014 | Puzzles

Playing With Food

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?

	Select Click for Answer
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

February 22, 2014March 2, 2014 | Puzzles

Black and White

loyd chess problem

By Sam Loyd. White to mate in two moves.

	Select Click for Answer
1. Qf8! and the queen will mate. From the Chicago Times, 1879.

February 21, 2014March 2, 2014 | Puzzles

Sweet Reason

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.)

	Select Click for Answer
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.

Click for Answer

February 19, 2014March 2, 2014 | Puzzles

The Pill Scale (Part 2)

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?

	Select Click for Answer
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.

Click for Answer

February 16, 2014February 23, 2014 | Puzzles

The Pill Scale (Part 1)

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?

	Select Click for Answer
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.

February 15, 2014February 23, 2014 | Puzzles