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Can You Solve The Diluted Wine Puzzle? Famous 16th Century Math Problem

A servant has a method to steal wine. He removes 3 cups from a barrel of wine and replaces it with 3 cups of water. The next day he wants more wine, so he does the same thing: he removes 3 cups from the same barrel (now with diluted wine) and replaces it with 3 cups of water. The following day he repeats this one more time, so he has drawn 3 times from the same barrel and has poured back 9 cups of water.

At this point the barrel is 50% wine and 50% water.

How many cups of wine were originally in the barrel?

Watch the video for a solution.

Can You Solve The Diluted Wine Puzzle?

Or keep reading for a solution.

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Answer To The Diluted Wine Puzzle

Suppose the barrel started with x cups of wine. We will keep track of how much wine is in the barrel and the concentration of wine in the barrel. Initially the barrel has x cups of wine and it is 100% wine (a concentration of 1).

Wine Amount (starting), Wine Concentration (starting)
x, 1

The servant first takes 3 cups of wine and replaces them with water. The amount of wine left is x – 3, and the concentration is the amount of wine, x – 3, divided by the total volume of liquid, which is x.

Wine Amount (step 1), Wine Concentration (step 1)
x – 3, (x – 3)/x

x – 3, 1 – 3/x

The servant then removes 3 cups from the barrel, and each cup contains a concentration 1 – 3/x of wine. The amount of wine left is x – 3 – 3(1 – 3/x)/x, and the concentration is the amount of wine, x – 3 – 3(1 – 3/x)/x, divided by the total volume of liquid, which is x.

Wine Amount (step 2), Wine Concentration (step 2)
x – 3 – 3(1 – 3/x), (x – 3 – 3(x – 3)/x)/x

x – 6 + 9/x, (x – 6 + 9/x)/x

x – 6 + 9/x, (1 – 3/x)2

The servant finally removes 3 more cups from the barrel, and each cup contains a concentration (1 – 3/x)2 of wine. The amount of wine left is x – 6 + 9/x – 3(1 – 3/x)2, and the concentration is the amount of wine divided by the total volume of liquid, which is x.

Wine Amount (step 3), Wine Concentration (step 3)
x – 6 + 9/x – 3(1 – 3/x)2, (x – 6 + 9/x – 3(1 – 3/x)2)/x

x – 9 + 27/x – 27/x2, (x – 9 + 27/x – 27/x2)/x

x – 9 + 27/x – 27/x2, (1 – 3/x)3

The final concentration should be equal to 50%, or 1/2.

(1 – 3/x)3 = 1/2

(1 – 3/x) = 1/21/3

21/3x – 3(21/3) = x

x(21/3 – 1) = 3(21/3)

x = 3(21/3)/(21/3 – 1) ≈ 14.54

Thus the original barrel contained approximately 14.5 cups of wine.

There is a shortcut to solving this problem! You can save many steps by noticing the concentration is 1 – 3/x after the first step.

Wine Amount (step 1), Wine Concentration (step 1)
x – 3, 1 – 3/x

The subsequent steps iterate the same process of removing 3 cups and then diluting the wine with water. Accordingly, the wine is diluted by the same percentage in each step. To find the new concentration, multiply by the factor 1 – 3/x.

Wine Concentration (step 2)
(1 – 3/x)2

Wine Amount (step 3), Wine Concentration (step 3)
(1 – 3/x)3

Now we can set the concentration equal to 1/2 and find the answer as before.

Sources

I read this problem in Famous Puzzles of Great Mathematicians by Miodrag S. Petkovic.

The puzzle appears in Niccolo Tartaglia’s work “General Trattato di Numeri” (1556).

StackExchange has the idea to multiply concentrations
http://puzzling.stackexchange.com/questions/28776/turning-wine-into-water/28781

See the article here:
Can You Solve The Diluted Wine Puzzle? Famous 16th Century Math Problem

Chess Puzzles – 365Chess.com

The best way to improve your chess is practicing! And the funniest way to practice is solving puzzles! Train your tactical abilities solving all kind of situations like the ones you’ll find over a chess board at a real game.

You can play in unrated mode, just for fun, without recording your solving history. Or you can have full tracking of your progress solving in Rated mode. If you play in this mode you’ll also obtain problems according to your current level.

Solve chess problems taken from real games. You should find the best continuation for you opponent’s moves.

Train your abilities facing chess studies from the most complete database of problems, created by the greatest composers.

Good news! Only a few minutes every day solving puzzles and you will boost your chess!

If you want to improve your game the fastest, easiest and most fun way, start solving puzzles right now.

One thing that is absolutely undisputed in chess training philosophy is: solve tactical chess puzzles regularly and you’ll get better and better everyday.

Train your abilities facing chess studies from the most complete database of problems, created by the greatest composers.

“My fascination for studies proved highly beneficial, it assisted the development of my aesthetic understanding of chess, and improved my endgame play” Vasily Smyslov

“I always urge players to study composed problems and endgames” Pal Benko.

Just a few samples of recently added games to our database.

Chess Composition

Chess Problem

Chess Composition

Go here to see the original:
Chess Puzzles – 365Chess.com

Can You Solve The Mind-Bending Self-Counting Sentence Riddle? Sunday Puzzle

In this sentence, the number of occurrences
of the digit 0 is __,
of the digit 1 is __,
of the digit 2 is __,
of the digit 3 is __,
of the digit 4 is __,
of the digit 5 is __,
of the digit 6 is __,
of the digit 7 is __,
of the digit 8 is __, and
of the digit 9 is __.

The challenge is to fill in the blanks with numerical digits (1, 2, etc.) so that the sentence is true.

There are 2 solutions. Can you find them?

Watch the video for a solution.

Can You Solve The Mind-Bending Self-Counting Sentence Riddle?

Or keep reading.

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Answer To The Mind-Bending Self-Counting Sentence Riddle

The answers can be found by experimentation. To start, each of the digits appears at least once, so fill in 1 for every line (except the digit 1 which is left blank at the moment).

TEST CASE 1
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is __,
of the digit 2 is 1,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

How many times does the digit 1 occur? The sentence already contains 10 occurrences of the digit 1, so let us try that value.

TEST CASE 2
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 10,
of the digit 2 is 1,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

This sentence is false, however, because the digit 1 appears 11 times in total (including the “1” in the number “10.”). We can try to adjust the sentence to 11 occurrences of the digit 1.

TEST CASE 3
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 11,
of the digit 2 is 1,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

But now this sentence is also false because it contains 12 occurrences of the digit 1 (the number “11” adds another “1” to the count). If we increase the count to 12, then the occurrences of the digit 2 would increase to 2, thereby reducing one of the occurrences of the digit 1. So let’s try keeping the occurrences of the digit 1 as 11, and increasing the occurrences of the digit 2 to be 2.

TEST CASE 4 = Solution One
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 11,
of the digit 2 is 2,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

This sentence is true and is the first solution to the puzzle. The digit 1 occurs 11 times, the digit 2 occurs 2 times, and every other digit occurs 1 time.

But there’s another solution this problem too! Let us adjust the above solution by considering if the digit 1 only has 9 occurrences to avoid having a double-digit number.

TEST CASE A
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 9,
of the digit 2 is 2,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

The sentence if false because there are 2 occurrences of the digit 9. If the digit 9 occurs 2 times, then that means the digit 2 has to occur 3 times total. And that also means the digit 3 occurs 2 times.

TEST CASE B
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 9,
of the digit 2 is 3,
of the digit 3 is 2,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 2.

This sentence is false; now there are only 7 occurrences of the digit 1. To fix the sentence, we can swap the occurrences of the digits 7 and 9. That is, write that the digit 7 occurs 2 times, and write that the digit 9 occurs 1 time.

TEST CASE C = Solution Two
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 7,
of the digit 2 is 3,
of the digit 3 is 2,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 2,
of the digit 8 is 1, and
of the digit 9 is 1.

This sentence is true and it is the second solution. It is in fact the unique solution if all the digits occur less than 10 times.

And there you have it: two true sentences that count themselves!

Sources and further reading
Douglas Hofstadter Metamagical Themas on Google Books
https://books.google.com/books?id=o8jzWF7rD6oC&lpg=PA390&ots=jRCg7rLufs&dq=The%20number%20of%200s%20in%20this%20sentence%20Metamagical%20Themas&pg=PA390#v=onepage&q&f=false

Math Central December 2002 Problem. Proof there are only 2 solutions
http://mathcentral.uregina.ca/mp/archives/previous2002/dec02sol.html

Self-descriptive sentences (and two solutions described)
http://www.cut-the-knot.org/ctk/SelfDescriptive.shtml

Unique solution if all occurrences less than 10
http://math.stackexchange.com/questions/19061/puzzle-digit-x-appears-y-times-on-this-piece-of-paper?rq=1

Sketch of proof that there are only 2 solutions
https://www.reddit.com/r/math/comments/2oh9s9/heres_a_maths_puzzle_my_friend_posted_on_facebook/cmn8f26/

Solving by iterative process
http://web.archive.org/web/20120428023510/http://www.lboro.ac.uk/departments/ma/gallery/selfref/index.html

Variations for self-counting sentences
http://lkozma.net/blog/self-counting-sentences/
http://lkozma.net/blog/self-counting-sentences-ii/

Autogram (self-counting with letters)
https://en.wikipedia.org/wiki/Autogram

Variation with letters (autogram)
http://www.braingle.com/news/hallfame.php?path=language/english/sentences/self.ref/self.ref.letters.p&sol=1

More here:
Can You Solve The Mind-Bending Self-Counting Sentence Riddle? Sunday Puzzle

Chess Problems, Puzzles, & Compositions – Chessopolis

135 Studies And Problems A whole career of composed problems from J.T. Sanderse.

Archives des problmistes In French. Problem site from Quebec.

Beautiful Chess Problems Only a few at the moment, but more promised.

Brunos Chess Problem of the Day Just like it sounds.

BYCA Chess Puzzles In addition to the collection of puzzles, features a puzzles league where you can participate in a friendly solving competition.

Chess Composition Microweb Featuring various sections on composed chess problems & problem-solving.

Chess Online Problems Problems galore. More added frequently.

Chess Problems Online A decent collection, intelligently displayed, from Mate in One to Mate in Six.

Chess Problems From Around The World Problems from everywhere.

Chess Problems Of 1001 Years Ago From the Chess Variants Homepage really old problems mainly of historical interest.

Chess Problems Unlimited Not quite unlimited, there are but several.

Chesspros Chess World Problems and instructive games.

Chess Puzzles by GMs Taken from the games of the greats.

Chess Puzzles Collection A nice collection taken from real games. Includes an index of the puzzles by theme.

CHEST Source code (in ANSI C) for a problem solving program.

Classic Chess Problems 2 and 3 movers from renowed composers.

Endgame Of The Day An endgame study a day. Also available as a mailing list.

Exeter Chess Club: Studies and Problems Intro to the world of problems.

GorFo Problem Chess Pages Has a problem of the month composed by the author, plus an archive of past problems.

Key Moves A new chess combination (from recent games) every week for your instruction and amusement.

Loiodices Chess Problems Lots of training positions to download.

Manolis Stratakis Chess Problems Page Pretty good page with plenty of problems and puzzles.

MateMaster Shareware software for solving chess problems.

Mater Will find any mate thats there, eventually.

Mat Plus The best chess problems from magazine Mat Plus.

Practical Chess Endgame Solve the endgame of the week.

Problemiste Chess Problem software.

Public Domain Chess Composition Books 19th Century (mostly) chess problem books in PDF format so you can print them out yourself. Nice!

Retractor Software for creating Retrograde Analysis chess problems.

The Retrograde Analysis Corner Retro problems and resources for retro enthusiasts.

Sack The King! A chess problem a day, with special animated interface to display the solution.

Solving Chess Tracking the world of competitive chess problem solving.

Torsten Lin Problem Chess Pages Problems and articles from noted composer. Some of the text in German.

Vincents Chess Problems Page Dozens of problems to pour over.

See the article here:
Chess Problems, Puzzles, & Compositions – Chessopolis

Grade 12 Geometry Problem From Australia. The Marshmallow Chocolate Dessert Volume. Sunday Puzzle

A marshmallow has the shape of a cylinder with a diameter of 5 and a height of 3.

A dessert is made with 24 marshmallows. There are two stacked layers, and each layer consists of 3 rows of 4 marshmallows each. The gaps between the marshmallows are filled with chocolate, as shown in the diagram.

marshmallow-area-problem-from-grade-12-australia-test-diagram

What is the volume of chocolate in the dessert? This problem was asked to grade 12 students in an Australian test. Give it a try, and watch the video for a solution.

Can You Solve This Grade 12 Geometry Problem From Australia? The Marshmallow Chocolate Puzzle

Or keep reading.

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Answer To The Marshmallow Chocolate Dessert Volume

The volume of chocolate is equal to the area of a cross section times the height.

Volume = (area of cross section) × (height)

The height of chocolate is equal to the height of 2 marshmallows, so the height is equal to 6.

Volume = (area of cross section) × 6

The area of a cross section is equal to the 6 gaps between the marshmallows, so the volume formula is directly related to the area of a single gap.

Volume = 6 × (area of a gap) × 6
Volume = 36 × (area of a gap)

What is the area of the gap? It is the area between four circles. As each marshmallow has a diameter of 5, each circle has a radius r = 5/2.

Now draw a square connecting the centers of the four circles. The side of the square, s, spans the radius of two adjacent circles, so the side has a length equal to the diameter, 5, of a circle.

The area of the gap is then the area of the square minus the area of 4 quarter-circles. But 4 quarter circles together equal a single circle. Therefore the area of the gap is the area of the square minus the area of a circle.

marshmallow-problem-area-of-gap

area of gap = s2 – πr2
area of gap = 25 – 25π/4

This can be substituted back into the volume equation to arrive at the answer.

Volume = 36 × (area of a gap)
Volume = 36 × (25 – 25 π/4)
Volume = 900 – 225 π ≈ 193.1

Sources
http://www.smh.com.au/nsw/hsc-2016-mixed-verdict-on-maths-exam-from-students-but-formula-reference-sheet-a-bonus-20161021-gs7inz.html

ATAR Notes
http://atarnotes.com/forum/index.php?topic=168134.0

The formula volume = (area of cross section) x height is an example of Cavalieri’s principle
https://en.wikipedia.org/wiki/Cavalieri%27s_principle

Continued here:
Grade 12 Geometry Problem From Australia. The Marshmallow Chocolate Dessert Volume. Sunday Puzzle

The Day After Tomorrow Is Yesterday Riddle

This problem is popular on Facebook and social media.

“When the day after tomorrow is yesterday, then today will be as far from Wednesday as today was from Wednesday when the day before yesterday was tomorrow. What day is today?”

Can you figure it out? Watch the video for a solution.

The Day After Tomorrow Is Yesterday Riddle – Can You Solve It?

Or keep reading.

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Answer To The Day After Tomorrow Is Yesterday Riddle

The puzzle can be re-written as follows, to emphasize the word “today” refers to 3 different days of the week.

When the day after tomorrow is yesterday, then ‘today’ will be as far from Wednesday as *today was from Wednesday when the day before yesterday was tomorrow. What day is today?

The question is written on today and we want to solve for the day of the week for today. Let’s write x for the day of the week of today.

Now we can solve for the other days of the week.

The day ‘today’ takes place when the day after tomorrow would be referred to as yesterday. Relative to today, tomorrow is x + 1, and the day after tomorrow is x + 2. One more day forward, x + 3, is when the day after tomorrow would be called yesterday. In summary, ‘today’ takes place 3 days after today.

Similarly, the day *today takes place when the the day before yesterday would be referred to as tomorrow. Relative to today, yesterday is x – 1, and the day before yesterday is x – 2. One more day backwards, x – 3, is when the day before yesterday would be called tomorrow. In summary, *today takes place 3 days before today.

The riddle also specifies ‘today’ is as far from Wednesday as *today. That is, both days need to be an equal distance from Wednesday.

Since ‘today’ and *today are an equal distance of 3 days from x, this implies x = Wednesday. That would mean ‘today’ is 3 days after Wednesday and *today would be 3 days before Wednesday.

Therefore, today is Wednesday.

Source: I saw this riddle several places, including Reddit Riddles:
https://www.reddit.com/r/riddles/comments/389uhu/when_the_day_after_tomorrow_is_yesterday/

Read this article:
The Day After Tomorrow Is Yesterday Riddle

The Rectangle Ratio, A Surprising Answer – Sunday Puzzle

Start with a square with length 1; label it R1. Create a new rectangle as follows: attach a rectangle of area 1 to the right side and then attach a rectangle of area 1 to the top side. The border of this shape is a new rectangle, R2.

rectangle-ratio-r1-to-r2

Starting with R2, we can attach another pair of rectangles of area 1 to form a new rectangle, R3.

rectangle-ratio-r2-to-r3

From Rn – 1, attach a rectangle of area 1 to the right side, and then a rectangle of area 1 to the top. The border of this shape is rectangle Rn.

rectangle-ratio-rn-1-to-rn

The problem is: what is the ratio of the length, Ln, to the width, Wn as n goes to infinity?

Watch the video for a solution.

Can You Solve The Rectangle Ratio Puzzle? The Surprising Answer

Or keep reading.

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Answer To The Surprising Rectangle Ratio

Rectangles added to the right side have a width equal to the previous rectangle’s width. So that the attached rectangle has area 1, the new length added must equal the reciprocal of the width of the previous rectangle. This observation provides a recursive formula for the length.

Ln + 1 = Ln + 1/Wn

rectangle-ratio-ln-recursive

Similarly, rectangles added to the top side increase the previous rectangle’s width by the reciprocal of the width of the intermediate rectangle formed from adding a rectangle to the side. The length of the intermediate rectangle is the length of the next rectangle, whose formula was derived above.

Wn + 1 = Wn + 1/Ln + 1

Wn + 1 = Wn + 1/(Ln + 1/Wn)

Wn + 1 = Wn + Wn/(LnWn + 1)

Note that LnWn is the area of rectangle Rn. The original square has area 1. In each step there are a pair of rectangles added, each with area 1, so the new rectangle has an area that is 2 larger than the previous rectangle. The rectangles will have areas 1, 3, 5, 7, etc, so the area of Rn equals 2n – 1.

Substituting this value gives the following recursive formula for the width:

Wn + 1 = Wn + Wn/(2n)

rectangle-ratio-wn-recursive

Now let’s calculate the ratio of the length to the width.

The ratio formula

We use the recursive formulas above to get the following:

Ln + 1/Wn + 1

= (Ln + 1/Wn)/(Wn + Wn/(2n))

= (LnWn + 1)/(Wn2[1 + 1/(2n)])

= (2n)(LnWn + 1)/(Wn2(2n + 1))

We can simplify this formula in two ways. First, LnWn is the area of rectangle Rn, which is 2n – 1.

Second, Wn2 equals the rectangle’s area divided by the ratio of its length to its width. That is, it equals (LnWn)/(Ln/Wn). The numerator is the area of the rectangle, which is 2n – 1.

So here is the simplified formula:

(2n)(2n – 1 + 1)/[(2n – 1)/(Ln/Wn)(2n + 1)]

= (Ln/Wn) * (2n)(2n)/[(2n – 1)(2n + 1)]

= (ratio L/W for previous rectangle) * [2n/(2n – 1)][2n/(2n + 1)]

We now have a recursive formula for calculating the ratio of the length to the width for the rectangle in step n.

The initial shape is a square with L1 = W1 = 1. The next rectangle (n = 2) has a ratio:

(1/1) * (2)(2)/[(2 – 1)(2 + 1)]

= (2/1)(2/3)

The next rectangle has the ratio that is (4/3)(4/5) multiplied by this previous ratio:

(2/1)(2/3)(4/3)(4/5)

The next rectangle has the ratio that is (6/5)(6/7) multiplied by this previous ratio:

(2/1)(2/3)(4/3)(4/5)(6/5)(6/7)

We can see the pattern continues, where each next rectangle has a ratio that is [2n/(2n – 1)][2n/(2n + 1)] multiplied by the previous ratio.

This leads to the general pattern for the ratio of rectangle Rn:

(2/1) · (2/3) · (4/3) · (4/5) · (6/5) · (6/7) · … · [2n/(2n – 1)][2n/(2n + 1)]

The product, as n goes to infinity, is known as the Wallis product, and it has a limiting value equal to π/2 ≈ 1.571.

The ratio of the length to the width of Rn converges to π/2, which is the area of a semi-circle with radius 1.

Proof Of Wallis Product

I made a separate post with a video to prove this result. Check them out:

The Wallis Product Formula Proof

Thanks to Patrons!

Kyle
Brian M. Mooney
Shrihari Puranik

You can support me and this site at Patreon.

Sources
I first read about this problem on Reddit Math Riddles
https://www.reddit.com/r/mathriddles/comments/53ubuo/fun_little_calculus_problem/

The problem on Reddit is from James Stewart’s Calculus in the section about integration by parts.
https://books.google.com/books?id=AavjDHGwGpIC&pg=SL1-PA146&dq=wallis+product&hl=en&sa=X&ved=0ahUKEwihidHA7avPAhXE8CYKHbvNBdI4FBDoAQhRMAk#v=onepage&q=wallis%20product&f=false

Physics Forum illustrates the pattern
https://www.physicsforums.com/threads/wallis-product.73921/

Wallis product
https://en.wikipedia.org/wiki/Wallis_product

See the article here:
The Rectangle Ratio, A Surprising Answer – Sunday Puzzle

Chess puzzle – Wikipedia

A chess puzzle is a puzzle in which knowledge of the pieces and rules of chess is used to solve logically a chess-related problem. The history of chess puzzles reaches back to the Middle Ages and has evolved since then.

Usually the goal is to find the single best, ideally aesthetic move or a series of single best moves in a chess position, which was created by a composer or is from a real game. But puzzles can also set different objectives. Examples include deducing the last move played, the location of a missing piece, or whether a player has lost the right to castle. Sometimes the objective is antithetical to normal chess, such as helping (or even compelling) the opponent to checkmate one’s own king.

Whereas the term chess puzzle refers broadly to any puzzle involving aspects of chess, a chess problem is an orthodox puzzle (see below) in which one must play and win or draw a game, starting with a certain composition of pieces on the chess board, and playing within the standard rules of chess.

Orthodox chess problems involve positions that can arise from actual game play (although the process of getting to that position may be unrealistic). The most common orthodox chess puzzle takes the form of checkmate in n moves. The puzzle positions are seldom similar to positions from actual play, and the challenge is not to find a winning move, but rather to find the (usually unique) move which forces checkmate as rapidly as possible.

Heterodox chess problems involve conditions that are impossible with normal play, such as multiple kings or chess variants, while fairy chess problems employ pieces not used in orthodox chess, such as the amazon (a piece combining the powers of the queen and the knight).

Chess puzzles can also be regular positions from a game (with normal rules), usually meant as training positions, tactical or positional, from all phases of the game (openings, middlegame of endings). These are known as tactical puzzles. They can range from a simple “Mate in one” combination to a complex attack on the opponent’s king. Solving tactical chess puzzles is a very common chess teaching technique. Although it is unlikely that the same position will occur in a game the student plays, the recognition of certain patterns can help to find a good move or plan in another position.

Some chess problems, like the Eight queens puzzle or the Knight’s Tour, have connections to mathematics, especially to graph theory and combinatorics. Many famous mathematicians have studied such problems, including Euler, Legendre, and Gauss. Besides finding a solution to a particular puzzle, mathematicians are usually interested in counting the total number of possible solutions, finding solutions with certain properties, and generalization of the problems to nn or rectangular boards.

Read more:
Chess puzzle – Wikipedia

Which Can Size Is Best? The Optimal Cylinder Riddle. Sunday Puzzle

This is a problem from Math Puzzles Volume 2 that I had not posted about before.

Imagine you want a cylindrical can to contain a fixed volume V. There are many possible shapes for the can that could hold a volume of V. The cylinder can either be wide and short (a large radius r and a small height h) or narrow and tall (a small radius r and a large height h).

best-can-sizes

Of all the different shapes, which one has the least surface area? That is, which can uses the least material to contain the volume? Solve for the ratio of the height to the radius.

Watch the video for a solution.

What’s The Best Soup Can Size? The Optimal Cylinder Puzzle

Or keep reading for an explanation.

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Answer To Which Can Size Is Best? The Optimal Cylinder Riddle

For a can with radius r and height h, the volume is equal to:

V = (area of circle)(height)

V = πr2h

The can’s surface area is equal to the sum of the area of two circles (the top and bottom of can) and the area of the side, which is a rectangle with one dimension h and the other dimension the circumference of the top of the can.

S = 2(area of circle) + (area side)

S = 2πr2 + 2πrh

For a fixed volume V, we can solve for the height of the can:

V = πr2h

h = V/(πr2)

We can substitute this into the surface area equation to express the surface area as a function of the radius.

S = 2πr2 + 2πrh

S = 2πr2 + 2V/r

The surface area can be made arbitrarily large by increasing the radius of the can, so there is no maximum value.

In order to find the minimum, we can take the derivative with respect to the radius r and set it equal to 0.

S‘(r) = 4πr – 2V/r2 = 0

V = 2πr3

Notice the second derivative is positive (4π + 4V/r3 > 0 for r > 0), so this is a minimum value.

This is the formula for the volume with minimum surface area. We know that for any can the volume is V = πr2h. We can therefore set those two equations equal to each other to solve for h.

r3 = πr2h

h = 2r

h/r = 2

This means the ratio of the can’s height to its radius should equal 2 to give a minimum surface area for the can.

Sources

I learned about this fact from Aha! Solutions by Martin Erickson. Link to book on Amazon: http://amzn.to/1sOwMnN

I also came across this problem from Datagenetics, which also did an analysis of actual cans. Interestingly many food cans from his pantry were pretty close to the optimal ratio, using only 2% more material or less than the optimal can size. Tuna was the exception as the can used 7% more than the optimal size.

http://datagenetics.com/blog/august12014/index.html

Original post:
Which Can Size Is Best? The Optimal Cylinder Riddle. Sunday Puzzle


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