Author Archives: Jim Stevenson

Pool Paving Problem

In my search for problems I decided to purchase Dan Griller’s GCSE problem book  mentioned in the Cube Roots Problem.  I am still a bit confused about the purpose of the GCSE exam and who it is for, since the other problems in Griller’s book are often as challenging or more so than the cube roots problem.  It is hard to believe students not pursuing college level degrees could solve these problems.  (Grades 8 and 9 referred to in the subtitle of the book must indicate something other than US grades 8 and 9, since the exams are aimed at 16 year-olds, not 13 and 14 year-olds.)

Supposedly the problems in Griller’s book are nominally arranged in increasing order of difficulty from problem 1 to problem 75.  However it seemed to me that there were challenging problems scattered throughout and the last problem was not all that much harder than earlier ones.  And many of them had a whiff of Coffin Problems—they seemed impossible at first (Problem 44: Construct a 67.5° angle!).  I don’t know how many problems are on the exam or how long the exam is, but anyone taking a timed exam does not have the leisure to mull over a problem.  The student only has a few minutes to come up with an approach and clever insights are rare under the circumstances.  Anyway, here is the last problem in the book.

“Problem 75.  A square pond of side length 2 metres is to be surrounded by twelve square paving stones of side length 1 metre.

(a)  The first design is constructed with a circle whose centre coincides with the centre of the pond.  Calculate exactly the total dark grey area for this design.

(b)  The second design is similar.  Calculate exactly the total dark grey area for this second design.”

Answer.

See the Pool Paving Problem for solutions.

The Squirrel Puzzle

For a change of pace, here is an early puzzle from Alex Bellos in The Guardian.

“Happy New Year guzzlers!  Today’s first problem concerns squirrels. Have a nibble—it’s not too hard a nut to crack.

The Squirrel King has buried the Golden Acorn beneath one of the squares in this 6x6 grid. Three squirrels—Black, Grey and Red—are each standing on a square in the grid, as illustrated.

(Note: for the purposes of today, squirrels can speak, hear, read, count and are perfect logicians. They can also move in any direction horizontally and vertically, not just the direction these cartoons are facing. They all can see where each other is standing, and the cells in the grid are to be considered squares.)

The Squirrel King hands each squirrel a card, on which a number is written. The squirrels can read only the number on their own card. The King tells them: ‘Each card has a different number on it, and your card tells you the number of steps you are from the square with the Golden Acorn. Moving one square horizontally or vertically along the grid counts as a single step.’ (So if the acorn was under Black, Black’s card would say 0, Grey’s would say 4, and Red’s 5. Also, the number of steps given means the shortest possible number of steps from each squirrel to the acorn.)

The King asks them: ‘Do you know the square where the Golden Acorn is buried?’ They all reply ‘no!’ at once.

Red then says: ‘Now I know!’

Where is the Golden Acorn buried? …”

See the Squirrel Puzzle

Sum Of Squares Puzzle

Yet another interesting problem from Presh Talwalkar.

“Two side-by-side squares are inscribed in a semicircle.  If the semicircle has a radius of 10, can you solve for the total area of the two squares? If no, demonstrate why not. If yes, calculate the answer.”

This puzzle shares the characteristics of all good problems where the information provided seems insufficient.

Answer.

See the Sum of Squares Puzzle for solutions.

Family Values

Here is a collection of puzzles from the great logic puzzle master Raymond Smullyan in a “Brain Bogglers” column for the 1996 Discover magazine.

  1. ELDON WHITE HAS FOUR DOGS. One day he put out a bowl of dog biscuits. The eldest dog came first and ate half the biscuits plus one more. Then the next dog came and ate half of what he found plus one more. Then the next one came and ate half of what she found plus one more. Then the little one came and ate half of what she found and one more, and that finished the biscuits. How many biscuits were originally in the bowl?
  2. Eldon once bought a very remarkable plant, which, on the first day, increased its height by a half, on the second day by a third, on the third day by a quarter, and so on. How many days did it take to grow to 100 times its original height?
  3. In addition to four dogs, Eldon has four children. The youngest, Betty, is nine years old; then there are twin boys, Arthur and Robert; and finally there’s Laura, the eldest, whose age is equal to the combined ages of Betty and Arthur. Also, the combined ages of the twins are the same as the combined ages of the youngest and the eldest. How old is each child?
  4. “How about a riddle?” asked Robert. “Very well,” said Eldon. “What is it that is larger than the universe, the dead eat it, and if the living eat it, they die?”

Answer.

See Family Values  for solutions.

Another Cube Slice Problem

This is a problem from the UKMT Senior Challenge for 2019.  (It has been slightly edited to reflect the colors I added to the diagram.)

“The edge-length of the solid cube shown is 2.  A single plane cut goes through the points Y, T, V and W which are midpoints of the edges of the cube, as shown.

What is the area of the cross-section?

A_√3_____B_3√3_____C_6_____D_6√3_____E_8”

Answer.

See Another Cube Slice Problem for solutions.

Serious Series

The following problem comes from a 1961 exam set collected by Ed Barbeau of the University of Toronto.  The discontinued exams (by 2003) were for 5th year Ontario high school students seeking entrance and scholarships for the second year at a university.

“If sn denotes the sum of the first n natural numbers, find the sum of the infinite series

.”

Unfortunately, the “Grade XIII” exam problem sets were not provided with answers, so I have no confirmation for my result.  There may be a cunning way to manipulate the series to get a solution, but I could not see it off-hand.  So I employed my tried and true power series approach to get my answer.  It turned out to be power series manipulations on steroids, so there must be a simpler solution that does not use calculus.  I assume the exams were timed exams, so I am not sure how a harried student could come up with a quick solution.  I would appreciate any insights into this.

Answer.

See Serious Series for a solution.

(Update 1/18/2021, 9/6/2024) Other Solutions
Continue reading →

Tree Trunk Puzzle

Here is another problem (slightly edited) from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).

“Holmes and I were walking along a sleepy lane in Hookland, making our way back to the inn at which we had secured lodgings after scouting out the estates of the supposed major, C. L. Nolan. Up ahead, a team of horses were slowly pulling a chained tree trunk along the lane. Fortunately it had been trimmed of its branches, but it was still an imposing sight.

When we’d overtaken the thing, Holmes surprised me by turning sharply on his heel and walking back along the trunk. I stopped where I was to watch him. He continued at a steady pace until he’d passed the last of it, then reversed himself once more, and walked back to me.

‘Come along, old chap,’ he said as he walked past. Shaking my head, I duly followed.

‘It took me 140 paces to walk from the back of the tree to the front, and just twenty to walk from the front to the back,’ he declared.

‘Well of course,’ I said. ‘The tree was moving, after all.’

‘Precisely,’ he said. ‘My pace is one yard in length, so how long is that tree-trunk?’

Can you find the answer?”

Answer.

See the Tree Trunk Puzzle for solutions.

The Triangle of Abū’l-Wafā’

I found an interesting geometric statement in a paper of Glen Van Brummelen cited in the online MAA January 2020 issue of Convergence:

“For instance, Abū’l-Wafā’ describes how to embed an equilateral triangle in a square, as follows: extend the base GD by an equal distance to E. Draw a quarter circle with centre G and radius GB; draw a half circle with centre D and radius DE. The two arcs cross at Z. Then draw an arc with centre E and radius EZ downward, to H. If you draw AT = GH and connect B, H, and T, you will have formed the equilateral triangle.”

So the challenge is to prove this statement regarding yet another fascinating appearance of an equilateral triangle.

See the Triangle of Abū’l-Wafā’

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