Here is another Quantum math magazine Brainteaser.
“Two wheels roll toward each other with identical angular velocity. At the moment of collision they contact each other at the same points that touched the ground before they began rolling. Could the radii of the wheels differ?”
See the Rolling Wheels Puzzle for solution.
For his Monday Puzzle in the Guardian Alex Bellos provided a seemingly impossible puzzle from the 1983 British teenager quiz show Blockbusters.
“In the much-missed student quiz show Blockbusters, teenagers would ask host Bob Holness for a letter from a hexagonal grid. How we laughed when a contestant asked for a P! Holness would reply with a question in the following style: What P is an area of cutting edge mathematical research and also a process in the making of an espresso? The answer is the subject of today’s puzzle: percolation.
Today’s perplexing percolation poser concerns the following Blockbusters-style hexagonal grid:
The grid above shows a 10×10 hexagonal tiling of a rhombus (i.e. a diamond shape), plus an outer row that demarcates the boundary of the rhombus. The boundary row on the top right and the bottom left are coloured blue, while the boundary row on the top left and the bottom right are white.
If we colour each hexagon in the rhombus either blue or white, one of two things can happen. Either there is a path of blue hexagons that connects the blue boundaries, such as here:
Or there is no path of blue hexagons that connects the blue boundaries, such as here:
There are 100 hexagons in the rhombus. Since each of these hexagons can be either white or blue, the total number of possible configurations of white and blue hexagons in the rhombus is 2 x 2 x … x 2 one hundred times, or 2100, which is about 1,000,000,000,000,000,000,000,000,000,000.
In how many of these configurations is there a path of blue hexagons that connects the blue boundaries?
The answer requires a simple insight. Indeed, it is the insight on which the quiz show Blockbusters relied.
For clarification: a path of hexagons means a sequence of adjacent hexagons that are the same colour.”
See the Blockbusters Problem for solution.
Here is a mind-numbing logic puzzle from Futility Closet.
“A puzzle by H.A. Thurston, from the April 1947 issue of Eureka, the journal of recreational mathematics published at Cambridge University:
Five people make the following statements:—
Which of these statements are true and which false? It will be found on trial that there is only one possibility. Thus, prove or disprove Fermat’s last theorem.”
Normally I would forgo something this complicated, but I thought I would give it a try. I was surprised that I was able to solve it, though it took some tedious work. (Hint: truth tables. See the “Pointing Fingers” post regarding truth tables.)
One important note. The author is a bit cavalier about the use of “Either …, or …”. In common parlance this means “either P is true or Q is true, but not both” (exclusive “or”: XOR), whereas in logic “or” means “either P is true or Q is true, or possibly both” (inclusive “or”: OR). I assumed all “Either …, or …” and “or” expressions were the logical inclusive “or”, which turned out to be the case.
See the Fermat’s Last Theorem Puzzle
This is a nice little puzzle from the late Nick Berry’s Datagenetics Blog.
“A quick little puzzle this week. (I tried to track down the original source, but reached a dead-end with a web search as the site that hosted it, a blogspot page under the name fivetriangles appears password protected, and no longer maintained). …
There are three identical triangles with aligned bases (in the original problem, it is stated they are equilateral, but I don’t think that really matters; Any congruent triangles will do, and I’m going to use isosceles triangles in my solving). If we say that one triangle has the area A, what is the area of the two shaded regions?”
See the Three Triangles Puzzle for solutions.
This is an initially mind-boggling problem from the 1995 American Invitational Mathematics Exam (AIME).
“Find the last three digits of the product of the positive roots of
”
See Log Lunacy for solution.
This puzzle from the Scottish Mathematical Council (SMC) Senior Mathematics Challenge seems at first to have insufficient information to solve.
“Ant and Dec had a race up a hill and back down by the same route. It was 3 miles from the start to the top of the hill. Ant got there first but was so exhausted that he had to rest for 15 minutes. While he was resting, Dec arrived and went straight back down again. Ant eventually passed Dec on the way down just half a mile before the finish.
Both ran at a steady speed uphill and downhill and, for both of them, their downhill speed was one and a half times faster than their uphill speed. Ant had bet Dec that he would beat him by at least a minute.
Did Ant win his bet?”
See the Close Race Puzzle for solutions.
(Update 1/2/2023) Alternative Solution from Oscar Rojas Continue reading
This is a provocative puzzle from the Maths Masters team, Burkard Polster (aka Mathologer) and Marty Ross as part of their “Summer Quizzes” offerings for 2013.
“In the picture the top curve is a semicircle and the bottom curve is a quarter circle. Which has greater area, the red square or the blue rectangle?”
See the Spiral Areas Puzzle for solutions.
This is a fun logic problem from the Mathigon math calendar for December 2022.
1. Alice said Bob did it.
2. Bob said Alice did it.
3. Carol said Alice didn’t do it.
4. Dan said it was either Alice or Carol.
Only one person is telling the truth. Who did it?
See Who Did It? for a solution.
Here is another Brainteaser from the Quantum magazine.
“King Arthur ordered a pattern for his quarter-circle shield. He wanted it to be painted in three colors: yellow, the color of kindness; red, the color of courage: and blue the color of wisdom. When the artist brought in his work, the king’s armor-bearer said there was more courage than wisdom on the shield. But the artist managed to prove that the proportions of both virtues were equal. Can you tell how? (A. Savin)”
This is another relatively simple problem, though it may look a bit daunting at first.
See Wisdom of Old
Alcuin of York (735-804) had a series of similar problems involving the distribution of corn among servants. Since the three propositions were the same format with only the numbers changing, I thought I would present them in a more concise form:
“Proposition
A certain head of household had a number of servants, consisting of men, women, and children, among whom he wished to distribute quantities, modia, of corn. The men should receive three modia; the women, two; and the children, half a modium.
(a) If the head of household has 20 servants and wished to distribute 20 modia of corn among them, let him say, he who can, How many men, women and children must there have been.
(b) If the head of household has 30 servants and wished to distribute 30 modia of corn among them, let him say, he who can, How many men, women and children must there have been.
(c) If the head of household has 100 servants and wished to distribute 100 modia of corn among them, let him say, he who can, How many men, women and children must there have been.”
I will give Alcuin’s solutions first, followed by my more expansive solutions that rely on our familiar symbolic algebra that was not available in Alcuin’s time.
