This is a somewhat challenging math cryptogram in a slightly different guise from the Canadian Math Society’s magazine, Crux Mathematicorum.
“But you can’t make arithmetic out of passion. Passion has no square root.” (Steve Shagan, City of Angels, G.P. Putnam’s Sons, New York, 1975, p. 16.)
On the contrary, show that in the decimal system
has a unique solution.
See the Passion Kiss Problem for solutions.
Here is a problem from the Quantum magazine, only this time from the “Challenges” section (these are expected to be a bit more difficult than the Brainteasers).
“Three circles with the same radius r all pass through a point H. Prove that the circle passing through the points where the pairs of circles intersect (that is, points A, B, and C) also has the same radius r.”
Indeed, I found this quite challenging. It took me several weeks to work out my approach and details.
This simple-appearing problem is from the 17 August 2020 MathsMonday offering by MEI, an independent curriculum development body for mathematics education in the UK.
“The diagram shows an equilateral triangle in a rectangle. The two shapes share a corner and the other corners of the triangle lie on the edges of the rectangle. Prove that the area of the green triangle is equal to the sum of the areas of the blue and red triangles. What is the most elegant proof of this fact?”
Since the MEI twitter page seemed to be aimed at the high school level and the parting challenge seemed to indicate that there was one of those simple, revealing solutions to the problem, I spent several days trying to find one. I went down a number of rabbit holes and kept arriving at circular reasoning results that assumed what I wanted to prove. Visio revealed a number of fascinating relationships, but they all assumed the result and did not provide a proof. I finally found an approach that I thought was at least semi-elegant.
See the Diabolical Triangle Puzzle
(Update 1/30/2021) New MEI Solution
This is an interesting problem from the 1977 Canadian Math Society’s magazine, Crux Mathematicorum.
“206. [1977: 10] Proposed by Dan Pedoe, University of Minnesota.
A circle intersects the sides BC, CA and AB of a triangle ABC in the pairs of points X, X’, Y, Y’ and Z, Z’ respectively. If the perpendiculars at X, Y and Z to the respective sides BC, CA and AB are concurrent at a point P, prove that the respective perpendiculars at X’, Y’ and Z’ to the sides BC, CA and AB are concurrent at a point P’.”
See the Twin Intersection Puzzle
Puzzles and Problems: plane geometry, Dan Pedoe, Crux Mathematicorum
This is another thoughtful puzzle from the imaginative mind of James Tanton (with slight edits).
“Three poles of height 1183 feet, 182 feet, 637 feet stand in the ground. Pick a pole and saw off all the taller poles at that height. Plant those tops in the ground too. Repeat until no more such saw cuts can be made. Despite choices made along the way, what final result is sure to occur? [Four poles, heights a, b, c, d ft?]”
See the Pole Leveling Puzzle for a solution.
James Tanton has come up with another imaginative concrete problem harboring a mathematical pattern.
“60 trees in a row. Their stars are yellow, orange, blue, Y, O, B, Y, O, B, … Their pots are orange, yellow, pink, blue, O, Y, P, B, O, Y, P, B, … Their baubles are mauve, pink, yellow, blue, orange, M, P, Y, B, O, M, P, Y, B, O, … Must there be an all yellow tree? All B? One with star = O, pot = O, baubles = M?”
See the Christmas Tree Puzzle for a solution.
Here is yet another problem from Presh Talwalkar. This one is rather elegant in its simplicity of statement and answer.
“Solve For The Angle – Viral Puzzle
I thank Barry and also Akshay Dhivare from India for suggesting this problem! This puzzle is popular on social media. What is the measure of the angle denoted by a “?” in the following diagram? You have to solve it using elementary geometry (no trigonometry or other methods). It’s harder than it looks. I admit I did not solve it. Can you figure it out?”
See the Shy Angle Problem for solutions.
Here is another delightful problem from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).
“In Sussex, Holmes and I ran into a pair of woodcutters named Doug and Dave. There was an air of the unreliable about them—not helped by a clearly discernable aroma of scrumpy—but they nevertheless proved extremely helpful in guiding us to a particular hilltop clearing some distance outside of the town of Arundel. A shadowy group had been counterfeiting sorceries of a positively medieval kind, and all sorts of nastiness had ensued.
The Adventure of the Black Alchemist is not one that I would feel comfortable recounting, and if my life never drags me back to Chanctonbury Ring I shall be a happy man. But there is still some instructive material here. Whilst we were ascending our hill, Doug and Dave made conversation by telling us about their trade. According to these worthies, working together they were able to saw 600 cubic feet of wood into large logs over the course of a day, or split as much as 900 cubic feet of logs into chunks of firewood.
Holmes immediately suggested that they saw as much wood in the first part of the day as they would need in order to finish splitting it at the end of the day. It naturally fell to me to calculate precisely how much wood that would be.
Can you find the answer?”
See the Loggers Problem for solutions.
This is a nifty little problem from the Quantum math magazine.
“Two ants stand at opposite corners of a 1-meter square. A barrier was placed between them in the form of half a 1-meter square attached along the diagonal of the first square, as shown in the picture. One ant wants to walk to the other. How long is the shortest path?”
See the Barrier Minimal Path Problem for solutions.
This is a fun logic puzzle from one of Ian Stewart’s many math collections. I discovered that the problem actually is basically one of Lewis Carroll’s examples from an 1896 book:
If all these statements are correct, do I avoid kangaroos, or not?
See Do I Avoid Kangaroos? for solutions.
