Here is a familiar puzzle from the Mathigon Puzzle Calendars for 2021.
“Given a line and two points A and B, which point P on the line forms the largest angle APB?”
See the Max Angle Puzzle
An excellent application of the solution to this puzzle can be found at Numberphile, where Ben Sparks explains an optimal rugby goal-kicking strategy.
This is another long historical story from Sam Loyd with a puzzle attached.
“NOTICING THE HIGH price recently paid at auction for an autograph of General Grant reminds me to say that I am the proud possessor of what I believe to be the last signature made by General Grant.
The story connected with it introduces a somewhat pretty problem, and induces me to pay a tribute to Grant’s mathematical ability, at the expense of the many who have no love for figures. I take occasion here to say that while journeying through life and jostling up against all manner of people, the fact has been impressed upon my mind that with few exceptions all successful men were those who endowed with a ready faculty for correct mental arithmetic. On the other hand, there is a class of never-do-wells who guess or jump at conclusions in a reckless way, and cannot even figure up how much to pay on the dollar when the inevitable smash comes.
I could mention a dozen incidents connected with great men as illustrating their aptitude for correct calculations, but this one will suffice to call attention to Grant’s aptitude for figures.
We all remember the story of how he figured his way into West Point, after that memorable journey for a pound of butter, when he heard of the chance for a competitive examination. Professor Agnell, the master of mathematics at West Point, with whom I used to play chess, used to say that “Grant had a great love for mathematics and horses.”
Grant did love a horse and could pick out the good qualities at a glance, and, oh, my! how he despised a man who would abuse a dumb animal!
My story turns upon an incident as told by Ike Reed, of the old horse mart of Johnson & Reed, who gave me the autograph from their sales book of 1884, as photographed in the picture. During the last term of his Presidency General Grant returned from his afternoon drive and in a humorous but somewhat mortified way told Colonel Shadwick, who kept the Willard Hotel, that he had been passed on the road by a butcher cart in a way that made his crack team appear to be standing still. He said he would like to know who owned the horse and if it was for sale.
The horse was readily found and purchased from an unsophisticated German for half of what he would have asked had he known the purchaser was the President of the United States. The horse was of light color and was none other than Grant’s favorite horse, “Butcher Boy,” named after the incident mentioned. Well, some years later, after the Wall street catastrophe, which impaired the finances of the Grant family, Butcher Boy and his mate were sent to the auction rooms of Johnson & Reed, and sold for the sum of $493.68. Mr. Reed said he could have gotten twice as much for them if he had been permitted to mention their ownership. But General Grant positively prohibited the fact being made known. “Nevertheless,” said Reed. “you come out two per cent, ahead, for you make 12 per cent, on Butcher Boy and lose 10 per cent, on the other.”
“I suppose that is the way some people would figure it out.” replied the General, but the way he laughed showed that he was better at figures than some people, so I am going to ask our puzzlists to tell me what he got for each horse if he lost 10 per cent on one and made 12 per cent on the other, but cleared 2 per cent on the whole transaction?
It may be mentioned incidentally that General Grant stated that he had presented one of the horses to Mrs. Fred Grant, and as shown in the receipt signed for her.”
See the Butcher Boy Problem
This is a fairly simple problem from Futility Closet, which is currently under a hiatus.
“Robert Bilinski proposed this problem in the April 2006 issue of Crux Mathematicorum. On square ABCD, two equilateral triangles are constructed, ABE internally and BCF externally, as shown. Prove that D, E, and F are collinear.”
See Line Work
This is a nice puzzle from the Maths Masters team, Burkard Polster (aka Mathologer) and Marty Ross as part of their “Summer Quizzes” offerings.
“In the picture does the green rectangle cover more or less than half of the [congruent] red rectangle?”
It is evident from the problem solution that the two rectangles are the same, so I made it explicit.
See the Covering Rectangle Puzzle
I found this problem from the 1981 Canadian Math Society’s magazine, Crux Mathematicorum, to be quite challenging.
“Proposed by Kaidy Tan, Fukien Teachers’ University, Foochow, Fukien, China.
An isosceles triangle has vertex A and base BC. Through a point F on AB, a perpendicular to AB is drawn to meet AC in E and BC produced in D. Prove synthetically that
Area of AFE = 2 Area of CDE if and only if AF = CD.”
See the Linked Triangles Problem
(Update 2/22/2023) Alternative Solution
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 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
Here is another typical sum puzzle from Presh Talwalkar.
“Solve the following sums:
_____1/(1×3) + 1/(3×5) + 1/(5×7) + 1/(7×9) + 1/(9×11) =
_____1/(4×7) + 1/(7×10) + 1/(10×13) + 1/(13×16) =
_____1/(2×7) + 1/(7×12) + 1/(12×17) + … =”
The only reason I am including this puzzle is that Talwalkar gets very excited about deriving a formula that can solve sums of this type. This gives me an opportunity to discuss the “formula vs. procedure” way of doing math.
See the Incredible Trick Puzzle
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.