# Pillar Wrapping Problem

This is a fun problem from the 1949 Eureka magazine.

“The following problems were set at the Archimedeans’ 1949 Problems Drive. Competitors were allowed five minutes for each question.  [This is problem #9.]

A pillar is in the form of a truncated right circular cone. The diameter at the top is 1 ft., at the bottom it is 2 ft. The slant height is 15 ft. A streamer is wound exactly five times round the pillar starting at the top and ending at the bottom. What is the shortest length the streamer can have?”

See the Pillar Wrapping Problem for solution.

# Moon Quarters Problem

This is a straight-forward problem from the Scottish Mathematical Council (SMC) Senior Mathematics Challenge.

“A circle has radius 1 cm and AB is a diameter.  Two circular arcs of equal radius are drawn with centres A and B.  These arcs meet on the circle as shown.  Calculate the shaded area.”

There are several possible approaches and the SMC offers two examples.

See the Moon Quarters Problem for solution.

# Max Angle Puzzle

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.

(Update 3/23/2023)  Solution Construction

# Butcher Boy Problem

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 for solution.

# Covering Rectangle Puzzle

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 for solutions.

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.”

(Update 2/22/2023, 6/9/2023) Alternative Solutions

# Rolling Wheels Puzzle

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.

# Blockbusters Problem

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.