Author Archives: Jim Stevenson

The Barrel of Beer

This is a great puzzle by H. E. Dudeney involving a very useful technique.

“A man bought an odd lot of wine in barrels and one barrel containing beer. These are shown in the illustration, marked with the number of gallons that each barrel contained. He sold a quantity of the wine to one man and twice the quantity to another, but kept the beer to himself. The puzzle is to point out which barrel contains beer. Can you say which one it is? Of course, the man sold the barrels just as he bought them, without manipulating in any way the contents.”

Answer.

See the Barrel of Beer for an answer.

Pool Party

Futility Closet offers another interesting puzzle:

“A billiard ball is resting on a table that measures 10 feet by 5 feet. A player hits it with no ‘English’ and it strikes four different cushions and returns to its starting point. University of Alberta mathematician Murray Klamkin asks: How far did it travel?”

After solving the problem myself, I verified that Futility Closet provides an answer, but without real justification. So I thought I would write up my solution.

Answer.

See Pool Party for a solution.

Three Jugs Problem Redux

I was sifting back through some problems posed by Presh Talwalkar on his website Mind Your Decisions, when I found another 3 Jugs problem, which was amenable to the skew billiard table solution from my earlier Three Jugs Problem. Here is his statement:

“A milkman carries a full 12-liter container. He needs to deliver exactly 6 liters to a customer who only has 8-liter and a 5-liter containers. How can he do this? No milk should be wasted: the milkman needs to leave with 6 liters of milk. Can he measure all amounts of milk from 1 to 12 (whole numbers) in some container?”

Answer.

I also believe I found a case where Talwalkar’s solution to the last question needs revision. See the Three Jugs Problem Redux for solutions.

Diluted Wine Puzzle

This was a rather intricate puzzle from Presh Talwalkar. I found his solution a bit hard to follow, so I tried for a clearer presentation.

“A servant has a method to steal wine. He removes 3 cups from a barrel of wine and replaces it with 3 cups of water. The next day he wants more wine, so he does the same thing: he removes 3 cups from the same barrel (now with diluted wine) and replaces it with 3 cups of water. The following day he repeats this one more time, so he has drawn 3 times from the same barrel and has poured back 9 cups of water. At this point the barrel is 50% wine and 50% water. How many cups of wine were originally in the barrel? ”

Answer.

See the Diluted Wine Puzzle for solutions.

Containing an Arc

This problem from Futility Closet proved quite challenging.

“University of Illinois mathematician John Wetzel called this one of his favorite problems in geometry. Call a plane arc special if it has length 1 and lies on one side of the line through its end points. Prove that any special arc can be contained in an isosceles right triangle of hypotenuse 1.”

My attempts were futile (maybe that is where the title of the website comes from). Maybe this qualifies for another Coffin Problem. But I did have one little comment about the Futility Closet solution. See Containing an Arc.

Three Coffin Problems

These are three “Coffin” Problems posed by Nakul Dawra on his Youtube site GoldPlatedGoof. (Nakul is extraordinarily entertaining and mesmerizing.) The origin of the name is explained, but basically they are problems that have easy or even trivial solutions—once you see the solution. But just contemplating the problem, they seem impossible. The idea was to kill the chances of the pupil taking an (oral) exam with these problems. I was able to solve the first two problems (after a while), but I could not figure out the third. See the Three Coffin Problems.

Four of a Kind

From Futility Closet we have another intriguing problem with what turns out to be a simple and elegant solution.

“If squares are drawn on the sides of a triangle and external to it, then the areas of the triangles formed between the squares each equal the area of the triangle itself.”

I originally assumed that the center triangle was a right triangle as suggested by the picture. But then I realized there was a solution that did not depend on that. See Four of a Kind.

Polygon Areas Problem

This is another problem from Futility Closet, though Futility Closet provides a “solution” of sorts. They provide a set of steps without explaining where they came from. So I thought I would fill in the gap. The problem is to find the area of an irregular polygon, none of whose sides cross one another, if we are given the coordinates of the vertices of the polygon.

Answer.

See Polygon Areas Problem for a solution.

Restructuring the US Senate

This subject admittedly has only a tenuous relationship to mathematics (via arithmetic), but perhaps it can join more mathematically challenging political topics like voting and gerrymandering. In any case, I was stimulated to consider the idea of reapportioning the US Senate by the % US population of each state by an 8 December 2018 article in the Atlantic by former Congressman John Dingell, who advocated abolishing the Senate. I thought this a bit too Draconian and considered the percent population idea as a better compromise. It turned out I was not alone in having this (obvious) thought: I just came across a more extensive 2 January 2019 Atlantic article by Eric Orts that concurs with my idea about reapportionment of the Senate, discusses the legal ramifications in more detail, and echoes the benefits I mentioned as well as others. See Restructuring the US Senate.

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