Conical Bottle Problem

I was astonished that this problem was suitable for 8th graders. First of all the formula for the volume of a cone is one of the least-remembered of formulas, and I certainly never remember it. So my only viable approach was calculus, which is probably not a suitable solution for an 8th grader.

Presh Talwalkar: “This was sent to me as a competition problem for 8th graders, so it would be a challenge problem for students aged 12 to 13. When a conical bottle rests on its flat base, the water in the bottle is 8 cm from its vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the height of the bottle? (Note “conical” refers to a right circular cone as is common usage.) I at first thought this problem was impossible. But it actually can be solved.”

Answer.

See the Conical Bottle Problem for solutions.

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Regiomontanus 1471 Problem

This is an old problem I had seen before. Here is David Wells’s rendition:

“Johannes Müller, named Regiomontanus after the Latin translation of Körnigsberg, his city of birth, later made famous by Euler, proposed this problem in 1471. … it is usually put in this form …: From what distance will a statue on a plinth appear largest to the eye [of a mouse!]? If we approach too close, the statue appears foreshortened, but from a distance it is simply small.”

I have added height numbers in feet for concreteness (as well as the mouse qualification, since the angles are measured from ground level). So the problem is to find the distance x such that the angle is maximal.

Answer.

See the Regiomontanus 1471 Problem for solutions.

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Parallelogram Problem

Catriona Shearer has come up with another challenging but elegant geometric problem. In some ways, it is similar to the famous Russian Coffin Problems that have an obvious solution—once you see it—but initially seem impenetrable. I really marvel at Catriona Shearer’s ability to come up with these problems.

“What’s the area of the parallelogram?”

Answer.

See the Parallelogram Problem for a solution.

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Who Gets To Be Called A Mathematician?

A mathematics friend of mine just sent me this link to a 2017 posting by “recovering mathematician” Junaid Mubeen that I found most apt. If you haven’t seen it yet, take a look.

“[M]athematics need not be situated at the extremes of established knowledge. We can all revel in problems whose solutions are known. Even when humankind has exhausted its capacity to extend its collective knowledge base, as individuals our ignorance is what keeps our mathematical instincts aflame. Problem solving lies between the boundaries of what we know and what we seek. This sweet spot is where we all—novices and experts alike—get to bend and twist what we know to forge new truths for ourselves. Who cares if our discoveries are already known to the rest of the world (or machines, for that matter)? The satisfaction of finding my own solution, of pushing through my own knowledge limits, is as enthralling as the pursuit of ‘new’ proofs promised by research mathematics. Let the machines come; mathematics does not belong to the omnipotent.”

This view echoes my feeling that we can enjoy performing music even if we can’t compose it. See Who Gets To Be Called A Mathematician.

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Straight and Narrow Problem

The following interesting behavior was found at the Futility Closet website:

“A pleasing fact from David Wells’ Archimedes Mathematics Education Newsletter: Draw two parallel lines. Fix a point A on one line and move a second point B along the other line. If an equilateral triangle is constructed with these two points as two of its vertices, then as the second point moves, the third vertex C of the triangle will trace out a straight line. Thanks to reader Matthew Scroggs for the tip and the GIF.”

This is rather amazing and cries out for a proof. It also raises the question of how anyone noticed this behavior in the first place. I proved the result with calculus, but I wonder if there is a slicker way that makes it more obvious. See the Straight and Narrow Problem.

(Update 3/25/2019) Continue reading →

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Counterfeit Coin in Base 3

Futility Closet presented a nifty method of solving the “counterfeit coin in 12 coins” problem in a way I had not seen before by mapping the problem into numbers in base 3. It wasn’t immediately clear to me how their solution worked, so I decided to write up my own explanation.

Futility Closet: “You have 12 coins that appear identical. Eleven have the same weight, but one is either heavier or lighter than the others. How can you identify it, and determine whether it’s heavy or light, in just three weighings in a balance scale? This is a classic puzzle, but in 1992 Washington State University mathematician Calvin T. Long found a solution ‘that appears little short of magic.’ ”

See Counterfeit Coin in Base 3.

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Missing Pages Puzzle

Setting aside my chagrin that the following problem was given to pre-university students, I initially found the problem to be among the daunting ones that offer little information for a solution. It also was a bit “inelegant” to my way of thinking, since it involved considering some separate cases. Still, the end result turned out to be unique and satisfying (Talwalkar’s Note 2 was essential for a unique solution, since the problem as stated was ambiguous).

“Kshitij from India sent me this problem from the 1994 India Regional Mathematics Olympiad.
‘A leaf is torn from a paperback novel. The sum of the numbers on the remaining pages is 15000. What are the page numbers on the torn leaf?’
Note 1: a ‘leaf’ means a single sheet of paper.
Note 2: the quoted problem is actual wording from the competition. But let me add an important detail: the book is numbered in the usual sequential way starting with the first page as page 1.”

Answer.

See the Missing Pages Puzzle for a solution.

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River Crossing

This is a riff on a classic problem, given in Challenging Problems in Algebra.

“N. Bank and S. Bank are, respectively, the north and south banks of a river with a uniform width of one mile. Town A is 3 miles north of N. Bank, town B is 5 miles south of S. Bank and 15 miles east of A. If crossing at the river banks is only at right angles to the banks, find the length of the shortest path from A to B.

Challenge. If the rate of land travel is uniformly 8 mph, and the rowing rate on the river is 1 2/3 mph (in still water) with a west to east current of 1 1/3 mph, find the shortest time it takes to go from A to B. [The path across the river must still be perpendicular to the banks.]”

Answer.

See the River Crossing for a solution.

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