Here is another challenging problem from the first issue of the 1874 TheAnalyst, which also appears in Benjamin Wardhaugh’s book.
“3. If a line make an angle of 40° with a fixed plane, and a plane embracing this line be perpendicular to the fixed plane, how many degrees from its first position must the plane embracing the line revolve in order that it may make an angle of 45° with the fixed plane?
—Communicated by Prof. A. Schuyler, Berea, Ohio.”
Part of the challenge is to construct a diagram of the problem. I used techniques for a solution that were barely in use when this problem was posed in 1874. The contrast between then and now is most revealing.
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.
I came across this problem in Alfred Posamentier’s book, but I remember I had seen it a couple of places before and had never thought to solve it. At first, it seems like magic.
In any convex quadrilateral (line between any two points in the quadrilateral lies entirely inside the quadrilateral) inscribe a second convex quadrilateral with its vertices on the midpoints of the sides of the first quadrilateral. Show that the inscribed quadrilateral must be a parallelogram.
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.]” See the River Crossing.
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. See Polygon Areas Problem.
I have always had a tenuous relationship with the concept of angular momentum, but recently my concerns resurfaced when I did my studies on Kepler, and in particular his “equal areas law” and Newton’s elegant geometric proof. I love the fact that a simple geometric argument, seemingly totally divorced from the physical situation, can provide an explanation for why the line from the Sun to a planet sweeps out equal areas in equal time as the planet orbits the Sun, solely under the influence of the gravitational force between them. However, modern physics books invariably cite the conservation of angular momentum as the “explanation.” I indicated before in my “Kepler’s Laws and Newton’s Laws” essay that this “explanation” irritated me. In this essay I go into detail about my reservations concerning this line of argument. See Angular Momentum.