YouTube Math-Puzzle and Math-Instruction Channels

[lpetrich]

So one gets possible x values -1/3, 1/2, 1, and 3. By symmetry, the other variables may also have these values.

Substituting in x and repeating this procedure to find y, I find

x = 1 -> y = 1
x = 1/2 -> y = 1/2
x = 3 -> y = 3 or y = -1/3
x = -1/3 -> y = 3

In each case, there is one value of z that completes the solution. (x,y,z) is

(1,1,1)
(1/2, 1/2, 1/2)
(3, 3, -1/3)
(3, -1/3, 3)
(-1/3, 3, 3)

 

[Swammerdami]

These are some of the solutions, but not all of them.

Yes, with characteristic lack of self-control I reacted prematurely and never even wrote down the operative equation 2X + 1/X = 3. Like all quadratics, this equation has TWO solutions.

If I'd made this blunder a few decades ago, my face would be red with shame; but now, most shameful is that I feel little shame, or at least not enough to compare with other shame sources. Perhaps on my next visit to the chamber of exotic services I should ask to be chained and flogged.

Although unsportsmanlike, could I be tempted to use my god-like super-moderator powers to expunge evidence of the blunder?

 

[lpetrich]

I've decided to generalize the previous problem to see if any further symmetry breaking would result, like x, y, and z all being unequal.

\( \displaystyle{ \frac{1}{x} + y + z = a } \\ \displaystyle{ x + \frac{1}{y} + z = a } \\ \displaystyle{ x + y + \frac{1}{z} = a } \)

The variables can have values a, -1/a, and (1/4)*(a +- sqrt(a^2-8))

There are five solutions. Three of them are permutations of {a, a, -1/a}, and the other two have all the variables equal to (1/4)*(a + sqrt(a^2-8)) or else (1/4)*(a - sqrt(a^2-8))

 

[lpetrich]

Terrific Toothpick Patterns - Numberphile - YouTube

Place a toothpick on a table. Now place a toothpick at each end, with the middle of each new toothpick touching the original toothpick. Repeat. One gets some interesting patterns as one goes.

Here is a page for visualizing what they look like: toothpick-like sequence exploration

Here is a paper on finding formulas for how many toothpicks in this sequence: [1004.3036] The Toothpick Sequence and Other Sequences from Cellular Automata

The Numberphile host discussed similar patterns, like using gull-wing toothpicks and shallow V-shaped ones. The latter ones make hexagonal patterns.

This is much like some cellular automaton, and the Numberphile host discussed the Ulam - Warburton one. Take a rectangular grid, with one cell on and all the other cells off. At each step, turn on each cell that has exactly one neighbor in the on state.

The host also discussed some variations: turn on if the number of "on" neighbors is odd, and use a hexagonal grid.

 

[AdamWho]

Micheal Penn has some great videos... but they can be a little long and more than once I fell asleep during a video.

 

[Swammerdami]

Here's a problem from a famous high school math contest. I won't give the YouTube link — Let's increase the chance one of you will have the great pleasure of solving it yourself!

Let S be a finite set of at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P ∈ S. The line rotates clockwise about the point P until the first time that the line meets some other point belonging to S. This point, Q. takes over as the next pivot, and the line now rotates clockwise about Q, until it next meets a point of S. This process continues indefinitely.

Show that we can choose a point P ∈ S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times.

Judging by contestants' success this was the 2nd-most difficult of the six problems in the contest. And yet with some insight, solution is straightforward with very little math knowledge required.

 

[Swammerdami]

Let S be a finite set of at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P ∈ S. The line rotates clockwise about the point P until the first time that the line meets some other point belonging to S. This point, Q. takes over as the next pivot, and the line now rotates clockwise about Q, until it next meets a point of S. This process continues indefinitely.

Show that we can choose a point P ∈ S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times.

Hints: Choose a grid so that the initial line is a longitude. Let N_W, N_E, N_S, N_P denote respectively the number of points to the West or East of the line, or on its Starboard or Port side. Note that the number of points is |S| = N_W + N_E + 1 = N_S + N_P + 1

What happens to N_P when a port-side point Q collides with the line astern of the current pivot P? What happens to N_S when a starboard-side point Q collides forward of the current pivot P? What's the situation after the line has rotated through 180°?

 

[Jarhyn]

Here's a problem from a famous high school math contest. I won't give the YouTube link — Let's increase the chance one of you will have the great pleasure of solving it yourself!

Let S be a finite set of at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P ∈ S. The line rotates clockwise about the point P until the first time that the line meets some other point belonging to S. This point, Q. takes over as the next pivot, and the line now rotates clockwise about Q, until it next meets a point of S. This process continues indefinitely.

Show that we can choose a point P ∈ S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times.

Judging by contestants' success this was the 2nd-most difficult of the six problems in the contest. And yet with some insight, solution is straightforward with very little math knowledge required.

So, I've been visualizing the problem since I got on the train this morning, and best I can tell it has to do with the fact that any finite set S with no collinear points, and you pick a line and a point and run the windmill, you will get a residue of unstruck points if you didn't pick well. If you pick one of those and bisect two struck points, it would cycle through a larger region until the line wipes the whole plane, on the "inside" of the shape.

 

[Jarhyn]

Here's a problem from a famous high school math contest. I won't give the YouTube link — Let's increase the chance one of you will have the great pleasure of solving it yourself!

Let S be a finite set of at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P ∈ S. The line rotates clockwise about the point P until the first time that the line meets some other point belonging to S. This point, Q. takes over as the next pivot, and the line now rotates clockwise about Q, until it next meets a point of S. This process continues indefinitely.

Show that we can choose a point P ∈ S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times.

Judging by contestants' success this was the 2nd-most difficult of the six problems in the contest. And yet with some insight, solution is straightforward with very little math knowledge required.

So, I've been visualizing the problem since I got on the train this morning, and best I can tell it has to do with the fact that any finite set S with no collinear points, and you pick a line and a point and run the windmill, you will get a residue of unstruck points if you didn't pick well. If you pick one of those and bisect two struck points, it would cycle through a larger region until the line wipes the whole plane, on the "inside" of the shape.

Something else occurred to me that you can always draw a shape around S. It will have a "most rounded perimeter" or a set of "perimeter" points.

Let A be the residue.

Continue with each nest until you have a set A that contains all remaining residue.

Pick P of any point in A such that P bisects the remaining points in A across it's horizon. If there is an odd number of remaining points in A, start the line anywhere past the opposing point from P in it's graph.

I think it can be any line off any point in A? But I want to also sweep all area.

 

[Tigers!]

I have found this site https://www.youtube.com/user/asistenacademy?app=desktop good for "easier".
As someone who is fascinated by maths and loves it but whose enthusiasm far exceeds their skill it is a great site.

 

[Swammerdami]

Jarhyn — Your solution may be correct. But you have not provided a detailed proof.

There is another solution, closely related to yours perhaps, but much simpler.