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# Monthly Archives: October 2019

# Lesz is More

• Matematyka jest najpotężniejszym intelektualnym wehikułem, jaki kiedykolwiek został skonstruowany, za pomocą którego uciekamy przed czasem, lecz nie ma powodu przypuszczać, że mogłaby kiedyś umożliwić tego rodzaju ucieczkę, jaką ucieleśnia pogoń za Absolutem. — Leszek Kołakowski

• Mathematics is the most powerful intellectual vehicle that has ever been constructed, by means of which we flee ahead of time, but there is no reason to suppose that it could someday enable the kind of escape embodied by the pursuit of the Absolute. — Leszek Kołakowski

# Koch Rock

The Koch snowflake, named after the Swedish mathematician Helge von Koch, is a famous fractal that encloses a finite area within an infinitely long boundary. To make a ’flake, you start with an equilateral triangle:

Koch snowflake stage #1 (with room for manœuvre)

Next, you divide each side in three and erect a smaller equilateral triangle on the middle third, like this:

Koch snowflake #2

Each original straight side of the triangle is now 1/3 longer, so the full perimeter has also increased by 1/3. In other words, perimeter = perimeter * 1⅓. If the perimeter of the equilateral triangle was 3, the perimeter of the nascent Koch snowflake is 4 = 3 * 1⅓. The area of the original triangle also increases by 1/3, because each new equalitarian triangle is 1/9 the size of the original and there are three of them: 1/9 * 3 = 1/3.

Now here’s stage 3 of the snowflake:

Koch snowflake #3, perimeter = 4 * 1⅓ = 5⅓

Again, each straight line on the perimeter has been divided in three and capped with a smaller equilateral triangle. This increases the length of each line by 1/3 and so increases the full perimeter by a third. 4 * 1⅓ = 5⅓. However, the area does not increase by 1/3. There are twelve straight lines in the new perimeter, so twelve new equilateral triangles are erected. However, because their sides are 1/9 as long as the original side of the triangle, they have 1/(9^2) = 1/81 the area of the original triangle. 1/81 * 12 = 4/27 = 0.148…

Koch snowflake #4, perimeter = 7.11

Koch snowflake #5, p = 9.48

Koch snowflake #6, p = 12.64

Koch snowflake #7, p = 16.85

Koch snowflake (animated)

The perimeter of the triangle increases by 1⅓ each time, while the area reaches a fixed limit. And that’s how the Koch snowflake contains a finite area within an infinite boundary. But the Koch snowflake isn’t confined to itself, as it were. In “Dissecting the Diamond”, I described how dissecting and discarding parts of a certain kind of diamond could generate one side of a Koch snowflake. But now I realize that Koch snowflakes are everywhere in the diamond — it’s a Koch rock. To see how, let’s start with the full diamond. It can be divided, or dissected, into five smaller versions of itself:

Dissectable diamond

When the diamond is dissected and three of the sub-diamonds are discarded, two sub-diamonds remain. Let’s call them sub-diamonds 1 and 2. When this dissection-and-discarding is repeated again and again, a familiar shape begins to appear:

Koch rock stage 1

Koch rock #2

Koch rock #3

Koch rock #4

Koch rock #5

Koch rock #6

Koch rock #7

Koch rock #8

Koch rock #9

Koch rock #10

Koch rock #11

Koch rock #12

Koch rock #13

Koch rock (animated)

Dissecting and discarding the diamond creates one side of a Koch triangle. Now see what happens when discarding is delayed and sub-diamonds 1 and 2 are allowed to appear in other parts of the diamond. Here again is the dissectable diamond:

Dia-flake stage 1

If no sub-diamonds are discarded after dissection, the full diamond looks like this when each sub-diamond is dissected in its turn:

Dia-flake #2

Now let’s start discarding sub-diamonds:

Dia-flake #3

And now discard everything but sub-diamonds 1 and 2:

Dia-flake #4

Dia-flake #5

Dia-flake #6

Dia-flake #7

Dia-flake #8

Dia-flake #9

Dia-flake #10

Now full Koch snowflakes have appeared inside the diamond — count ’em! I see seven full ’flakes:

Dia-flake #11

Dia-flake (animated)

But that isn’t the limit. In fact, an infinite number of full ’flakes appear inside the diamond — it truly is a Koch rock. Here are examples of how to find more full ’flakes:

Dia-flake 2 (static)

Dia-flake 2 (animated)

Dia-flake 3 (static)

Dia-flake 3 (animated)

Previously pre-posted:

• Dissecting the Diamond — other fractals in the dissectable diamond