The Koch Snowflake, a Boundary That Keeps Growing, math that surprises you lesson

Start with an equilateral triangle and keep adding smaller and smaller triangular bumps to its sides. Every step makes the boundary one third longer, so the distance around the shape grows past any number you can name. Yet the shape never leaves a fixed region of the plane. How can a boundary that keeps growing enclose a finite area? This page builds the Koch snowflake step by step, counts its segments and its added triangles, and shows that its perimeter is infinite while its area is exactly $\tfrac85$ of the area of the starting triangle.

Key terms

Koch snowflake

Definition The Koch snowflake starts from an equilateral triangle with side length $s$ , called stage $0$ . One step applies the following replacement to every segment of the current boundary at once. Divide the segment into three equal parts, build an equilateral triangle outward on the middle part, pointing away from the inside of the shape, and replace the middle part by the other two sides of that new triangle. Each segment becomes four segments, each one third as long, and the endpoints of the old segment stay where they are. The new triangles never overlap one another or the shape built so far, so every stage is a polygon that does not cross itself, and it contains all the earlier stages. The Koch snowflake is the limit of these stages. Its boundary is made of three Koch curves, one grown from each side of the triangle. Stage $n$ has the perimeter $P_n$ and the area $A_n$ given by

\begin{gathered} P_n = 3s \left(\tfrac{4}{3}\right)^n, \\[0.4em] A_n = A_0 \left(\tfrac{8}{5} - \tfrac{3}{5}\left(\tfrac{4}{9}\right)^n\right), \end{gathered}

where $A_0 = \tfrac{\sqrt{3}}{4} s^2$ is the area of the starting triangle. As $n$ grows, $P_n$ grows without bound while $A_n$ increases toward $\tfrac85 A_0$ . So the snowflake has an infinite perimeter and a finite area, which is no contradiction, because length and area respond differently to small added pieces. Every stage has a finite perimeter as well as a finite area, and only the limit has an infinitely long boundary.

In plain words Keep putting smaller copies of the same triangular bump on every straight piece of the boundary. The edge gets one third longer at every step, but the bumps shrink so quickly that each step adds less area than the step before.

Example Take $s = 1$ , so the starting perimeter is $3$ and $A_0 = \sqrt{3}/4 \approx 0.433$ .

Stage	Segments	Segment length	Perimeter	Area
$0$	$3$	$1$	$3$	$A_0$
$1$	$12$	$\tfrac13$	$4$	$\tfrac43 A_0$
$2$	$48$	$\tfrac19$	$\tfrac{16}{3}$	$\tfrac{40}{27} A_0$
$3$	$192$	$\tfrac{1}{27}$	$\tfrac{64}{9}$	$\tfrac{376}{243} A_0$

Stage $1$ is a six-pointed star, made of the triangle and three new triangles with side $\tfrac13$ , each with $\tfrac19$ of the area of the original. By stage $3$ the perimeter is about $7.11$ , more than twice the starting perimeter, while the area is about $1.547 A_0$ , already close to the limit $1.6 A_0$ .

Koch curve

Definition The Koch curve is what the same replacement produces from a single segment. Start with a segment of length $L$ and apply the step to every segment, again and again. Stage $n$ is a path of $4^n$ segments, each of length $L/3^n$ , joining the two original endpoints, and these paths approach one continuous curve, the Koch curve. It consists of four copies of itself, each scaled down by the factor $\tfrac13$ . It lies inside the isosceles triangle that has the original segment as its base and base angles of $30$ degrees, and it has no tangent line at any of its points. Helge von Koch published it in 1904 as an example of a continuous curve without a tangent anywhere, built by elementary geometry. Each side of the starting triangle of the snowflake grows into one Koch curve, and the three curves meet at the corners of the original triangle.

In plain words The Koch curve is one side of the snowflake on its own. It stays within a low triangle over the segment it started from, yet it is so crinkled that no piece of it, however small, is a straight segment or a smooth arc.

Example With $L = 1$ and the segment from $(0, 0)$ to $(1, 0)$ , stage $1$ is the path through the points $(0, 0)$ , $\bigl(\tfrac13, 0\bigr)$ , $\bigl(\tfrac12, \tfrac{\sqrt{3}}{6}\bigr)$ , $\bigl(\tfrac23, 0\bigr)$ and $(1, 0)$ , made of four segments of length $\tfrac13$ with total length $\tfrac43$ . Stage $2$ has $16$ segments of length $\tfrac19$ and total length $\tfrac{16}{9}$ . The highest point of every stage after stage $0$ is the tip $\bigl(\tfrac12, \tfrac{\sqrt{3}}{6}\bigr)$ , about $0.289$ above the base, which is also the top corner of the containing triangle. The four copies are visible already at stage $2$ , where each of the four segments of stage $1$ carries a bump one third the size of the first one.

Infinite perimeter

Definition The length of a curve is defined with paths of straight segments. Mark finitely many points along the curve in order, join consecutive marks by straight segments, and add up the lengths of those segments. The length of the curve is the smallest number that none of these totals exceeds. A curve has infinite length when no such number exists, that is, when the totals can be made larger than any number. For the Koch snowflake with side $s$ , stage $n$ has $3 \cdot 4^n$ segments of length $s/3^n$ , so its perimeter is

P_n = 3s\left(\tfrac{4}{3}\right)^n.

Since $\tfrac43 > 1$ , these perimeters grow without bound. The corners of every stage are never moved by later steps, so they all lie on the boundary of the snowflake, and the stage $n$ polygon is one of the marked paths for that boundary. Its length $P_n$ is therefore at most the length of the boundary, for every $n$ , so the boundary has infinite length. Each stage itself has a finite perimeter, so the infinite perimeter belongs to the limit only.

In plain words Every step multiplies the distance around the shape by $\tfrac43$ , and multiplying again and again by a number larger than $1$ eventually beats any fixed number. An ant walking along the finished boundary at a steady speed would never complete one lap.

Example For $s = 1$ the perimeters $3$ , $4$ , $\tfrac{16}{3} \approx 5.33$ , $\tfrac{64}{9} \approx 7.11$ and $\tfrac{256}{27} \approx 9.48$ grow slowly at first. Stage $12$ has perimeter about $94.7$ and stage $13$ about $126.3$ , so the $13$ th step is the first to push the perimeter past $100$ , and stage $21$ is the first with a perimeter past $1000$ . Meanwhile the whole snowflake stays inside a circle of radius $1/\sqrt{3} \approx 0.577$ around the center of the triangle.

Finite area

Definition A region has finite area when its area is a real number rather than infinitely large. The region enclosed by the Koch snowflake has area $\tfrac85 A_0$ , where $A_0$ is the area of the starting triangle. At step $k$ , for $k \ge 1$ , one new triangle is added on each of the $3 \cdot 4^{k-1}$ segments of the previous stage. Its side is $s/3^k$ , so its area is $A_0/9^k$ , because multiplying every length by a number $c$ multiplies every area by $c^2$ . The new triangles never overlap one another or the earlier shape, so step $k$ adds the area

\begin{aligned} &3 \cdot 4^{k-1} \cdot \frac{A_0}{9^k} \\[0.4em] &\quad = \frac{A_0}{3} \left(\frac{4}{9}\right)^{k-1}. \end{aligned}

These additions form a geometric series with first term $A_0/3$ and ratio $\tfrac49$ , and since $\tfrac49$ lies between $0$ and $1$ , the series converges, as explained in Zeno’s Achilles and the Tortoise. Its sum is

\frac{A_0/3}{1 - 4/9} = \frac{3}{5} A_0,

so the snowflake has area $A_0 + \tfrac35 A_0 = \tfrac85 A_0$ . Every stage has a smaller area than this, since each stage leaves out all the later additions.

In plain words The number of new triangles is multiplied by $4$ at each step, but each new triangle has $\tfrac19$ of the area of a triangle from the step before, so each step adds $\tfrac49$ as much area as the previous one. Additions that shrink by a fixed factor smaller than $1$ add up to a finite total.

Example The sum can be found by shifting the series. Call the total added area $T$ , so

\begin{aligned} T &= \tfrac13 A_0 + \tfrac49 \cdot \tfrac13 A_0 \\[0.4em] &\quad + \left(\tfrac49\right)^2 \cdot \tfrac13 A_0 + \cdots. \end{aligned}

Multiplying by $\tfrac49$ moves every term one place along, so $\tfrac49 T$ is the same series without its first term $\tfrac13 A_0$ . Subtracting gives $T - \tfrac49 T = \tfrac13 A_0$ , so $\tfrac59 T = \tfrac13 A_0$ and $T = \tfrac35 A_0$ . This shortcut is valid only because $T$ is a finite number. With ratio $2$ instead of $\tfrac49$ , the same steps would give the absurd answer $T = -\tfrac13 A_0$ for a sum of positive terms that grows without bound.

There is also a simple bound. The snowflake lies inside the regular hexagon whose corners are the three corners of the starting triangle and the three tips of the stage $1$ star, because each of its three Koch curves stays inside the triangle built on its side with base angles of $30$ degrees, and the top corner of that triangle is a tip of the star. That hexagon has area $2A_0$ . For $s = 1$ the snowflake’s area is

\tfrac85 \cdot \tfrac{\sqrt{3}}{4} = \tfrac{2\sqrt{3}}{5} \approx 0.693,

less than the hexagon’s area $\tfrac{\sqrt{3}}{2} \approx 0.866$ .

Check yourself

Start the snowflake from an equilateral triangle with side $3$ . Find the number of segments, their length and the perimeter at stage $2$ , and the area of the finished snowflake.

Show the answer

Stage $2$ has $3 \cdot 4^2 = 48$ segments, each of length $3/9 = \tfrac13$ , so its perimeter is $48 \cdot \tfrac13 = 16$ . The formula gives the same value, $3 \cdot 3 \cdot \bigl(\tfrac43\bigr)^2 = 16$ .

The starting triangle has area $A_0 = \tfrac{\sqrt{3}}{4} \cdot 9 = \tfrac{9\sqrt{3}}{4} \approx 3.90$ , so the finished snowflake has area

\frac{8}{5} \cdot \frac{9\sqrt{3}}{4} = \frac{18\sqrt{3}}{5} \approx 6.24.

Tripling the side triples every perimeter and multiplies every area by $9$ .

Change the rule so that every new triangle points inward, into the shape, and its inside is removed instead of added. What happens to the perimeter and to the area?

Show the answer

Each segment still becomes four segments one third as long, so the perimeter is still $P_n = 3s\bigl(\tfrac43\bigr)^n$ and grows without bound.

The area now goes down by the same amounts that the outward rule adds, $\tfrac13 A_0$ at step $1$ , then $\tfrac{4}{27} A_0$ , and so on, but only because the removed triangles never overlap one another. That condition holds here. The curve grown inward from each side stays inside the triangle formed by that side and the center of the starting triangle, these three triangles do not overlap, and inside each of them the removed triangles are mirror images of the added triangles of the ordinary snowflake. So the area approaches

A_0 - \tfrac35 A_0 = \tfrac25 A_0.

At step $1$ the three removed triangles have their tips exactly at the center, so what is left is three pieces that meet only at that point.

With side $1$ , after how many steps is the area of the snowflake within $1$ percent of its limit, and what is the perimeter at that stage?

Show the answer

The area still missing after $n$ steps is

\tfrac85 A_0 - A_n = \tfrac35 \left(\tfrac49\right)^n A_0,

and one percent of the limit is $0.016 A_0$ . After $4$ steps the gap is about $0.0234 A_0$ , which is $1.46$ percent of the limit, and after $5$ steps it is about $0.0104 A_0$ , which is $0.65$ percent. So $5$ steps are needed.

At stage $5$ the perimeter is $3 \cdot \bigl(\tfrac43\bigr)^5 = \tfrac{1024}{81} \approx 12.64$ , more than four times the starting perimeter $3$ , and it keeps growing by a third with every further step while the area hardly changes.

The Koch Snowflake, a Boundary That Keeps Growing

Key terms

Koch snowflake

Koch curve

Infinite perimeter

Finite area

Check yourself

How this lesson connects

Used later in