When people say to each other something like "there is a universe hidden in your eyes" or "I'm lost in your eyes, my spirits rise to the skies", they are not exaggerating. There is actually a mathematical truth in these sayings!
Today, we will prove the following mathematical fact:
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| there exists a polygon in the human eye whose perimeter is equal to the circumference of the earth |
In a circle of diameter 1 cm (about the size of the iris of the human eye), there exists a polygon whose perimeter is equal to the circumference of the earth (about 40,000 km)!
We will construct that polygon as follows. First, in the middle of the circle, we draw a small equilateral triangle whose perimeter is equal to 1 cm.
Next, we divide each side of the triangle into three equal segments. Use the middle segment as a base, construct an equilateral triangle which points outward. What we obtain is a new polygon that has 12 sides.
Again, we divide each side of the new polygon into three equal segments and use the middle segment to construct an equilateral triangle, we obtain a new polygon with 48 sides.
Keep doing like this, at each step, we divide each side of the polygon into three equal segments and use the middle segment to construct an equilateral triangle. What we obtain is a polygon that has the shape of a snowflake. In mathematics, we call it Koch's snowflake because this construction was originally described by the mathematician Koch in 1904.
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| Koch's snowflake |
We will show that, with this construction, after a number of steps, we will obtain a Koch's snowflake polygon whose perimeter is longer than the circumference of the earth!
Indeed, at each step of the construction, the perimeter of the polygon is increased by a factor of $\frac{4}{3}$.
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| the perimeter of the polygon is increased by a factor of $\frac{4}{3}$ |
Since we start with an equilateral triangle with perimeter $1$ cm,
- after the first step, the polygon with 12 sides has perimeter equal to $\frac{4}{3} \approx 1.3$ cm
- after the second step, the polygon with 48 sides has perimeter equal to $\frac{4}{3} \times \frac{4}{3} \approx 1.7$ cm
- in the next step, the perimeter is $\frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \approx 2.3$ cm
- after $n$ steps, we obtain a Koch's snowflake polygon with perimeter equal to $\left( \frac{4}{3} \right)^n$ cm
We make the following two observations:
- After $10^{11}$ construction steps, the perimeter of the Koch's snowflake polygon is longer than the circumference of the earth $$\left( \frac{4}{3} \right)^n cm > 100,000 ~km \mbox{ when we choose } n = 10^{11}$$
- No matter how many construction steps we have performed, the Koch's snowflake polygons always lie inside the following triangle $ABC$. So we can be sure that the Koch's snowflake polygons never grow outside of the human eye!
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| Koch's snowflake polygons always lie inside triangle $ABC$ |
We have now proved the following surprising fact
Inside a circle of diameter 1 cm (about the size of the iris of the human eye), there exists a polygon whose perimeter is longer than the circumference of the earth (about 40,000 km), longer than the circumference of the sun (about 4,400,000 km), and even longer than the circumference of the whole universe (if the universe is bounded)!!!
Let us stop here for now. Hope to see you again next time.
When I look in your eyes
I see the wisdom of the world in your eyes
I see the sadness of a thousand goodbyes
When I look in your eyes
And it is no surprise
To see the softness of the moon in your eyes
The gentle sparkle of the stars in the skies
When I look in your eyes
In your eyes
I see the deepness of the sea
I see the deepness of the love
The love I feel you feel for me
Autumn comes, summer dies
I see the passing of the years in your eyes
And when we part there'll be not tears, no goodbyes
I'll just look into your eyes
Those eyes, so wise, so warm, so real
How I love the world your eyes reveal
Leslie Bricusse
Homework.
1. Use the binomial identity, to prove that for any $x > 0$,
$$(1 + x)^n > 1 + n x$$
Then show that the perimeter of the Koch's snowflake polygon obtained after $n$ construction steps satisfies
$$\left( \frac{4}{3} \right)^n > \frac{n}{10}$$
And with $n = 10^{11}$, show that
$$\left( \frac{4}{3} \right)^n cm > 100,000 ~km$$
2. Use induction to prove that the Koch's snowflake polygons always lie within the triangle $ABC$ as follows.
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4. After $n$ construction steps, what is the area of the Koch's snowflake polygon? Show that the area is finite.