## The Pythagorean tree is a fractal generated by the following procedure:

Start with a unit square. Then, calling one of the sides its base (in the animation, the bottom side is the base):

Attach a right triangle to the side opposite the base, with the hypotenuse coinciding with that side and with the sides in a $3\text – 4\text – 5$ ratio. Note that the smaller side of the triangle must be on the ‘right’ side with respect to the base (see animation).

Attach a square to each leg of the right triangle, with one of its sides coinciding with that leg.

Repeat this procedure for both squares, considering as their bases the sides touching the triangle.

The resulting figure, after an infinite number of iterations, is the Pythagorean tree.

It can be shown that there exists at least one rectangle, whose sides are parallel to the largest square of the Pythagorean tree, which encloses the Pythagorean tree completely.

Find the smallest area possible for such a bounding rectangle, and give your answer rounded to $10$ decimal places.

### This problem assumes some familiarity with sequences and series, geometry and limits.

We start by looking at the ‘tree’ formed by a $3-4-5$ right triangle and iteratively adding $3-4-5$ triangles as specified in the problem statement.

Let’s denote the areas of the squares, as we go up from the base, as $A_0, A_1, A_2, …$. The area of the base square, $A_0$, is 1 (as it is a unit square).

By looking at the construction, we notice that the next square, $A_1$, has a side length of 24 units. Why? Consider the right triangle formed with the base square. Given the Pythagorean relation, we know that the length of the hypotenuse, which is the side of $A_1$, is $\sqrt{3^2 + 4^2} = 5$ units. This line is divided into three parts by $A_1$ and the square attached to the other leg of the right triangle which we call $A’_1$. The side lengths of $A_1$ is the same as the length of the longer (4 units) part, so $A_1$ is $5/3 \times 4 = 20/3$ units. Therefore, the area of this square (since area = side length^2) is $(20/3)^2 = 400/9$.

Using the same reasoning, the area of $A_2$ is $(20/3)^2 * (3/5)^2 = 144/25$.

We notice a pattern — to get from each $A_n$ to $A_{n+1}$, we multiply by $(3/5)^2$. This is because with each iteration up the tree, the side length of next square shrinks by a ratio of $3/5$.

Now that we have recognized this geometric progression, we use the formula for the sum of a geometric series:

$S = A_0 / (1- r) = 1 / (1 – (3/5)^2) = 1 / (1 – 9/25) = 25/16 = 1.5625$

So, $S$, the sum of all areas of the squares in the Pythagorean tree, is the smallest area a rectangle around this tree can have, because this rectangle must contain all these squares.

But wait! The problem wants the answer to 10 decimal places, so $1.5625$ would not be correct. For the problem: in fact, the sum of the areas is so close to the real value that we can use any standard numerical method to get $1.5625000000$. That’s the final answer.

##### More Answers:

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