The Elegant Solution to a Simple but Complex Geometry Problem

Imagine you have an isosceles triangle with width W and height H. You start by fitting the largest circle possible inside this triangle. Above that, you fit the largest circle possible in the remaining open space. You continue this process infinitely. The question is, what is the sum of the circumferences of all these circles?

At first glance, this problem might seem like it will have a complex solution. However, there is a simple and elegant way to solve it. This article shall provide a step-by-step guide to understanding and solving this fascinating puzzle.

Understanding the Geometry

Consider an isosceles triangle with a base of width W and height H. The first task is to fit the largest possible circle (inscribed circle) into this triangle. The radius of this circle can be determined using the formula for the radius of the inscribed circle in a triangle:

where A is the area of the triangle and P is the perimeter of the triangle.

The area A of the isosceles triangle is:

The perimeter P can be found using the base W and the two equal sides, which can be calculated as:

Therefore, the radius r of the inscribed circle is:

The next step is to find the radius of the second inscribed circle, which is in the space above the first one. This new circle is also inscribed in a smaller similar triangle, which is formed by the original triangle and the first inscribed circle. The height of this smaller triangle is H - 2r and the base is W - 2r.

The radius of the second inscribed circle, r2, is given by:

Sum of the Circumferences

Now, we need to sum the circumferences of all the circles. The circumference of the first circle is:

The circumference of the second circle is:

In general, the circumference of the nth circle is:

The sum of the circumferences is, therefore:

To find the sum of the circumferences, we need to sum the series of the radii. Notice that the process of fitting smaller circles into the remaining space is a geometric progression. Each subsequent circle has a radius that is a fraction of the previous one.

The ratio between consecutive radii is given by the similarity of the triangles. The height of each new triangle is reduced by 2 times the radius of the previous circle, which is a fraction of the original height. Therefore, the radius of each subsequent circle is 1/4 of the previous one.

This implies that the radii of the circles form a geometric series with the first term r and the common ratio 1/4. The sum of an infinite geometric series with the first term A and common ratio r (where |r|

Substituting A r and r 1/4, we get:

Therefore, the sum of the circumferences is:

Given the initial radius of the first circle, r, we have a simple and elegant solution to the problem. The sum of the circumferences of all the circles is 8πr/3.

Conclusion

This problem is a reminder that sometimes, in mathematics, the solutions to seemingly complex problems are remarkably simple. By understanding the geometric properties and the properties of the series, we can arrive at a solution with ease. The elegant solution to the sum of the circumferences of all the circles is 8πr/3, which is a testament to the beauty and simplicity of mathematics.

Keywords

- isosceles triangle

- inscribed circle

- infinite series