# Two Paths

Which of these two paths made of semicircles is shorter?

## Problem

Image

In the diagram, AE is divided into four equal parts.

Semicircles have been drawn with AE,

AD and DE as diameters.

This has created two new paths from A to E, an upper path and a lower path.

Which path is shorter?

This problem is taken from the UKMT Mathematical Challenges.

## Student Solutions

Let the diameter of the smallest semicircle be $d$. Then $d$ is equal to one of the equal parts of AE, and the diameters of the other semicircles are $3d$ and $4d$, as shown below.

Image

**Finding the lengths of the paths in terms of $d$**

The upper path is half of the circumference of a full circle with diameter $4d$.

The circumference would $\pi\times4d=4\pi d$, so the length of the path is $2\pi d$.

The lower path is the distance around a semicircle with diameter $3d$ and the distance around a semicircle with diameter $d$.

The distance around the medium semicircle is $\frac12\pi\times3d=\frac32\pi d$.

The distance around the smallest semicircle is $\frac12\pi\times d=\frac{1}{2}\pi d$.

So the total length of the lower path is $\frac32\pi d+\frac12\pi d=\frac42\pi d=2\pi d$.

So the two paths are of equal length.

**Considering formulae for lengths of semicircles**

Both paths are made up of semicircles. The circumference of a circle is proportional to its diameter (multiply by $\pi$), so the distance around a semicircle is also proportional to the diameter (multiply by $\frac12\pi$).

So the length of the lower path is $\frac12\pi\times d+\frac12\pi\times 3d=\frac12\pi\times (3d+d)=\frac12\pi\times4d$. But that is equal to the length of the upper path! So the two paths are of equal length.

*Will this work for other pairs of paths whose horizontal distance is equal?*