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Two Paths

Age 14 to 16
ShortChallenge Level Yellow starYellow star
Secondary curriculum
  • Problem
  • 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.
 

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?

You can find more short problems, arranged by curriculum topic, in our short problems collection.

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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