# Extend a Circumference by 1m

This question is related to the Extending the length of a curcumference by 1 meter problem.

But instead of making a larger circle as in the example above, the loose string is hung from a nail on a wall - like a circular picture frame.

Consider a taut string around the circumference of a cylindrical picture frame of radius R. Then add 1m to the string and put the string around the cylinder and then "hang" the cylinder like a picture frame from a nail on a wall. The string will form an apex at the nail, with the two tangents from the cylinder which then go around the major arc of the circle.

Question: what is the height of the apex from the top of the circle?

• See this figure (you need to find an expression for h in terms of R): dropbox.com/s/uqky1xbv4r6lbnw/Circle.gif?dl=0 – Khalid Jun 15 '15 at 13:31
• Khalid: Your right angles look a bit fat... – TonyK Jun 15 '15 at 13:46
• Tony - yes, I know... it was a very quick sketch, but i guessed everyone would realise that the tangents were at right angles to the radii. – Khalid Jun 15 '15 at 14:20

Let $P$ be the peak of the string, let $C$ be the centre of the circle, and let $T$ be a point of tangency. Let $\theta=\angle PCT$.

The total length of the extended string is $2\pi R+1$. This length is also $R(2\pi-2\theta)+2R\tan\theta$. It follows that $$\tan\theta-\theta=\frac{1}{2R}.$$ This is a transcendental equation with no closed form solution. However, if $R$ is large, such as in the classical problem where $R$ is the equatorial radius of the Earth, the angle $\theta$ is very small. Then we can use the approximation $\tan\theta\approx \theta+\frac{\theta^3}{3}$ to get a good numerical approximation of $\theta$.

• Should that read: This length is also R(2π − 2θ) + 2Rtanθ ? Is there no expression for the height h without resorting to trigonometry? I was looking for a simple GEOMETRIC solution. – Khalid Jun 15 '15 at 14:41
• @Khalid: thank you, corrected. The rest of the post already used the correct expression. Because of the nature of the equation, there is no geometric solution, though there may be a good geometric approximate solution. – André Nicolas Jun 15 '15 at 15:45

Building on the answer from André Nicolas, we have $\tan\theta-\theta\approx \frac{\theta^3}{3}$, so $\theta \approx \left(\frac{3}{2R}\right)^{1/3}$.

But what we want is the height of $P$ above the surface, which is $h = R(\sec\theta-1) \approx \frac12R\theta^2$.

This gives

$$h \approx \left(\frac{9R}{32}\right)^{1/3}$$

The radius of the earth at the equator is about $6378$km, which gives a value for $h$ of about $121.5$ metres.

• I was hoping for an exact geometric solution, but as Andre suspects, I may have to make do with the approximations given above. I'm sure that the maths needed to demonstrate that a geometric solution does not exist is beyond me. But thank you Tony and Andre for your help. – Khalid Jun 15 '15 at 16:49
• @Khalid: I'm not sure I know what you mean by a "geometric solution". Do you mean a construction using ruler and compass? That is unlikely to be possible, because all such lengths are algebraic. But if you allow constructions using circle and string, well, the solution is in the question... – TonyK Jun 15 '15 at 16:53
• I guess what I mean is a solution that does not require the use of any of the trigonometric functions. – Khalid Jun 16 '15 at 8:26