# Thread: Show that length of arc is between that of line segements

1. ## Show that length of arc is between that of line segements

This is probably a really stupid question, but how (without resorting to calculus in any way) would you show that the length of the arc AC is between the lengths of the line segments AB and AD? ($\displaystyle \sin x < x < \tan x$ is not sufficient).

2. ## Re: Show that length of arc is between that of line segements

Start by taking angle x to be measured in radians. Then the length of arc AC is exactly x. What you want to show is that $\displaystyle sin(x)\le x\le tan(x)$. Notice that you have two right triangles- one, triangle OAB, with hypotenuse of length 1, the other, triangle OAD, with a leg of length 1.

3. ## Re: Show that length of arc is between that of line segements

Yes, I know all of that - I drew the diagram. I'm trying to find an (hopefully intuitive) geometrical argument that the length of the arc is between the lengths of the two line segments.

$\displaystyle x \ge \sin x$ because $\displaystyle \sin x$ is the shortest distance between the line $\displaystyle OD$ and the point $\displaystyle A$. But I can't think of a good reason why the arc must be shorter than $\displaystyle AD$. There are certainly curves within the triangle $\displaystyle \triangle ABD$ that are longer than $\displaystyle AD$, so how can I argue (without reference to $\displaystyle x$ and $\displaystyle \tan x$) that the arc $\displaystyle AC$ is shorter than the line segment $\displaystyle AD$?

4. ## Re: Show that length of arc is between that of line segements

Originally Posted by Archie
But I can't think of a good reason why the arc must be shorter than $\displaystyle AD$. There are certainly curves within the triangle $\displaystyle \triangle ABD$ that are
longer than $\displaystyle AD$, so how can I argue (without reference to $\displaystyle x$ and $\displaystyle \tan x$) that the arc $\displaystyle AC$ is shorter than the line segment $\displaystyle AD$?
The largest triangle given is half of an equilateral triangle. Flip it over upwards to get the full equilateral triangle. Line segment OA is an altitude of that triangle.
Extend the green arc around to make a circle. There are five more equilateral triangles of the same size that are adjacent to each other, all having point O as a
vertex, and beginning with a triangle that shares a side with line segment OD directly below.

The six equilateral triangles overlayed on the green circle divide the circle into six congruent sectors. The altitudes of each of these equilateral triangles divide
each sector in to two smaller congruent sector pieces. The green arc in the problem is associated with one of these smaller sector pieces. There are 12 of them.

The length of each green arc for each smaller sector piece, radius equals 1, is (2*pi*1)/12 = pi/6 ~ $\displaystyle 0.5236,$ and so that is what arc AC's length is approximately.

The outside figure is a regular hexagon. Look at one of its sides. But line segment AD equals half the length of one of its sides. Triangle OAD is a right triangle,
and it's an equal half of an equilateral triangle. Then it's a 30-60-90 triangle. Because the altitude equals 1, then the shortest side equals 1/sqrt(3). That shortest

The shortest side of triangle OAD is line segment AD, and its length equals $\displaystyle \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.5774.$

Therefore, line segment AD is longer than arc AC.

5. ## Re: Show that length of arc is between that of line segements

Originally Posted by greg1313
The largest triangle given is half of an equilateral triangle.
I'm pretty sure it isn't and I'm absolutely sure it's not in general since $\displaystyle x$ is going to head towards zero.

6. ## Re: Show that length of arc is between that of line segements

I realised that I was working with the wrong diagram. This is the one (which make life much easier).