11
$\begingroup$

When travelling along a outer arc between A and B you have two choices, either diverting onto the inner circular arc or carrying on the outer circular arc, as shown below:

two arcs joined at their extremities by paths of length a that are parallel to the radius

You start on the outer arc, and can either take the 'bridging' path labelled a, to get to the inner arc, walk that inner arc path, and then go from the inner arc back onto the outer arc, via a similar 'bridging' arc. Or you may continue on the outer arc.

The angle of the arc is ϴ (Theta) and the total radius of the outer arc is r, the radially parallel paths are of length a.

At which length a is the diversion via the inner arc (including the two 'bridging' paths) shorter than continuing on the outer arc?

$\endgroup$
2
  • 3
    $\begingroup$ In defense of this question (since I see it has two close votes) - our canonical math puzzle vs math problem policy states three criteria for what distinguishes math puzzles, and I believe that two of them are met here: "Clever or elegant solution, often an "aha" moment" and "Unexpected or counterintuitive result." $\endgroup$
    – bobble
    Aug 11 at 22:37
  • $\begingroup$ The question is interesting for $a>r$. $\endgroup$
    – Servaes
    Aug 12 at 6:26
8
$\begingroup$

The diversion to the inner arc is optimal whenever

$\theta > 2$
(Does not depend on $a$!)

Reasoning

The length of the outer arc is just $r \theta$.
The length of the inner arc is $(r-a)\theta$ while the two additional straight lines adds $2a$.
Hence, the diversion via the inner arc is strictlyoptimal whenever $$ r \theta > (r-a) \theta + 2a$$ $$ \Rightarrow a \theta > 2a \Rightarrow \theta > 2$$

Note that for $a=0$, the paths coincide and for $a<0$ (if you want to consider that), the condition is reversed. For $\theta=2$ the lengths are the same.

What if $a>r$?

This was asked by Servaes in the comments so I thought I'd answer here. Firstly, we have to interpret what to do when we take the first bridge of length $a$ and it seems to me the most natural thing to do would be to follow the arc around the origin in the same direction (clockwise/anticlockwise). In this case, the condition for this being the optimal path becomes $$ r\theta > (a-r)\theta + 2a$$ $$\Rightarrow a < \frac{2\theta}{\theta+2} r$$ Notice that because $a>r$ we need, at least $\theta > 2$ but after this point the optimality condition actually does depend on $a$.

$\endgroup$
7
  • $\begingroup$ YES! Exactly. Did you also find this unintuitive at first glance? $\endgroup$
    – Pureferret
    Aug 11 at 13:38
  • 2
    $\begingroup$ @Pureferret It is a little surprising $\endgroup$
    – hexomino
    Aug 11 at 13:44
  • 2
    $\begingroup$ I think a good intuition is that ϴ can be thought of as defining the shape of the curve (how long it is) while r can be thought of as a scaling factor. As a increases, the inner path is scaled down at a rate similar to the growth of the ends connecting the inner and outer path. When ϴ is great enough, the effect of the scaling of the radii outpaces the static rate of change of a. $\endgroup$
    – hazilnut
    Aug 11 at 13:49
  • $\begingroup$ What about $a>r$? $\endgroup$
    – Servaes
    Aug 12 at 6:26
  • 1
    $\begingroup$ @Servaes yes, I've added it in already $\endgroup$
    – hexomino
    Aug 12 at 10:23
2
$\begingroup$

Answer:

ϴ must be greater than 2 radians.

Proof:

Assuming ϴ is measured in radians, the outer path's length is equal to $$r\theta$$ and the inner path's length is equal to $$2a+(r-a)\theta$$ Since we want to find where the inner path is smaller than the outer path, we solve the inequality: $$2a+(r-a)\theta<r\theta$$ $$(2-\theta)a+r\theta<r\theta$$ $$(2-\theta)a<0$$ Since it makes little sense for a to be less than or equal to zero, we can divide both sides by a to get: $$2<\theta$$

$\endgroup$
1
$\begingroup$

Other answers assume the bridging path is at right angles to the two arcs, which is not explicitly stated in the problem description or diagram (although it is the most obvious inference...). [Ok, it was - I missed the "radially parallel", but the intentional misinterpretation of the problem led me down an interesting path anyway]

If instead we allow any straight line path that connects the two arcs,

The detour becomes more efficient for (outer) arc angles somewhat smaller than the answer given when assuming perpendicular paths, and depends on the ratio of the inner to outer radius.

In the following diagram,

diagram referred to in following comments
The pink lines are at right angles to the arcs, but the purple lines represent the shortest path that goes at least some way along the inner green arc.
In "practical" terms, one could imagine that the blue green and red represent fences, so you have the option of staying outside the blue fence, or going inside at which point you must go all the way to the green fence to use the single gate in the red fence.

At the transition point where both routes are equal,

The purple and blue lines are precisely equal.
Length of blue line is equal to $r\theta$
Length of purple line is equal to $2\sqrt{(r\sin\frac{\theta}2)^2 + (r\cos\frac{\theta}2 - rq)^2}$, where $0 \le q \le 1$ is the ratio between inner and outer arc, which using the fact that $\sin^2 = 1 - \cos^2$ can be easily simplified to $2r\sqrt{1 + q^2 - 2q\cos\frac{\theta}2}$
So we "simply" find the value of $\theta$ for which $\theta = 2\sqrt{1 + q^2 - 2q\cos\frac{\theta}2}$
$\frac{\theta^2}4 = 1 + q^2 - 2q\cos\frac{\theta}2$

However, that's probably not solveable algebraically...

but a numeric plot showing the relationship between the ratio of inner to outer circle, and the resulting angle at which it becomes more efficient to move to the inner arc using a tangent line (or a straight line to the midpoint of the arc if shorter) is shown below:
enter image description here
For $q=0$ where the path is perpendicular to the outer arc, this does indeed agree with the answers making the assumption that the path is always perpendicular, as $\theta = 2$ in that case. At the other limit, when $q=1$ the two arcs have zero distance between them, so a "transition" to the inner arc is always at least as good as staying on the outer one.

$\endgroup$
2
  • $\begingroup$ In the question I state "the radially parallel paths are of length a." is that not equivalent to them being at right angles to the arcs? Let me know how to clear up the phrasing (but please keep your answers, it's interesting!) $\endgroup$
    – Pureferret
    Aug 11 at 15:40
  • 1
    $\begingroup$ Edited to acknowledge I'd not properly read that part... although if I'm rules-lawyering, I could certainly argue that any straight line is parallel to at least one of the possible radii of a circle (albeit not necessarily within the arc specified). Stating that the transition path was "at right angles" would probably have made the intent even clearer, but I knew perfectly well I was misinterpreting the intent... $\endgroup$
    – Steve
    Aug 11 at 15:50
1
$\begingroup$

Another way of thinking about it...

You can tell that the choice to take the inner arc doesn't depend on $a$, since that would imply an optimal radius $r' < r$. Intuitively, such a radius can't be defined, since there's nothing to base it on -- $r$ is already a free variable.

In other words, if $a_{1}$ where $0 < r - a_{1} < r$ is better than the outer ring, then scale the problem by $(r - a_{1}) / r$. In this new problem, $0 < r' - a'_{1} < r'$ where $r' = r - a_{1}$ so $0 < r' - a'_{1} < r - a_{1} < r$. That is, if it makes sense move inward by $a_{1}$, then it makes sense to move inward again by $a'_{1}$, and then $a''_{1}$, an on and on arbitrarily close to $a = r$.

Then...

Clearly the optimal path does depend on $\theta$. With a small angle, it's easy to see that $2a > r\theta$ for at least some $a$, and we already know that some is enough. At the same time, it's obvious that $2r < \pi r$ when $a = r$ and $\theta = \pi$.

The latter inequality is the hint pointing to the formal answer. Since we know that effectively, either $a = 0$ or $a = r$, we'll always be comparing $r\theta$ with $2r$... or $\theta$ with $2$.

$\endgroup$
1
  • 1
    $\begingroup$ Welcome to the site, Randal! I'm Randal Thor :-) $\endgroup$ Aug 12 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.