There is a circular area with radius 1 km. And there is a tunnel, which is just under the surface, but invisible - unless you dig. It is known that the tunnel goes under the area (at least touches it at one point), it is straight and infinitely long (in both directions).
You have a plow and can dig along some lines with it. When you plow and cross the tunnel you will find it. How much (how long) and where do you have to plow to guarantee that you will find the tunnel?
You are allowed to plow outside of the area as well as inside. You can take the plow out of the ground and move it over the ground without plowing.

For example, you could choose to plow just along the perimeter, and then your result would be $2\pi\approx6.28\ \text{km}$. The task is to make this number as small as possible.

I don't know any good approach, but two people told me that at least $4.83\ \text{km}$ is possible to achieve, and one told that less than $4.8\ \text{km}$ is also possible.

  • $\begingroup$ "at least touches it at one point" means that the tunnel can be tangent to the circle? $\endgroup$
    – leoll2
    Apr 13, 2015 at 16:28
  • $\begingroup$ @leoll2, yes, that's correct. $\endgroup$
    – klm123
    Apr 13, 2015 at 16:30
  • $\begingroup$ Wait, so you're saying that something like this can happen? If so, I don't think it's possible you can get it smaller than the circumference since you must reach every point on the perimeter. $\endgroup$
    – Allan
    Apr 13, 2015 at 16:54
  • $\begingroup$ @Allan, yes, and it is possible: puzzling.stackexchange.com/a/11973/28 $\endgroup$
    – klm123
    Apr 13, 2015 at 16:57
  • 1
    $\begingroup$ Now I want to know what other restrictions for this problem make interesting puzzles too . What if you can plow only within the circle (and the boundary)? Is it simpler in that case? $\endgroup$
    – Simd
    Jan 20 at 9:17

8 Answers 8


4.8264 km

Opaque circle solution with a length of ~4.8264 km

Plough a $120°$ section of the circle, extending both ends in a straight line by a distance of $\sqrt{1/3}$ km to meet the vertices of a hexagon enclosing the circle. Two more lines of length $1$ km and $\sqrt{1/3}$ km are ploughed as shown to cover paths through the other half of the circle. The total length ploughed is equal to:

$$\frac{2\pi}{3} + 3 \times \sqrt{\frac13} + 1 = 4.82644591 \ \text{km}$$

(This puzzle is related to the opaque square problem discussed in Professor Stewart's Casebook of Mathematical Mysteries)

  • 12
    $\begingroup$ It annoys me that even though the puzzle is symmetric, the answer is not. $\endgroup$
    – Etheryte
    Apr 14, 2015 at 0:17
  • 7
    $\begingroup$ What software do y'all use to draw these shapes? $\endgroup$
    – JLee
    Apr 14, 2015 at 0:36
  • 3
    $\begingroup$ @JLee I used Inkscape $\endgroup$
    – r3mainer
    Apr 14, 2015 at 9:51
  • 2
    $\begingroup$ 6 sqrt(1/3) lengths, pointed towards the center from each of the hexagon's vertices, would not be able to see a North-South tunnel in the center. $\endgroup$ Apr 14, 2015 at 22:05
  • 3
    $\begingroup$ To the people who find it strange that the solution to a symmetrical problem is non-symmetrical: This is not unusual. Take magic squares for example, there are multiple 3x3 magic squares, each of which is non-symmetrical by itself. But the saving grace is that the SET of all solutions is symmetrical, i.e. for any magic square with 9 in top left, there will be corresponding ones with a 9 in the other 3 corners. The same is true here - any rotation/reflection of the above diagram is also valid. For any symmetrical problem, the set of all solutions is symmetrical, that's the rule. $\endgroup$ Dec 29, 2016 at 14:37

4.8205 km (or 4.8189 km with slightly more work).

Let $ABCDE$ be a circumscribed regular pentagon around the circle, let $M$ be the point of tangency between segment $BC$ and the circle, and let $N$ be the point of tangency between segment $DE$ and the circle. Let $X$ be the midpoint of line $BE$.

Plow the arc of the circle connecting $M,N$, and plow the line segments $AX, BM, NE$. The total length is

$\frac{4\pi}{5} + |AX| + 2|BM| = \frac{4\pi}{5} + \frac{3\sqrt{5}-5}{2} + 2\sqrt{5-2\sqrt{5}} = 4.82046...$.

Solution to the opaque circle problem based on a regular pentagon

To get 4.8189...

Instead of using a regular pentagon use a slightly irregular circumscribed pentagon. If the pentagon is symmetric around $AX$, with the angle at $C$ equal to $\pi-x$ and the angle at $B$ equal to $\frac{\pi-y}{2}$, then the total length comes out to

$2x + \cos(x+y)\sec(y)-\sec(x+2y) + 2\tan(y)$,

and plugging in $x \approx 1.259557, y \approx 0.6432556$ we get $4.8189...$.

We can also get a lower bound on the total length. First note that we have to plow at least one point with each possible $x$ coordinate between $1$ and $-1$, so the length of the projection of our plow path onto the $x$-axis is at least $2$. Thus the original length is at least 2 km.

Slightly more sophisticatedly, we can combine the fact that the projection onto the $x$-axis has length at least $2$ with the fact that the projection of the length onto the $y$-axis has at least $2$ to see that the total length is at least $2\sqrt{2}$ km: the average of the two projections of any straight line of length $l$ onto the $x$ and $y$ axes is at most $\frac{l}{\sqrt{2}}$, and the total of these averages over all the line segments we plow must be at least $2$.

Taking this to its logical conclusion, if we try averaging the projection of a straight line segment onto a uniformly random direction, we get $\frac{2l}{\pi}$, so the total length plowed must be at least $\pi$ km.

  • $\begingroup$ I considered the solution of his kind, with a regular polygon with 2 sides cut away and a second line starting from the missing vertex to the centre. Doing the full analysis one gets: $L(n) =2\pi (n-3)/n + 2 \cdot \tan(\pi/n) \cdot [1 + \sin(\pi/n)]$ where $L(n)$ is the length in km. This function has a minimum at 5, where it has value: $L(5) = 2 \sqrt{5 - 2 \sqrt{5}} \cdot \left[ 1 + \sqrt{5/8 - \sqrt{5}/8} \right] + 4\pi/5 \simeq 4.82046$ beating even some 3 segment answers. I don't have a figure, just imagine his with a pentagon instead of an hexagon. $\endgroup$
    – 12345ieee
    Apr 14, 2015 at 0:05
  • 4
    $\begingroup$ Could you add a picture? $\endgroup$
    – klm123
    Apr 14, 2015 at 6:35
  • 1
    $\begingroup$ @klm123 I went ahead and added one $\endgroup$
    – r3mainer
    Apr 16, 2015 at 9:26

4.8179 km

I have made a little improvement on @squeamish ossifrage answer where it just requires a little optimization on it, resulting;


with the minimization of the equation.

First of all, using squeamish ossifrage's idea, I have found every angle is equal to each other except the arc one, so I asked myself why it has to be that way;

enter image description here

The red line shows how we are supposed to plow, which is found earlier but not with the optimal angle. So I suspected that the length of the arc and the rest is not optimal even though it seems they are.

Then I defined $|GJ|$ and $|HL|$ in terms of $a$ as below (let me know if you want to show me how I found these equations for each);

$|HL|=2\cdot \sec{a} \cdot sin^{2}a$

To find $|HL|$, I put another illustration below:

enter image description here

As you see, we already know that $|AI|=|AH|=\sec{a}$ and the angle $\measuredangle {HAI}=2a$ so;

$|AL|=\sec{a} \cdot \cos{2a}$


$|HL|=|AH|-|AL|=\sec{a}-\sec{a} \cdot \cos{2a}=\sec{a}(1-\cos{2a})=2\cdot \sec{a} \cdot sin^{2}a$

and lastly,


To show that, I put a zoomed picture on that area;

enter image description here

  1. $\measuredangle {FAM}=3a$ as a result $\measuredangle {MFA}$ becomes $\frac{\pi}{2}-3a$.
  2. $|AF|=|AG|$ and $\measuredangle {FAG}=2a$ so $\measuredangle {GFA}=\frac{\pi}{2}-a$
  3. $\measuredangle {GFM}=\measuredangle {GFA}-\measuredangle {MFA}=2a$

Since we know $|GF|=2\tan{a}$ then;

$|GJ|=2\tan{a}\cdot \sin{2a}=4\sin^2{a}$

Moreover, we now that the tangent lines values are $tana$ already. As a result, we define the total length of the plow;

$P(a)=2\pi-8a+2\tan{a}+2\cdot \sec{a} \cdot \sin^{2}a+4\sin^{2}a$ $P(a)=2\pi-8a+2\tan{a}+2\sin^{2}a\left (2+ \sec{a} \right )$

so we need to minimize this function; as a result mentioned at the beginning, the angle $a$ becomes;

$0.4867$ as radian or $27.89$ as degree.

Note: The illustration is made with the result angle $a$.

  • $\begingroup$ So what is this equation about? $\endgroup$
    – klm123
    Dec 26, 2016 at 5:13
  • $\begingroup$ @klm123 I try to explain it, let me know if you have any question. $\endgroup$
    – Oray
    Dec 26, 2016 at 8:31
  • $\begingroup$ thanks for such a big job! I've got HL just like you (though I don't understand why do you need to write sqrt from square), but my result for GJ is not that simple: wolframalpha.com/input/… $\endgroup$
    – klm123
    Dec 26, 2016 at 9:48
  • $\begingroup$ @klm123 lol I have no idea why I leave it as that, let me fix it :) $\endgroup$
    – Oray
    Dec 26, 2016 at 9:56
  • 2
    $\begingroup$ Think of the all those centimeters you're saving! Real pleasure to see this solution bloom. Gorgeous diagrams. $\endgroup$
    – humn
    Dec 26, 2016 at 12:10

Combining my second guess with CarpetPython's answer (green line is what to plow):
$4.87\ \text{km}$
enter image description here

Second guess:
Ok, I think I can get it down to $5.196\ \text{km}$

If the circle is inscribed in a hexagon, first draw 3 of the spokes (1.155 km each). From the tips of the spokes, imagine it creates the equilateral triangle. From the center of each face, draw a line to the closest corner. Those lines are 1/2 the length of the first set of spokes. So, $4.5 \times 1.155 = 5.196\ \text{km}$ The red lines below are the plow lines. enter image description here

First guess:
If you plow an X across the circle, with leg lengths of $\sqrt{2}\ \text{km}$, you can get your plowing down to $5.656\ \text{km}$ (or imagine that the circle in in a square and you plow both diagonals).

If the tunnel is on the tangent of the circle (say, at the very top) it will barely hit each of the ends of the X.

  • $\begingroup$ Does this include the distance travelled between ends of the X lines? $\endgroup$ Apr 13, 2015 at 16:31
  • $\begingroup$ @IanMacDonald, it must not, only plowing distance is matter. So 5.6km is correct distance for this example, but there are much room for improvement. $\endgroup$
    – klm123
    Apr 13, 2015 at 16:32
  • 6
    $\begingroup$ Interesting note: Wolfram provides a 2 arc solution with a total length of around $4.819$ $\endgroup$
    – Tryth
    Apr 13, 2015 at 22:41
  • 7
    $\begingroup$ Even more interesting note: After scouring the internet, I found a reference which gives a 3-arc diagram with approximate length $4.7999$. It was not easy to find =( $\endgroup$
    – Tryth
    Apr 13, 2015 at 23:17
  • 3
    $\begingroup$ @justhalf Here you go. $\endgroup$
    – Tryth
    Apr 14, 2015 at 11:18

4.8736 km

I have modified my answer when I observed that a central blocking element would result in less plowing. I wrote a short program to try a range of angles (as my calculus and geometry are too rusty) to find the best angle of 30 degrees.

EDIT: Note that JonTheMon posted his 4.87 answer first while I was still programming, so he gets the credit for first with this answer.

new plow diagram

 A  Plow Distance
 19 4.91250640267
 20 4.90597796428
 21 4.90001860583
 22 4.89463591311
 23 4.88983859219
 24 4.88563652529
 25 4.88204083336
 26 4.87906394597
 27 4.87671967936
 28 4.87502332331
 29 4.87399173788
 30 4.87364346116   <- best
 31 4.87399882899
 32 4.8750801084
 33 4.87691164606

First Solution: 5.14 km

First plow the southern half of the circumference, then from each end plow 1 km north (making a U shape). The total distance will be pi+2 or 5.14 km.

plow diagram


5.217 km

I used inscribed (green) and circumscribed (black) squares to help. The original circle is shown in blue. The plow paths are shown in red. The black square has a side of length $2$ and the green square has a side length of $\sqrt2$. The plow path lengths are as follows:

A = $\sqrt2 = 1.414$

B = $\sqrt2\div2 = 0.707$

C = $\sqrt{\sqrt2\div2} = 0.841$

Total Plow Length $ = 5.217$

Even if my math is right, this is still not the optimal answer according to OP's source. I hope that it at least throws out some new ideas, though.



NOTE: Not a complete answer. The purpose of this answer is to possibly offer an idea that someone else can turn into a formal answer.

I was thinking that a shape like this could offer potential. Ignore the red circle. It is just added in for visual effect. Also, the plowed lines obviously won't be tapered like they are in the illustration. enter image description here

Things to consider:

  1. Experiment with the length of the lines
  2. Experiment with the best angle of the lines, relative to the tangent of the circle at that point
  3. Experiment with the number of lines used.
  4. Experiment with adding a small circle to the center, if needed.
  • 3
    $\begingroup$ It's hard to see how this would be less than the perimeter answer. $\endgroup$
    – Jiminion
    Apr 13, 2015 at 19:01
  • 2
    $\begingroup$ I was thinking that the angles would force a bisecting line to hit either one side or the other (or, if close, that a small center circle might help). $\endgroup$
    – JLee
    Apr 13, 2015 at 19:06
  • 1
    $\begingroup$ I think this is the best answer. But there is an optimal number of dashes that make sense. $\endgroup$
    – Jiminion
    Apr 14, 2015 at 14:53
  • $\begingroup$ @Jiminion I suspect that it could, but I am not familiar enough (yet) with engineering or CAD software to test it. It's on my bucket list. Maybe you could do it first and post the answer, and I'd surely upvote it! $\endgroup$
    – JLee
    Apr 14, 2015 at 15:00

I think the 4.83 solution is this:

Circle Solution

The long line to plow along is equal to the hypotenuse of a square with each side being two length, or

sqrt(2^2 + 2^2) = sqrt(8) = 2.828

and the shorter line is the diameter of the circle which is 2. This makes the total length of plowing 2.828 + 2 or 4.828. Note that the lines are perpendicular to each other.

  • 5
    $\begingroup$ Put a nearly vertical tangent at the left, slightly turned like this /: it doesn't hit the long leg; the diameter can only hit an exact tangent at its end point, no other tangents. $\endgroup$
    – user66554
    Apr 13, 2015 at 17:23

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