It'd be of the greatest interest to have not only a rigorous solution, but also an intuitive insight onto this simple yet very difficult problem:

Let there exist some tower which has the shape of a cylinder and whose radius is A. Further, let this tower be surrounded by a walking lane whose width is B. Now, there are two individuals who are on the walk; what is the probability that they're able to see each other?

Peter Mortensen
  • 627
  • 5
  • 11
  • 594
  • 4
  • 11
  • 5
    Where on the walking lane are the people standing? On the edge farthest from the tower? – Franklin Pezzuti Dyer May 25 '17 at 21:26
  • 2
    Are we assuming each location on the walkway is equally likely for each traveler? – eyeballfrog May 25 '17 at 21:26
  • 2
    the walking line is a circle on the ground or a spiral from bottom to top ? Is the tower higher than peoples' size ? Is it opaque ? – zwim May 25 '17 at 21:26
  • 4
    Do we care about field of vision for people, or do we consider that if they are not geometrically obstructed they can see 360° ? – zwim May 25 '17 at 21:36
  • 2
    Are you assuming that both people are randomly distributed in an annulus of inner radius A and outer radius A+B? – David Quinn May 25 '17 at 21:51
  • 1
    Are the two walkers going at the same speed? Is their speed constant? – Ian Miller May 26 '17 at 01:29
  • At work I would avoid the high likelihood of human error and go for a numerical solution. All you need is the formula for a line between two points and then the distance of that line from the origin. A spreadsheet would do a good job of it. Sorry to be so un-mathematical. – richard1941 May 31 '17 at 17:17

2 Answers2



By symmetry, one may assume that the first individual is located on the horizontal radius on the left.

The surface he can see is the portion of the lane cut by the tangents to the tower. The area of this surface, $R(r)$, can be computed with a little bit of trigonometry, as the sum of an annular sector, two right triangles and two segments. We can use a reduced area, i.e. the fraction of the whole ring which is visible.

enter image description here

Then you need to compute the average area of this zone for all positions on the radius. As we assume a model of uniform distribution, the positions must be weighted by the distance to the center, as longer circumferences have higher probabilities (said differently, the element of area in polar coordinates has a factor $r\,dr$).


The aperture of the annular sector is $2\alpha=2\arccos\dfrac Ar$.

Then the equation of a tangent:


The tangency point is given by

$$t_t=r\sin A=\sqrt{r^2-A^2},$$

$$(x_t,y_t)=\frac Ar(-A,\sqrt{r^2-A^2}),$$

and the intersection with the outer circle, by

$$t_i=r\sin\alpha+\sqrt{r^2\sin^2\alpha+r\sin\alpha((A+B)^2-r^2)},$$ $$(x_i,y_i)=(-r,0)+t_i(\sin\alpha,\cos\alpha).$$

A triangle has height $B$ and basis $t_i-t_t$, and a segment has radius $A+B$ and aperture angle


Computing the integral looks like a tremendous task.

With $S$ the area of the annulus,

$$SR(r)=\frac S\pi\arccos\frac Ar+B\sqrt{r^2-A^2+\sqrt{r^2-A^2}((A+B)^2-r^2)}+(A+B)^2\left(\beta-\sin\frac{\beta}2\right)$$ where

$$\beta=\arctan\frac{\left(\sqrt{r^2-A^2}+\sqrt{r^2-A^2+\sqrt{r^2-A^2}((A+B)^2-r^2)}\right)\dfrac Ar}{-r+\left(\sqrt{r^2-A^2}+\sqrt{r^2-A^2+\sqrt{r^2-A^2}((A+B)^2-r^2)}\right)\dfrac{\sqrt{r^2-A^2}}r}\\+\pi-\arccos\frac Ar.$$


  • 12
    Note to editors: The choice to put "Sigh" at the end is stylistic and has nothing to do with math or the quality of the post. Transparent attempts to earn reputation simply by deleting it and putting holier-than-thou comments in the edit summary will be rejected. – The Count May 26 '17 at 15:05
  • @DavidQuinn: no, I am using the reduced area, which is dimensionless. This is said in the text and apparent from the formulas. –  May 26 '17 at 19:10

I will answer the version of the problem where each person will be on the outside edge of the lane. You can view the problem as two concentric circles of radii $A$ and $A+B$, with two points on the outer circle. For the version of the problem where the people can stand anywhere on the lane, see Yves's excellent answer.

If the two people are standing as far apart as possible while still being able to see each other, then their line of vision is tangent to the circle of radius $A$. Since they are standing on the outside of the lane, there will be a right triangle formed by the tangent point, the first person, and the center of the circles; the hypotenuse is $A+B$ and one leg is $A$. Thus the angle subtended by the two people on the outer circle is $2 \arccos \frac{A}{A+B}$. By construction this is the largest angle the two people can subtend if they want to still see each other.

If the first person is fixed (e.g., if you condition on the first person's location) and the second person is uniform along the outer edge of the walk, then the probability that the second person is within angle $2 \arccos \frac{A}{A+B}$ of the first person is $\frac{1}{2\pi} \cdot 4 \arccos \frac{A}{A+B}$.

  • 81,252
  • 5
  • 56
  • 122