I'm sure a lot of you are acquainted with Alhazen's Billiard problem, which involves finding the point on the edge of a circular billiard table at which a cue ball at a given point must be aimed in order to canon off the edge of the table and hit another ball at a second given point.

While reading on this math problem, I thought of my favorite Marvel superhero, Captain America (Cap'). You may also know that Cap' has a special ability in which he is able to throw his shield in way so that it will ricochet off certain objects and return to him. Let's examine a scenario that places him in the context of the Billiard problem:

Cap' is standing at point C in an empty circular room. (Point C is not at the centre of the room, because if so he'd be able to throw the shield in any direction and have it bounce off the wall and return to him, since any trajectory would be the normal to the tangent at the point where the shield hits the wall. C is also not on the edge of the circle because that would lead to a trajectory based on inscribed polygons.)

At point C, to have the shield return to him with 1 richochet, all Cap' needs to do is similarly throw the shield along the line of symmetry of the room that passes through C, so we'll constrain his aim to a point not on that line of symmetry, which means we'll be looking for 2 ricochets or more.

These are the scenarios I'd like you all to consider:

Where should Cap' aim to have the shield return to him in:

  • 2 ricochets
  • 3 ricochets
  • 4 ricochets
  • n ricochets?

Edit: From the answers I've read and my own working, I believe there is a pattern for trajectories with either odd or even bounces. It would be nice if someone could algebraically find a generalized way to compute the coordinate of the ricochet point given the number of times we want the shield to bounce off the wall before returning.

Note: For sake of mathematical analysis, let's assume the shield (a point mass) always moves in a straight line, does not lose kinetic energy and obeys the Law of reflection (angle of incidence = angle of reflection.)

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    $\begingroup$ For $2$ ricochets, consider the following: Take the room to be the unit circle, and note that we can always spin the room so that Cap stands on the $x$-axis; say, at point $C(c,0)$. "All you need to do" is find a point $T$ (say, above the $x$-axis) on the circle such that the the ricochet from $C$ at $T$ goes straight down (that is, perpendicular to the $x$-axis). By symmetry, the shield will ricochet at $T^\prime$ (the reflection of $T$ in the $x$-axis) and head directly for $C$. Finding $\angle COT$ in terms of $c$ is relatively straightforward. $\endgroup$
    – Blue
    Commented Aug 17, 2015 at 8:53
  • $\begingroup$ For a circular room, think about your regular polygons and polygrams. $\endgroup$
    – Neil W
    Commented Aug 17, 2015 at 9:14
  • $\begingroup$ @Blue yes, for 2 ricochets the obvious is to somehow create an isosceles triangle with $C$ at one vertex and 2 other vertices at the edge of the circle. But how would one go about finding $T$? $\endgroup$
    – Razorlance
    Commented Aug 17, 2015 at 10:11
  • 1
    $\begingroup$ Also for fun, consider that he may throw his shield with a back-handed throw, trace a regular star polygonal pattern (see en.wikipedia.org/wiki/Star_polygon), and catch it on the forehand side. For example, it may trace the path a 5-pointed star (Schläfli symbol {5,2}), but that has a fairly large center compared to ricochet points. A {7,3} star has a tighter center. $\endgroup$ Commented Aug 17, 2015 at 11:15
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    $\begingroup$ @Razorlance: It's not necessary to rotate the room. Nor is it necessary to assign the room a unit radius, or center it at the origin, or even draw it in the $xy$-plane. But doing so helps, by reducing the problem to its essence; in this case, it spotlights that the only relevant parameter (other than $n$) is Cap's distance from the room's center, relative to the room's radius ... the value I call $c$. It may be "very easy" to work with a parameterized line of symmetry, but it's very very very easy to work with one that aligns with the $x$-axis. I'll take any advantage I can get. :) $\endgroup$
    – Blue
    Commented Aug 17, 2015 at 11:46

1 Answer 1


This is basically a long comment about OP's linked solution to the $2$-ricochet case: You're right, it's a "mess". :)

Don't feel too bad, however. It took me a couple of tries to find a clean approach. A first attempt involved an irreducible cubic; the second, a cubic with an extraneous linear factor. However, a little perseverance and geometric insight finally got me to just the core quadratic relation.

Below is an illustration of (half of) the $2$-ricochet case, where Captain America stands at point $C$, distance $c$ (with $0 < c \leq 1$) from the center $O$ of a unit-radius room:

enter image description here

Here, $T$ is the first ricochet point (the second being the reflection of $T$ in $\overleftrightarrow{OC}$), and necessarily-acute $\theta$ is both the angle of incidence and reflection of the shield bouncing off of the wall at $T$. Similar right triangles in the figure give us this relation:

$$\frac{\cos\theta}{c+\sin\theta} = \frac{\sec\theta}{2c} \quad\to\quad 2 c \cos^2\theta = c + \sin\theta \quad\to\quad 2 c \sin^2\theta+\sin\theta- c = 0$$ Solving for $\sin\theta$ gives $$\sin\theta = \frac{1}{4c}\left( -1 \pm \sqrt{ 1 + 8 c^2 }\right) \qquad\to\qquad \sin\theta = \frac{1}{4c}\left( -1 + \sqrt{ 1 + 8 c^2 }\right)$$ Where the acuteness of $\theta$ allows us to replace "$\pm$" with "$+$". $\square$

(Sanity check: When Cap stands at the wall, so that $c=1$, the formula gives $\sin\theta = 1/2$. That is, $\theta = 30^\circ$, just as one would expect when $C$ and $T$ are vertices of an inscribed equilateral triangle.)

  • $\begingroup$ Very elegant solution. +1 $\endgroup$
    – wltrup
    Commented Aug 21, 2015 at 12:31
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    $\begingroup$ Very nice. Out of curiosity, what application did you use to draw the diagram? $\endgroup$
    – Razorlance
    Commented Aug 22, 2015 at 10:26
  • 1
    $\begingroup$ This is not the only solution for $2$ bounces, at least if $c$ is sufficiently large. Here is another one: If $c>1/2$ then rotate an inscribed equilateral triangle until it passes through $C$. Now fire along the triangle. This gives two further solutions which are not symmetric in the $x$-axis. $\endgroup$ Commented Aug 22, 2015 at 23:10
  • $\begingroup$ Sorry, of course that's 3 bounces. $\endgroup$ Commented Aug 22, 2015 at 23:11
  • 1
    $\begingroup$ @Razorlance: When Sean describes symmetry with respect to "the $x$-axis", he means symmetry with respect to "the line joining $C$ to the origin" (which only happens to align with the $x$-axis in the convenient model). So, an asymmetric shield path in this sense would remain asymmetric upon rotating the room, the path, and the Cap'n. The point here is that shield path is a chain of chords, and it need not be the case (for $n>2$) that this Cap-line meets any of those chords at their midpoints or endpoints the way it does for $n=2$. $\endgroup$
    – Blue
    Commented Aug 23, 2015 at 9:19

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