Solution to the conversion-kicking problem

Remember Regiomontanus’s rugby problem? Expressed in geometrical terms, the task was this: given two points A and B (the goalposts) and a third point T (the location of the try) lying on the same straight line, find a point P on the line through T perpendicular to ABT, such that the angle \angle APB is maximised.

Schematic of the problem. A and B are the goalposts and T is the point where the try was scored. P and P' are possible locations from which to take the kick. Note that the angles APB and AP'B are different!

Schematic of the problem. A and B are the goalposts and T is the point where the try was scored. P and P’ are possible locations from which to take the kick. Note that the angles APB and AP’B are different!

The key to solving this geometrically is a lovely little circle theorem. If we draw any circle through two points A and B and pick a third point C on the arc AB, then the angle \angle ACB is the same whatever point C we pick. Conversely, if we are given two points AB and fix an angle \alpha, then the locus of all the points C for which \angle ACB = \alpha is, up to reflection, the arc of a circle passing through A and B. There’s a nice interactive demonstration here, and a sketched proof here.

(You may be familiar with the special case of this theorem in which AB is the diameter of a circle: pick any other point C on the circumference and \angle ACB will be a right angle. This is Thales’ theorem: it’s said that it’s one of the oldest theorems in mathematics, and that Thales was so chuffed when he proved it that he went straight out and sacrificed an ox in gratitude to the gods. I guess most of us feel like that some days.)

Knowing that this theorem is the tool for the job, the rest is almost easy. Given two fixed points A and B, the smaller the circle is the larger the angle \angle ACB will be. (This is “obvious”, and can be proved by taking C to be the point on the circle furthest from the midpoint of AB and considering isosceles triangles.) So to find P in our original problem, we want to find the smallest circle through A and B such that one of the points on the circle also lies on the dashed line through T. But that just means we need the circle through AB which is tangent to the dashed line. We construct this circle (doing so is a limiting case of the famous Problem of Apollonius), and take the kick from the point P at which it touches the line.

The solution to the problem. The circle ABP is tangent to the line TP'; the point P is then the point on TP' that maximises the angle APB.

The solution to the problem. The circle ABP is tangent to the line TP’; the point P is then the point on TP’ that maximises the angle APB.

Neat, huh? (Mind you, according to the official laws of rugby, the kicker only has one minute to take the kick once s/he has declared an intention to do so… so I guess rugby players must be really good at doing plane geometry problems at speed!)

(DP)

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2 Responses to Solution to the conversion-kicking problem

  1. nigelmottram says:

    But what if I find the point P and then realise I can’t kick that far?

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