WoW Mathematics: Breach of Contract

Ever signed a contract later regretted? Unfortunately, they are all too common during the Blood-Queen Lana'thel encounter, as she regularly binds raid members into Pacts of the Darkfallen. The law in Icecrown Citadel cares not of requirements of intention and genuine consent, so surely it will not mind a breach? What is the most expedient and mathematically sensible way to do so?

Summary:

• To remove Pact of the Darkfallen the quickest way possible in the 10-man Blood-Queen Lana'thel Encounter, the raid members should run to the midpoint of the interval they form.
• To remove Pact of the Darkfallen in the quickest way possible in the 25-man Blood-Queen Lana'thel encounter:
• If the raid members trace an acute triangle, they should run to its circumcentre.
• If the raid members trace a right-angled or obtuse triangle, they should run to the midpoint of the longest side.
• The circumcentre can be determined as:
• The point of intersection of any two diameters, if given the circumcircle.
• The point of intersection of the perpendicular bisectors of the sides, if not given the circumcentre (as per the above image).

The given:

• All affected raid members start running at the same time.
• All affected raid members run at the same speed. That means no movement-impairing effects or speed boosts!

The detail:
Ideally, each of the two or three affected raid members should run the same distance. If they do not, one will be waiting while the other(s) is/are still running, consuming precious time. By definition, the midpoint of an interval is equidistant (the same distance) from the endpoints, so for the 10-man encounter, the two affected should meet halfway.

For the 25-man encounter, there is a second dimension to deal with! The point where all three have to run the same distance to reach must still be found. By definition, the centre of a circle is equidistant from all points on the circumference. The three would therefore be looking for the circumcircle they form three points of:

Because of the distortions caused by perspective, the above ellipse does not exactly correspond with the circle drawn on the plane of the ground! Also, because of perspective, the particle effect does not exactly appear behind the ground trace! For convenience, the ground trace (in translucent red) was added. Be sure to base the circumcircle on it!

One way of determining the circumcentre is by constructing the three perpendicular bisectors of the sides and finding their point of intersection, as demonstrated previously. However, if given the circumcircle (like in this example), and especially when looking at the plane from the oblique ground view, an alternative is to construct at least two diameters and find their point of intersection. It can be approximated by the major and minor axes of the apparent ellipse (which approximate two diameters on the actual circle):

The three affected then simply meet at this point:

This holds perfectly well for an acute triangle. However, for a right-angled triangle, two of the paths lie on the hypotenuse, and the one on the right-angle vertex simply crosses to its midpoint.

This principle degenerates when the three affected form an obtuse triangle. Here, the circumcentre lies outside the triangle, and the one on the obtuse vertex needs to cross the longest side to get to it. If this principle is still followed in this case, all three are not running the shortest distance possible! To do so, they all should meet at the midpoint of the longest side, similar to the right-angled triangle case. This case should still be the easiest to handle overall, because the two on the ends of the longest side simply run at each other.