# (Improved; Kindly assist )CSI Forensic Maths

In a square room with 7 m-long sides, five people have shot at random people. They all had two bullets and killed two people in the crowd. Other people in the room were not hurt. CCTV recordings show the positions of the shooters, of the victims and of the bystanders just before the tragedy. The family of the victims now want to find out who killed whom. You’re in charge of the investigation” can you provide an answer? Is there a unique solution? We assume that all individuals can be represented as circles of radius r = 30 cm.

Table 1: The coordinates of the suspects and the victims $$\begin{array}{lll} \mbox{Suspect} & x & y \\ A & 3.448069 & 1.825537 \\ B & 5.660685 & 6.548274 \\ C & 4.940687 & 4.281677 \\ D & 4.673141 & 1.664502 \\ E & 0.75885 & 2.935295 \\ \end{array}$$

$$\begin{array}{lll} \mbox{Victims} & X1 & Y1 \\ 1 & 0.878583 & 1.301652 \\ 2 & 5.003333 & 0.768736 \\ 3 & 1.736526 & 4.021814 \\ 4 & 0.81187 & 5.252127 \\ 5 & 3.607172 & 0.302864 \\ 6 & 0.765227 & 0.474429 \\ 7 & 3.862082 & 6.516331 \\ 8 & 5.482858 & 3.218573 \\ 9 & 3.884144 & 2.628101 \\ 10 & 1.80347 & 2.135908 \\ \end{array}$$ Steps I followed: The map of the room is scaled (7m square). The origin is at one corner (0, 0). I assigned the positions of the suspects, victims and bystanders the coordinates (x, y). Using the distance formula; I plugged the point of the victims and suspects (shooters) into that formula to attain the equation of a circle. Max distance btn the points (h, k), and (x, y) is d and can be achieved by the distance formula: d= √(( x_2 )-x_1 )^2 + (y_2- y_1 )^2 Dist. d can be equated to the radius of the circle, r, and by squaring all sides of the equation, I get; r^2=(x-h)^2 + (y- k)^2
This is an example of how I got the distance For shooter A and victim 6, the min distance, d is = √(( x_2 )-x_1 )^2 + (y_2- y_1 )^2= √(( 0.765227)-3.448069)^2 + (0.474429- 1.825537)^2 = √((-2.682842 ) )^2 + (-1.351108)^2 = √9.023134 = 3.003853 m The shorter the distance btn the shooter and the victim, the higher the chance of the victim being shot. This implies that for two victims shot by 1 shooter, is the ones nearest to him, i.e. the one whose, d, is smaller for 2 victims. Using the equation of a circle in the given room, the suspects can be identified with certainty.

I calculated the distances for each suspect manually and came up with the result summarized in the table below: Table 2: suspects and victims they killed and their shortest distance d, (m) btn them $$\begin{array}{lll} \mbox{Suspect} & Victims killed & Distances d, between suspects and victims\\ A (3.448069, 1.825537) & 5 (3.607172, 0.302864) and 6 (0.765227, 0.474429) & 1.530963m and 3.003853m\\ B (5.660685, 6.548274) & 7 (3.862082, 6.516331) and 10 (1.80347, 2.135908) & 1.79886m and 5.860638m \\ C (4.940687, 4.281677) & 4 (0.81187, 5.252127), and 8 (5.482858, 3.218573) & 4.241333m and 1.193372m \\ D (4.673141, 1.664502) & 2 (5.001333, 0.768736), and 9 (3.884144, 2.628101) & 0.954685m and 1.245407m \\ E (0.75885, 2.935295) & 1 (0.878583, 1.301652), and 3 (1.736526, 4.021814) & 1.638925m and 1.461634m \\ \end{array}$$

Please help me solve this using Mathematica. You are also free to point out places that I should improve.

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"The shorter the distance btn the shooter and the victim, the higher the chance of the victim being shot. This implies that for two victims shot by 1 shooter, is the ones nearest to him, i.e. the one whose, d, is smaller for 2 victims." The fact that a certain event is more likely to occur does not imply that it must have occurred. As far as we know from the data, it could have happened that all shooters managed to shoot victims far away from them. –  Austin Mohr Apr 23 '12 at 4:47
This is math.stackexchange.com/questions/135485/… with alleged improvements and serious formatting problems, and no response to my question as to the source of the problem. –  Gerry Myerson Apr 23 '12 at 4:49
@AustinMohr Thank you for pointing that out. I came up with that assumption because i could not think of any other way. If there is another angle you can look at it kindly let me know. If there is none, then take it as an assumption and help me solve using Mathematica. Thanks again –  Niga Apr 23 '12 at 4:52
@GerryMyerson Was given in class by our lecturer –  Niga Apr 23 '12 at 4:54
@Niga If you assume the victims fill the entire circle that represents them, then it does follow that you can't shoot through a circle without harming that person. Perhaps you can deduce some facts like "Shooter X only has line of sight to Victims Y and Z, so he must have shot them." –  Austin Mohr Apr 23 '12 at 4:58