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During the last country fair, you have seen a weird arrow-shooter machine.
Its shape is a regular polygon with $N$ sides of length $1$ and it fires one bolt from each vertex at the same time.
The arrows have the same length $A$, same speed; also, suppose that they are unidimensional (no width).
At the beginning, the arrows are arranged so that each arrow's tail coincides with its vertex, and the angle of fire is variable for every bolt (of course, the arrow must be fired inside the polygon).
This picture may be helpful to understand the initial conditions:
enter image description here

Assuming that every angle is determined randomly with equal probability, what is the chance that arrows won't collide after being shot?

Notes:

  • Two or more arrows collide when the segments that represent them have a common point (they intersect).
  • You can imagine the arrows as very thin sticks, with a fixed length $A$.
  • $N>2$ and $0<A<1$ are parameters, while the firing angles (all different) are random variables.
  • The arrows are shot in a straight line. All have the same speed. Their motion is uniform rectilinear. The direction is only determined by the (random) angle.
  • Although the arrows are fired inside the polygon, they may collide out of it.

HINT:

Divide et impera! Simplify the problem, then "add parts" to your solution.

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  • $\begingroup$ Comments removed. Please direct longer discussions to this chat room. Thank you! $\endgroup$ – Aza Mar 28 '15 at 8:37
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    $\begingroup$ Voting the close because this seems to be a rather involved and technical math problem. There isn't any indication of a nice solution or a puzzley insight. $\endgroup$ – xnor Mar 28 '15 at 19:34
  • $\begingroup$ Maybe with these types of puzzles, the OP should assure us that there is a puzzly way to solve it. $\endgroup$ – Lopsy Mar 28 '15 at 21:23
  • $\begingroup$ I'm also of the idea that this is a math-based problem, and therefore off-topic here. $\endgroup$ – Marco Bonelli Mar 29 '15 at 16:54
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    $\begingroup$ My intuition says that the collision events depend on each other in horrible ways. To clarify, what we want is the probability that none of the $N$ arrows ever collide? $\endgroup$ – Lopsy Mar 30 '15 at 16:31
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Edit: Mostly rewritten, with all the math done up to the integral, and results from a Monte Carlo simulation.

Problem definition

As I interpret the problem, what we have is a number of inequalities that must be satisfied by $ N $ random variables, $\theta_i$, $i=1..N$. Define the angles such that when they are $0$, all the arrows are pointed counterclockwise around the polygon. For reference, let's also let $X_i$ be the vertex of the polygon corresponding to $\theta_i$.

A general approach to solving the problem is:

  1. Given a value of one angle, $\theta_i=\alpha$, find the range of $\theta_j$ which will lead to a collision.
  2. The probability of a collision at that angle, $P(C_{i,j}|\theta_i=\alpha)$ is simply the collision range divide by the total allowed range. The probability of no collision is $P(\neg C_{i,j}|\theta_i=\alpha) = 1 -P(C_{i,j}|\theta_i=\alpha) $
  3. To find the total probability for a collision between arrows $i$ and $j$, integrate over the allowed range of angles $\theta_j$, and divide by the allowed range. A priori, we don't know if this can be done analytically.
  4. The total number of pairs $(i,j)$ is the binomial $\binom{N}{2}=\frac{N(N-1)}{2}$. Among these, we only have $\lfloor\frac{N}{2}\rfloor$ different probabilities, because any vertices separated by the same distance will have the same probability. However, we cannot simply multiply all the probabilities together to get the answer, because the outcomes are not independent. This means we will also have to evaluate the probability of any combination of collisions. To calculate these, we can multiply the conditional probabilities inside the integrals... Making it even less likely than an analytical solution exists.

It is my opinion that the math gets too complicated for this to be considered a "puzzle." It is my suspicion that any clever solution that bypasses the math is wrong, either by incorrectly simplifying the geometry, or by assuming the collisions are independent events. However, I'm having fun, so I'll go ahead and finish my answer.

What is a collision?

For any two arrows $i$ and $j$, we can draw lines along their paths as defined by the angles $\theta_i$ and $\theta_j$. Unless the lines are parallel, there will be exactly one intersection point. Call that point $X_{ij}$. If the two arrows are to collide, then that will be the point where it happens. A collision occurs when some part of both arrows is present at the intersection point at the same time. Because all the arrows travel at the same speed, and are fired at the same time, they are each always the same distance from their respective points of origin. Except in very rare cases, the distance to $X_{ij}$ from the two starting vertices will be different. However, if the tail of one arrow has not yet passed when the head of the other arrow arrives, then there is a collision. The limiting case is when the very head of arrow $i$ hits the very tail of arrow $j$, or the very head of arrow $j$ hits the very tail of arrow $i$. In the first case, $\overline{X_iX_{ij}}$, the distance from $X_i$, where arrow $i$ started, to the intersection point $X_{ij}$is longer than $\overline{X_jX_{ij}}$, the distance from $X_j$ to $X_{ij}$, by the length of an arrow, $A$. As an equation,

$$\overline{X_iX_{ij}}-\overline{X_jX_{ij}}=A \tag1 \label{dist1}$$

In the second case, the opposite is true: This distance from $X_j$ to $X_ij$ is longer than the distance from $X_i$ to $X_{ij}$ by the length of an arrow, or

$$\overline{X_jX_{ij}}-\overline{X_iX_{ij}}=A \tag{2} \label{dist2}$$

Flipping equation $\eqref{dist2}$ around and combining the two, we get

$$\overline{X_iX_{ij}}-\overline{X_jX_{ij}}=\pm A\tag3 \label{dist3}$$

Fun with Hyperbolas

As it happens, such an equation defines a hyperbola with foci at $X_i$ and $X_j$. Define a coordinate system with the origin centered between $X_i$ and $X_j$, with the positive x-axis passing through $X_j$ (and thus, the negative x-axis passing through $X_i$). The equation of a hyperbola is: \begin{align} {{x^2}\over{a^2}}-{{y^2}\over{b^2}}&=1 \tag{4} \label{hyp1} \end{align} The difference in lengths from the two foci to the points on the hyperbola is $\pm {2a}$, so for us: \begin{align} 2a&=A\\ a&={A\over2} \tag{5} \label{a} \end{align} The distance between the foci is $2c$, where $$c^2 = a^2 + b^2$$ For our situation, then, \begin{align} 2c &= \overline{X_iX_j}\\ c &= {\overline{X_iX_j}\over 2} \tag{6} \label{c}\\ b^2 &= c^2 - a^2\\ &= \left({\overline{X_iX_j}\over 2}\right)^2-\left({A\over2} \right)^2 \tag{7} \label{b} \end{align}

The eccentricity for such a hyperbola, which we will end up using later, is: \begin{align} \epsilon &= {c\over a}\\ &={{\overline{X_iX_j}/2}\over{A/2}}\\ &={{\overline{X_iX_j}}\over{A}} \tag{8} \label{epsilon}\\ \end{align}

Here is a figure:

Hyperbola showing range of angles which cause a collision

In this figure, $\overline{X_iX_j}=1$ and $A=0.3$, which makes the hyperbola relatively easy to see. The hyperbola is the two blue curves, and the dashed lines are the asymptotes, which the hyperbola arms approach as they extend toward infinity. The green line represents the path of an arrow emanating from $X_i$, and the two red lines represent the two possible paths of an arrow from $X_j$ which will hit the other arrow head-to-tail or tail-to-head. Notice that they intersect the green line at points which are on the hyperbola. The shaded red area represents intermediate angles between the two extremes, where the arrows will collide at some intermediate point, i.e.:

$$-A\le \overline{X_iX_{ij}}-\overline{X_jX_{ij}}\le A \tag{9} \label{dist4}$$

There are a few things we can observe based on this picture, regarding the angles of the two arrows and the asymptotes. All lines, except those parallel to an asymptote, will intersect the hyperbola twice. However, in our problem, the paths of the arrows are really rays, not lines, and so if the intersections are "behind" the starting point, we don't want to consider them. The location of the intersections is different depending on how the path of the arrow corresponds to the the asymptotes. The angle the asymptotes make with the x-axis is given by \begin{align} \DeclareMathOperator{\arcsec}{arcsec} \phi=\arcsec\epsilon \tag{10} \label{phi} \end{align}

For the moment, let's call the angle between the green line and the positive x-axis $\alpha$ (counter-clockwise positive), and the angle made by the red line(s) and the negative x-axis $\beta$ (clockwise positive). Later on we will relate these to $\theta_i$ and $\theta_j,$ but we're not there yet.

  1. If $\alpha \ge \phi$ or $\alpha \le -\phi$, the path of the arrow does not intersect the right arm of the hyperbola. In this case, equation $\eqref{dist1}$ has no solution. However, it may still possible for the arrows to collide, as long as there is an intersection with the left arm. In this case, we add another condition, which is that the two arrow paths do not diverge: \begin{align} \alpha \ge \phantom{-}\phi&\implies\quad \phantom{-}\pi - \beta > \alpha \tag{11} \label{limit1}\\ \alpha \le -\phi&\implies\quad -\pi - \beta < \alpha \tag{12} \label{limit2} \end{align}
  2. If $\alpha \ge \pi - \phi$ or $\alpha \le -\pi + \phi$, the path of the arrow also does not intersect the left arm of the hyperbola, in which case equation $\eqref{dist2}$ also has no solution, and it is impossible for the arrows to collide.

One last bit of math we want to get from the Wikipedia page on hyperbolas is the equation in polar coordinates centered at the focus, which is useful in, for instance, orbital mechanics. It actually only lists the equation for the focus and the near arm, but we can get the equation for the far arm by an analogous derivation, which I won't type out.

\begin{align} r_{near} &= \frac{a(\epsilon^2-1)}{\epsilon\cos\theta+1} \tag{13} \label{rnear}\\ r_{far} &= \frac{a(\epsilon^2)}{\epsilon\cos\theta-1} \tag{14} \label{rfar} \end{align}

Note that, as I have defined $\alpha$ and $\beta,$ these equations apply to either one of them. Since the near arm for $\alpha$ is the far arm for $\beta$ and vice-versa, and we know that, at any given point the difference in the distances is $2a$, we can use these formulas for either focus. Given $\alpha$, then, we can solve for two angles $\beta_+$ and $\beta_-$, representing the intersection with the right and left arms of the hyperbola, respectively. In order to have a result which is valid on the range$[-\pi,\pi]$, we use the tangent half-angle substitution, $\cos\theta=\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$:

\begin{align} r_{\alpha} &= r_{\beta+} +2a\\ \frac{a(\epsilon^2-1)}{\epsilon\cos\alpha-1} &= \frac{a(\epsilon^2-1)}{\epsilon\cos\beta_++1}+2a\\ \left[\left(\epsilon^2+1\right)+2\epsilon\cos\alpha\right]\cos\beta_+&=\left(\epsilon^2+1\right)\cos\alpha +2\epsilon\\ \left[\left(\epsilon^2+1\right)+2\epsilon\tfrac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)}\right]\tfrac{1-\tan^2(\beta_+/2)}{1+\tan^2(\beta_+/2)}&=\left(\epsilon^2+1\right)\tfrac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)} +2\epsilon\\ &\ \ \vdots\\ \tan(\beta_+/2) &= \pm \frac{\epsilon-1}{\epsilon+1}\tan(\alpha/2)\\ \beta_+ &= \pm 2 \arctan\left(\frac{\epsilon-1}{\epsilon+1}\tan(\alpha/2)\right)\tag{15} \label{betaleft}\\ \phantom{x}\\ r_\alpha &= r_{\beta-} -2a\\ &\ \ \vdots\\ \beta_- &= \pm 2 \arctan\left(\frac{\epsilon+1}{\epsilon-1}\tan(\alpha/2)\right)\tag{16} \label{betaright} \end{align}

In both cases, we want the positive solution, which corresponds to $\alpha$ and $\beta$ having the same sign, which is required for a collision.

Polygon geometry

The interior angles of a regular $N$-gon are given by $$\theta_0={{N-2}\over{N} }\pi \tag{17} \label{theta0}$$ and we therefore have $$ 0 \le \theta_i \le\theta_0 \tag{18} \label{thetarange}$$

The radius of our polygon is given here: $$R = {L\over {2\sin{\pi\over N}}} \tag{17} \label{radius}$$

For any $i$, $j$, with $j=i+k, 1\le k \le N-1$, draw a polygon containing the points $X_i .. X_j$. There are $k+1$ such points. As such the interior angles of this polygon must add up to $(k-1)\pi$. For the $k-1$ points between $X_i$ and $X_j$, these interior angles are also interior angles of the regular $N$-gon, and so they are each equal to $\theta_0$. The other two angles are similar, and therefore each equal to

\begin{align} \delta &= {1\over 2}\left[(k-1)\pi - (k-1)\theta_0\right] \\ &= {{k-1}\over N}\pi \tag{18} \label{delta} \end{align}

$\overline{X_iX_j}$ is a chord of the circle circumscribing the $N$-gon, whose subtended angle is $k{{2\pi}\over N}$; its length (here) is: \begin{align} \overline{X_iX_j} &= 2R\sin\left(\frac{k\pi}{N}\right)\\ &=L\frac{\sin\left(\frac{k\pi}{N}\right)}{\sin\left(\frac{\pi}{N}\right)} \tag{19} \label{XiXj} \end{align}

We are now in a position to relate the parameters of the hyperbola in the previous section to the parameters and random variables in the problem. First, from $\eqref{epsilon}$ and $\eqref{XiXj}$,

$$\epsilon=\frac{L}{A}\frac{\sin\left(\frac{k\pi}{N}\right)}{\sin\left(\frac{\pi}{N}\right)} \tag{20} \label{epsilon2}$$

And then, referring to the diagram,

enter image description here

\begin{align} \alpha &= \theta_i-\delta \tag{21} \label{alpha}\\ \beta &= \theta_0- \theta_j - \delta \tag{22} \label{beta}\\ -\delta&\le(\alpha,\beta)\le \theta_0 - \delta \tag{23} \label{anglerange} \end{align}

Solving the integral?

Neglecting for the moment the conditions in $\eqref{limit1}$ and $\eqref{limit2}$, the integral we want is

\begin{align} P(C_{i,j})&=\int_{-\delta}^{\theta_0 - \delta}P(C_{i,j}|\alpha)\mathrm{d}\alpha\\ &=\int_{-\delta}^{\theta_0 - \delta}\frac{\beta_+(\alpha)-\beta_-(\alpha)}{\theta_0}\mathrm{d}\alpha\\ &=\frac{1}{\theta_0}\left(\int_{-\delta}^{\theta_0 - \delta}\beta_+(\alpha)\mathrm{d}\alpha-\int_{-\delta}^{\theta_0 - \delta}\beta_-(\alpha)\mathrm{d}\alpha\right)\\ \end{align}

The question here is whether we can evaluate: $$\int 2 \arctan\left(\frac{\epsilon\pm1}{\epsilon\mp1}\tan(\alpha/2)\right)\mathrm{d}\alpha$$ Or, more generally: $$\int \arctan(c\tan x)\mathrm{d}x$$

Mathematica manages this, but the result is not anything pretty, and it does not simplify it into anything that I would consider elegant. In order to solve the full problem analytically, we also need to find joint probabilities like: $$\int \arctan\left(c_1\tan (x -d_1)\right)\arctan\left(c_2\tan (x -d_2)\right)\dots\mathrm{d}x$$

At this point, I give up on the analytic approach.

Monte Carlo

We have everything we need to determine whether a given set of random variables will lead to a collision. To find the probabilities, let's just run a lot of tests. Here are the results for $N\in[3,8]$. For each graph, values of $\frac{A}{L}$ were considered on the range $[0,1]$ in steps of $0.02$, with one million trials per point, 51 million trials per graph. Each graph took 6–7 minutes to generate on my laptop.

Probability distribution for N=3

Probability distribution for N=4

Probability distribution for N=5

Probability distribution for N=6

Probability distribution for N=7

Probability distribution for N=8

The general pattern is that the probability drops from 1 to 0 fairly quickly as $\frac{A}{L}$ goes from 0 to 1. The entire distribution also drops a little bit as $N$ increases. Of course, we could have said that from the beginning.

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  • $\begingroup$ lol this is made simultaneously $\endgroup$ – Abr001am Mar 27 '15 at 22:24
  • $\begingroup$ Now let's see if both our work checks out... $\endgroup$ – brendan Mar 27 '15 at 22:31
  • $\begingroup$ well you did calculate final value which i didnt , but you followed same steps as me , you did even visited same reference of polygon radius .... are we twins ? lol $\endgroup$ – Abr001am Mar 27 '15 at 22:34
  • $\begingroup$ That's a pretty complicated formula... :S $\endgroup$ – Joe Z. Mar 28 '15 at 0:17
  • $\begingroup$ I've thought about my final solution on the way home, and realized that it's wrong. It is a lower bound on the probability, though. $\endgroup$ – brendan Mar 28 '15 at 1:50
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  • I have settled a little simulation with 6 arbitrary arrow releasers with two changeable parameters (coordinate of eventual collision and length of the arrow A) , to detect where the arrows would meet head to tail or contrarily .

First off we must separate cases ,

when the arrow is unleached from any vertex , the angle $\theta$ which it makes with any side is not the same as that it makes with other side .

So when the polygone is n sided , the sum of its angles = $(n-2) \pi$ so each angle is $\frac{(n-2)\pi}{n}$

enter image description here

  • Update: since the arrow can hit another one or be hit in its tail the difference is fixed . Simulated scenario works fine because the computation is up to a distant system server .

in this picture lets calculate the parameters.

$ ac=\frac{x}{2cos\theta}$

$ ad= ac-A$

$ tan\delta = \frac{ac*sin\theta}{x-ad*cos\theta} $

$$ \delta_1(\theta,x) = arctg \frac{ac*sin\theta}{x-(ac-A)*cos\theta}$$ $Cos\delta_2 = \frac {x-(Cos\theta*(ac+A+r))}{ac+r}$ $$ \delta_2(\theta,x) = arccos \frac{x-(ac+A+r)*cos\theta}{ac+r}$$

  • Update3: after a looong struggle with a third degree polynomial equation i found $r$ as following:

$Cos(e)=\frac{\epsilon}{A+r}$

$Sin(e)=\frac{(A+r)^2-\epsilon^2}{A+r}$

$(A+r)^2-\epsilon^2=ac^2-(ac+r-\epsilon)^2$

$cos(\pi-(2\theta+e))=-cos(2\theta)cos(e)+sin(2\theta)sin(e)=\frac{ac+r-\epsilon}{ac}$

r =

  • $(\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - \frac{1}{(2*cos(2\theta) - 2/ac*(A - ac))}*(sin(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + ((\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - \frac{1}{(2*cos(2\theta) - 2/ac*(A - ac))}*(cos(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + \frac{1/6}{(cos(2\theta) - 1/ac*(A - ac))^2}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^2 - (\frac{1/9}{(cos(2\theta) - 1/ac*(A - ac))^2}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^2 + \frac{1}{(3*cos(2\theta) - 3/ac*(A - ac))}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac))))^3)^{1/2} + \frac{1/6}{(cos(2\theta) - 1/ac*(A - ac))^2}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^{1/3} + \frac{1}{(3*cos(2\theta) - 3/ac*(A - ac))}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)) + (\frac{1/9}{(cos(2\theta) - 1/ac*(A - ac))^2}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^2 + \frac{1}{(3*cos(2\theta) - 3/ac*(A - ac))}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac))))/(\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - 1/(2*cos(2\theta) - 2/ac*(A - ac))*(sin(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + ((\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - \frac{1}{(2*cos(2\theta) - 2/ac*(A - ac))}*(sin(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + \frac{1/6}{(cos(2\theta) - 1/ac*(A - ac))^2}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^2 - (\frac{1/9}{(cos(2\theta) - 1/ac*(A - ac))^2}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^2 + 1/(3*cos(2\theta) - 3/ac*(A - ac))*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac))))^3)^{1/2} + 1/6/(cos(2\theta) - 1/ac*(A - ac))^2*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^{1/3}$

Now lets calculate $x$ in function of $k$

$x=L=1$ for the nearest vertex

the radius of the circle which crosses all vertexes of regular polygon of l =length of one of n sides is $\frac{l}{2sin(\pi/n)}$

the distance between two vetexes belonging to that circle of radius = r is $2r sin\beta/2 $ where $\beta$ is the angle formed by the centeroid and two vertexes

when $\beta=2\pi/(n-k+1)$ and $r=\frac{1}{2sin(\pi/N)}$

$x=2r sin\beta/2 = \frac{sin(\pi/(N-k+1))}{sin(\pi/N)}$

this relation works fine for k=1 , x= 1

Now lets calculate the probability

  • Update2 : removing intersections

$ P1= \int_{0}^{\frac{(n-2)\pi}{n(n-3)}} \frac{1}{\frac{(n-2)\pi}{n}} ( |\delta_1(\theta,1) - \delta_2(\theta,1)|+(1-\frac{|\delta_1(\theta,1) - \delta_2(\theta,1)|}{\frac{(n-2)\pi}{n}})|\delta_1(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)})| + (1-(\frac{(1-((1-\frac{|\delta_1(\theta) - \delta_2(\theta)|}{\frac{(n-2)\pi}{n}})|\delta_1(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)})|)}{\frac{(n-2)\pi}{n}}))|\delta_1(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)}) - \delta_2(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)})|+ ... *|\delta_1(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/2)}{sin(\pi/n)}) - \delta_2(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/2)}{sin(\pi/n)})| )d\theta$

$ P1= \int_{0}^{\frac{(n-2)\pi}{n(n-3)}} \frac{1}{\frac{(n-2)\pi}{n}} ( |\delta_1(\theta,1) - \delta_2(\theta,1)|+(1-\frac{|\delta_1(\theta,1) - \delta_2(\theta,1)|}{\frac{(n-2)\pi}{n}})|\delta_1(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)})| + (1-(\frac{(1-((1-\frac{|\delta_1(\theta,1) - \delta_2(\theta,1)|}{\frac{(n-2)\pi}{n}})|\delta_1(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)})|)}{\frac{(n-2)\pi}{n}}))|\delta_1(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)}) - \delta_2(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)})|+ ... |\delta_1(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/2)}{sin(\pi/n)}) - \delta_2(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta)| ,\frac{sin(\pi/2)}{sin(\pi/n)})d\theta$

.... until $P(n-1)$

each $\delta_n$ is writen in function of $x_n$ and last $\theta$

holy sh** this doesnt seem to end yet , we must calculate the common probability and subtract it from P , following this rule: $P1 \bigcup P2 = P1+P2-P1 \bigcap P2$ enter image description here

so probability of the arrows to be in ollision is

$P = p_1+ ... + p_{n-1} - P_1 \bigcap ....\bigcap P_{n-1}$

The global form of the probability for any arrow i to be hit by atleast one among n-1-i arrows coming from counter-clockwise direction is:

  • $U_m= \frac{(1-U_{m-1})}{\frac{(n-2)\pi}{n}} | \delta_1(| \theta-\frac{(n-2)m\pi}{n(n-3))} |,\frac{sin(\pi/(n-m))}{sin(\pi/n)}) - \delta_2(| \theta-\frac{(n-2)m\pi}{n(n-3))} |,\frac{sin(\pi/(n-m))}{sin(\pi/n)}) |$

  • $$Pr_i= \int_{0}^{\frac{(n-2)\pi}{n}}(\sum_{k=0}^{n-2-i}U_k) d\theta$$

where $i = 0 .. n-2$ and $U_{-1} is Idiomatically =0$

The general probability for any two arrows atleast to be in collision is:

  • $Pr=Pr_0+(1-Pr_0)Pr_1+(1-(1-Pr_0)Pr_1)Pr_2 ..... Pr_{n-2}$

probability to be not is just

  • 1-Pr

Note:

  • Now i think i did subtly detracted the pobability range to more convenient values , eventhough it is more complicated but more accurate , now any one is usual with primitive calculation is welcome to assist here.

-this answer-appendix depends on the meaning of the word 'chance' used by the op

  • for $n-1$ arbitrary values of $\theta$ this following table represents the condition for a non-eventual clash

$$ \begin{array}{|c|c|c||c|c|c|} \hline angle &\theta_1& \theta_2& \theta_k& ...& \theta_{n-1}\\\hline \notin [value_1,value_2]= & [\delta_1(\theta_2,1),\delta_2(\theta_2,1)]& .& .& .& .\\\hline for\ any\ \theta_k& \frac{(n-2)(k-2)\pi}{n(n-3)} \pm[ \delta_1(| \theta_k-\frac{(n-2)(k-2)\pi}{n(n-3)} |,\frac{sin(\pi/(n-k+2))}{sin(\pi/n)}) , \delta_2(| \theta_k-\frac{(n-2)(k-2)\pi}{n(n-3)} |,\frac{sin(\pi/(n-k+2))}{sin(\pi/n)})]& .& .& .& .\\\hline for\ last\ \theta_k &\frac{(n-2)(n-3)\pi}{n(n-3)} \pm[ \delta_1(| \theta_{n-1}-\frac{(n-2)(n-3)\pi}{n(n-3)} |,\frac{sin(\pi/2)}{sin(\pi/n)}) ,\delta_2(| \theta_{n-1}-\frac{(n-2)(n-3)\pi}{n(n-3)} |,\frac{sin(\pi/2)}{sin(\pi/n)})] & .& .& .& .\\\hline \end{array} \qquad$$

  • this is my second simulation version with 6 arbitrary arrow releasers with two changeable parameters (track of the arrow thrown with a specific angle $\theta$ and length of the arrow A) , to detect all probable impacts with other lower and higher bounds of ather 5 arrows' possible gradients.

To be continued ...

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  • $\begingroup$ whoever downvoted me be sure either you are misunderstanding these notations ,or your method is different than this but , both cases , you are doing a mistake. $\endgroup$ – Abr001am Mar 30 '15 at 13:22
  • $\begingroup$ I know you're getting close to the end, but I've been looking through your math, and found an error early on you might want to know about. On your triangle $abc$, you've defined $ad=ac-A$, but this isn't actually the condition "head of one arrow hits tail of the other" because $bd\ne ac$. For head to touch tail, you need $ad=bd-A$ (and $ad=bd+A$). I'm not sure if your subsequent simulation has fixed this issue for you or not. $\endgroup$ – brendan Mar 30 '15 at 18:18
  • $\begingroup$ @brendan about your remark , yes i assumed a conscious proportionnal fault that the difference is 2A , -A for a head-tail collision , 0 fo head-to-head and A for T-H clash , despite the lower bound is right, the higher is nearly erronousbut ....... i ensure that my simulation works fine because if you know how the system does work , you insert data and let the sever calculate the result for you , and thats the secret of its slowness of reaction $\endgroup$ – Abr001am Apr 1 '15 at 10:15

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