take common speed v(t) integrable function on 0,+infinity (e.g -t as in the first case or -10*t*cos(t) )let V(t) = integral_0^t v(s) dslet f_n = abs(V(t) + (n+1)/2)) + (n-1)/2f_i-1 = abs(f_i - i/2)+(i-2)/2 for i = n .. 2are the trajectories for ants n..1 take opposites for ants -1..-n

<t>MatthewM, basically I did the same eliminating some states as below:this is the version without counting the draws as win for player1:p = 1/4 * T H pm1/4 * T T p1/4 * H T pp1/16 * HH HT 11/16 * HH HH 11/16 * HT HT p1/64 * HTH HHH 01/64 * HTH HHT pp1/64 * HTT HHH 01/64 * HTT HHT pp=pp/4+pm/4+p/4+1...

<t>pm is basically the probability of player1 winning the game if in the first round p1 got a T and p2 got an h (so p1 had a bad start and has slightly less chance to win )the problem is to find the probability of player1 getting hh before player2 gets hhh. for me the game stops as soon as player1 g...

<t>p = proba of winning for player 1pp = proba of winning for player 1 if starting with TH,.T (e.g. H,T or THTH,TTTT etc)pm = proba of winnign for palyer 1 if starting with .T,TH (e.g. T,H or THTHT,HTTTH etc)Equation pp=1/2+pm/4+p/4,pm=pp/4+p/4+p/8+pp/16+2/16,p=pp/4+pm/4+p/4+1/8+p/16+pp/64+p/64 solv...

<t>givenstudent A writes astudent B writes bteacher writes smin < smaxproblem: find sum sanswersA1:if a > smin => s = smax qedelse a < sminB1:if b > smin => s = smax qedelse b < sminA2:from B1: a + b < smin + aif smax > smin + a => s = smin qedelse a > smax - sminB2:from A1: a + b < smin + bif smax ...

<t>the problem is more clear if you put it likethe guy who knows kills himself only at midnight and the others discover the dead in the morning.put yourself in their place:if you see 0 whites you know you are the one and you kill yourself at midnight 1. if you see 1 white and no one dies at 1 then y...

(1,1),(1,2),(4,1)

say (0,x_0) is the focal, by analytic geometry (and maybe a bit of trigonometry) you see that a solution should verify something likeand if you fix an initial condition outside the origin then this should have a unique solution on a neighbourhood of it.

4 beats 5:last four would get 6 points at least as the sum of the matches done just between them.i.e second has at least 6 points. cannot be 7 or 6.5 as 1st is strictly greater than 2nd. so it is 6. i.e last 4 has lost all the games against first 4. i.e. 5 lost to 4.

<t>a is the path clockwise around the nail Aa' is the path counterclockwise around the nail Ab is the path clockwise around the nail Bb' is the path counterclockwise around the nail Be is elementary path hanging out in the airnote that aa' = a'a = ebb' = b'b = eyou can represent a way of hanging by ...

starter needs max 4 steps to get a triangle and it works for 5 points as well

you can free the central cube in 5 cuts: cut the on diagonal to 2 identical pieces than free the 4 remaining faces of the central half cubes simultanously at 4 cuts and put back the halves togather..as you say there is a way to do it at 6 but there might still be a way to do it at 5...

5.each cut can be applied to all existing pieces by stacking all on top of the other and going through all at once. this multiplies then number of pieces by 2. as 27 <2^5 you need only 5 cuts