@Carl Love   Thank you for your detailed explanation!

@Carl Love  Thank you for your helpful comment! The effectiveness of your package (0.1 seconds) for solving such problems is striking. But you wrote that your method does not directly prove the uniqueness of the solutions found. And how much time require such proof?

@Markiyan Hirnyk  Well, and if  t = 10^8 .. 10^8+20 ?

@Markiyan Hirnyk   My code has an important advantage over  dsolve/events  method, because it uses the explicit equation of the trajectory. For example, try to find the trajectory by using  dsolve/events  in the time range  t=10000 .. 10020, if the initial conditions are given for t=0 .

@Markiyan Hirnyk Of course, some variants are only for presentation of result. For the calculations the form  A  can be used. 

@Carl Love  Because the chairs with numbers  1  and  8  are next to each other.

@Dira 

PartialSum1:=proc(n::posint)

local S, k;

S:=0:

for k to n do

S:=S+4*(-1)^(k-1)/(2*k-1);

od;

evalf(S);

end proc:

@Timm  

restart;

b:=c-a:

[a+b+2, a+2+b, b+2+a];

               [c+2, c+2, c+2]

@J4James   The same can be done in one line without calling  plots  package:

restart;

plot([[seq([n, sin((1/10)*n)], n=0..30)], [seq([n, cos((1/10)*n)], n=0..30)]], color=[red, green], style=[line, point]);

@litun    mul(s-a[i], i={$1..100} minus {2});

@Markiyan Hirnyk   I did not notice the obvious: with respect to  f(x)  we have the usual quadratic equation.

@Carl Love  Thanks for the useful information on more intelligent   i  &^ n mod m  command .

@acer  Thank you very much for the detailed analysis of the situation!

@Carl Love   The center of the described circle can be not only inside but also outside the triangle. You have considered only the first case.

@Markiyan Hirnyk   I observed that  if ahead as a factor there is the empty symbol, then it "pulls" a forward all numeric constants, for example

``*sin(x)*sqrt(2)+sqrt(2)*cos(x);