# ten ants lie on the real line

Question:

Ten ants are on the real number line. At time $t=0$, the $k-th$ ant starts at the point $k^2$ and travelling at uniform speed, reaches the point $(11-k)^2$ at time $t=1$. The number of distinct times at which at least two ants are at the same location is:

Options:
1. $45$
2. $11$
3. $17$
4. $9$

My method:- i found out the velovity of the k th particle and it turned out to be $11(11-2k)$

so the first ant move 99 m second moves 11 m and so on so the first ant crosses 9 other ants second crosses 8 and so on therefore the answer is $9+8+7...+1=45$

but the given answer is $17$

Note:-this question was asked in the exam and i want to challenge its key.So can anybody tell what the correct answer is?

• The question is not asking how many crosses happen between the ants, why are you assuming your $45$ crossings all happen at different times? – Alessandro Codenotti Nov 10 '16 at 14:36
• no i have taken it to be crossing between 0 to 1 sec – MathMan Nov 10 '16 at 14:38
• as the 10th ant moves to the left the first ant moves to the right and it has to overtake or cross evrybody – MathMan Nov 10 '16 at 14:38

The timetable of ant $A_k$ $(1\leq k\leq 10)$ is given by $$x_k(t)=(1-t)k^2+t(11-k)^2\ .$$ Two ants $A_k$ and $A_l$ with $l>k$ could in principle meet at a certain time $t$ which is the solution of the equation $$x_k(t)=x_l(t)\ \tag{1}$$ A priori this solution could lay in the exterior of the $t$-interval $[0,1]$. In this case the ants would not meet during the experiment.
Now analyze how many different admissible values for $t$ you can get by solving equations of the type $(1)$.
• @MathMan, you are correct that all $45$ pairs of ants will meet one another at some time or other. The point is, some of these pairwise encounters happen simultaneously. The problem wants you to count the number of distinct times the encounters occur. – Barry Cipra Nov 10 '16 at 15:13