2023 SSMO Team Round Problems/Problem 8

Problem

Three rabbits run away from the origin at the same speed and constant velocity such that the angle between any two rabbits' directions is $120^\circ$ . After $12$ seconds, a hunter with a speed $\sqrt{7}$ times that of the rabbits runs from the origin. Let the minimal time in seconds needed for her to meet (and subsequently) catch all three rabbits be $a + b\sqrt{c}$ . Find $a + b + c$ .

Solution

WLOG let the rabbits’ speed be $1$ arbitrary unit per second and the hunter’s speed be $\sqrt{7}$ units per second. Obviously, the first rabbit is able to be caught if the hunter runs in a straight line. Let $t_1$ be the amount of time necessary. Then, by considering the distance from the origin, we create the following equation: $(t_1+12)=\sqrt{7}t_1$ Thus $t_1=2+2\sqrt{7}$ .

In order to catch the next rabbit, the hunter must run in a straight line to a point on the rabbit’s path such that they meet at the same time. She can choose either of the remaining rabbits (due to symmetry, it does not matter). Let $t$ be the amount of time necessary, and let $a$ be the distance already run by the rabbit (in other words, the distance from that rabbit to the origin, which should be the same distance as between the hunter and the origin). Then we create a triangle with sides $\sqrt{7}t,t+a,a$ and one angle (between sides of length $a,t+a$ ) of $120^\circ$ . Thus, by using the Law of Cosines: $\begin{aligned} \sqrt{7} t & = \sqrt{(t + a)^{2} + a^{2} - 2 a (t + a) \cos 120^{\circ}} \\ = \sqrt{t^{2} + 2 a t + a^{2} + a^{2} - 2 a (t + a) \cdot (- \frac{1}{2})} \\ = \sqrt{t^{2} + 2 a t + 2 a^{2} + a t + a^{2}} \\ = \sqrt{t^{2} + 3 a t + 3 a^{2}} \end{aligned}$ Thus $7t^2=t^2+3at+3a^2$ , so $2t^2-at-a^2=0$ . Solving this quadratic yields $t=\frac{a\pm3a}{4}=a\text{ or }-\frac{1}{2}a$ Obviously $t$ is nonnegative, so $t=a$ . Thus the time it takes to run to the second rabbit is the same as the distance from the origin to begin with. We can find this value easily; since $t_1=2+2\sqrt{7}$ and the rabbit was given $12$ seconds to begin with, we conclude that the rabbit ran the sum, or $14+2\sqrt{7}$ units. Thus it also takes this many seconds for the hunter to catch the second rabbit. However, the second rabbit initially is also $14+2\sqrt{7}$ units from the origin, but it is able to run another $14+2\sqrt{7}$ units by the time the hunter catches it. Thus the hunter is $28+4\sqrt{7}$ units from the origin by the time of the second catch. Then it takes $28+4\sqrt{7}$ seconds for the hunter to catch the third rabbit.