Johnny and you were playing a game. Johnny was standing 10m away from the centre of a circular tree with a radius of 1m, while you were standing on the opposite side of the tree, also 10m away from the centre. The game was that if you could reach the tree without being seen, you would win, but if Johnny could see you without his line of sight being blocked by the tree, he would win. What was the minimum constant speed you had to run at to win the game, if Johnny was running at 5m/s?
Source: puzzledquant.com
My approach: I understand it is sort of geometry based but I can't form the question in my mind. After time t, I get to 10-kt distance from tree. Now how exactly is Johnny running, I have an intuition it's something to do with tangents.
This question has indeed to do with tangents.
Let us take line $J_0M_0$ as the abscissas axis ($J_0,M_0$= initial positions at time $0$ of Johnny and Me) with the center of the tree as origin. In this way, we can take $J_0=(-10,0)$ and $M_0=(10,0)$. Let $(T)$ be the disk with center $O(0,0)$ and radius $1$ representing the tree.
Consider the family of concentric disks $(D_t)$ with radius $t$ centered at $J_0$. $(D_t)$ is the locus of positions where Johnny can be at time $t$.
Now, consider the 2 common external tangents to disk $D_t$ and disk $(T)$ (line $A_tB_t$ and its symmetrical line wrt horizontal axis). Let $N_t(f(t),0)$ be the point of intersection of these tangents. This defines a shadow area (kind of triangular) where I must be at time $t$, this shadow area shrinking while $t$ increases.
Comments : For the sake of simplicity, let us consider without loss of generality that the speed of Johnny is $1$ m/s.
for values of $t \le 1$, Johnny cannot see me because the tangents do not intersect on the positive part of the axis.
for values of $t$ with $1 \le t < 2$, the tangents' intersection point $S_t$ has an abscissa $>10$ ; therefore, again, I am not seen.
Let us consider now the case where $t \le 2$.
If we denote by $f(t)$ the abscissa of $S_t$. The similarity of right triangles $J_0A_tS_t$ and $OB_tS_t$ allows to write :
$$\frac{10+f(t)}{f(t)}=\frac{t}{1}$$
giving :
$$f(t)=\frac{10}{t-1} \ \ \text{for} \ \ t \ge 2$$
The problem is now to find a function $g(t)=at+10$ accounting for the uniform speed "$|a|$" I have to take ($a<0$ because I move backwards on the axis), in order - in particular - that the curve of $g$ (a line) is always below the curve of $f$ (a hyperbola).
Can you take it from here ?
Remark : at the end, don't forget to multiply the obtained speed by $5$...