A frog can “jump around” a point in the following way: if a frog at point $M$ jumps around point $N$, it lands on the point of rotation of $M$ that is $120^o$ anti-clockwise about $N$ (i.e. if the point of landing is $M'$, then $NM = NM'$ and $\angle MNM' = 120^o$ in directed angle). A frog is initially at point $X$. $A$, $B$, $C$ are points such that $X$, $A$, $B$ and $C$ are on the same plane and $XA = 10$, $AB = 15$. The frog first jumps around $A$, then around $B$, then around $C$, and then continues to jump over around $A$, $B$ and $C$ alternately. After $420$ jumps, the frog return the its original position at point $X$. The maximum possible value of $XC$ is $p$ and the minimum possible value of $XC$ is $q$. Enter $p^2 - q^2$.
Problem
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Tags: geometry, combinatorics, combinatorial geometry, geometric transformation, rotation
natmath
10.01.2024 02:07
Position everything on the complex plane so that $A,B,C$ are associated with complex numbers $a,b,c$ respectively. Consider the functions $f_A,f_B,f_C$ that map points to their jump arounds points $A,B,C$ respectively. We have that
$$f_A(x)=\omega(x-a)+a=\omega x+(1-\omega) a$$Note that
$$(f_C\circ f_B\circ f_A)(x)=\omega(\omega (\omega x + (1-\omega) a)+(1-\omega) b)+(1-\omega) c$$$$(f_C\circ f_B\circ f_A)(x)=x + \omega^2(1-\omega) a)+\omega(1-\omega) b)+(1-\omega) c$$We can see that this is just displacing $x$ by some constant. If this function application is supposed to return $x$ to itself after $140$ iterations, it must be the case that this constant is $0$. I.e. $a,b,c$ must be an equilateral triangle (and this condition is sufficient). So $AC=15$. Triangle inequality gives us that $5 \leq XC\leq 25$, so our answer is $600$
ap246
10.01.2024 17:39
@above, why must $a,b,c$ form an equilateral triangle? The only solution I see is $a = b = c$
natmath
10.01.2024 20:56
You can prove that the only solutions to $$a+b\omega+c\omega^2=0$$are those that have $a,b,c$ as vertices of an equilateral triangle (either in counterclockwise or clockwise order idr).