# Pythagorean motion

At the same moment two particles start respectively from vertices $B$ and $C$ of triangle $ABC$ which has a right angle at $C$. The particles move at constant speeds and arrive at vertex $A$ at the same moment.

If the time of travel is $(AB/AC)^{3/2}$ and the length of $BC$ is equal to the sum of the constant speeds, determine the time of travel.

let $\tau = (\frac{AB}{AC})^{3/2}$ and $BC=v_1+v_2$
We know that $\tau_1=\tau_2=\tau$ $\rightarrow$ $\frac{AB}{v_1}=\frac{AC}{v_2}$ $$\frac{v_1}{v_2}=\frac{AB}{AC}=\tau^{2/3}\hspace{1 pt}...(1)$$
then $AB=v_1.\tau$ and $AC=v_2.\tau$ $$\sqrt{AB^2-AC^2}=v_1+v_2$$ $$\tau.\sqrt{v_1^2-v_2^2}=v_1+v_2$$ square both sides $\rightarrow \tau^2.(v_1-v_2)=v_1+v_2$ $$v_1(\tau^2-1)=v_2(\tau^2+1)$$ $$\frac{v_1}{v_2}=\frac{\tau^2+1}{\tau^2-1}$$ from the (1) we know that : $$\frac{\tau^2+1}{\tau^2-1}=\tau^{2/3}$$ arrange a little bit : $$\tau^{8/3} -\tau^2-\tau^{2/3}-1=0 \hspace{2 pt}...(2)$$ set $\tau^{2/3}=p$, and eq (2) become $p^4-p^3-p-1=0$, factor and we get : $(p^2+1)(p^2-p-1)=0$
clearly $p \neq \pm i$ . And from the other , we get $p=\frac{1}{2}(1 \pm \sqrt{5})$. Because $\sqrt{5} > 1$ , the only solution for $p$ is :
$$p=\frac{1}{2}.(1+\sqrt{5})$$ $$\tau=(\frac{1}{2}.(1+\sqrt{5}))^{3/2} \approx 2.05817$$