Problem 533

Suppose that the length of the track is $2L \text{ m}$, that Arun’s constant speed
is $a\text{ m/s}$, and that Bella’s
constant speed is $\$b\text{}$
m/s}$.

When Arun and Bella run over the same interval of time, the ratio of the
distances that they run is equal to the ratio of their speeds.

Consider the interval of time from the start to when they first meet. In
the diagram, $A$ is Arun’s starting
point, $B$ is Bella’s starting point,
and $P$ is this first meeting
point.

[[IMAGE0]]

Since Arun has run $100\text{ m}$
and together they have covered half of the length of the track, then
Bella has run $(L - $100)\text{}$
m}$.

Thus, $\$\dfrac{a}{b} = \dfrac{100}{L-100}\$.$

From their first meeting point $P$ to
their second meeting point, which we label $Q$, Bella runs $150\text{ m}$.

[[IMAGE1]]

Over this time, Arun runs from $P$
to $B$ to $Q$.

Since Bella runs $150 - 100 = $50\text{}$
m}$past$A$, then$QB = (L - $50)\text{}$ m}$(because$AB = $L\text{}$ m}$and$AQ = $50\text{}$ m}$) and so Arun runs$(L - $100)\text{}$ m} + (L - $50)\text{}$ m}$
which is equal to $(2L - $150)\text{}$
m}$.

Thus, over this second interval of time, $\dfrac{a}{b} = \dfrac{2L-150}{150}$.

Equating expressions for $\dfrac{a}{b}$ and solving, $$\begin{align*} \dfrac{100}{L-100} & = \dfrac{2L-150}{150} \\ 100 \cdot 150 & = (L-100)(2L-150) \\ 15\,000 & = 2L^2 - 350L + 15\,000 \\ 350L & = 2L^2\end{align*}$$ Since $L \neq 0$, then $2L = 350$, and so the total length of the
track is $350\text{ m}$.

Checking, if the length of the track is $350\text{ m}$, then half of the length is
$75\text{ m}$.

This means that from the start to $P$, Arun runs $100\text{ m}$ and Bella runs $75\text{ m}$.

Also, from $P$ to $Q$, Bella runs $150\text{ m}$ and Arun runs $200\text{ m}$.

Note that $\$\dfrac{100}{75} = \dfrac{200}{150}\$$ so these numbers are consistent with the given
information.
Using exponent laws, the following equations are equivalent:
$$\begin{align*} 4^{1 + \cos^3 \theta} & = 2^{2 - \cos\theta} \cdot 8^{\cos^2\theta} \\ (2^2)^{1 + \cos^3 \theta} & = 2^{2 - \cos\theta} \cdot (2^3)^{\cos^2\theta} \\ 2^{2 + 2\cos^3 \theta} & = 2^{2 - \cos\theta} \cdot 2^{3\cos^2\theta} \\ 2^{2 + 2\cos^3 \theta} & = 2^{2 - \cos\theta + 3\cos^2\theta} \\ 2 + 2\cos^3 \theta & = 2 - \cos\theta + 3\cos^2\theta \\ 2\cos^3\theta - 3\cos^2\theta + \cos\theta & = 0 \\ \cos\theta(2\cos^2 \theta - 3 \cos\theta + 1) & = 0\\ \cos\theta(2\cos \theta - 1)(\cos\theta - 1) & = 0\end{align*}$$ and so $\$\cos \theta$ =
0$or$$\cos \theta$ = 1$ or
$\cos \theta = \frac{1}{2}$.

Since $\$0\degree \leq \theta \leq 360\degree$$, the solutions are$$\theta = 90\degree, 270\degree, 0\degree, 360\degree, 60\degree, 300\degree\$.$

Listing these in increasing order, the solutions to the original
equation are $$\theta = 0\degree, 60\degree, 90\degree, 270\degree, 300\degree, 360\degree$$