Swimming and Running

$f(x) = (3-\sin(2\pi x))\sin\left(\pi x - \frac{\pi}{4}\right) - \sin\left(3\pi x +\frac{\pi}{4}\right) \\$ $I = \{ 0\leq x \leq 100: 0\leq f(x) \leq 2\sqrt2 \}$

Let $\pi x = \theta\\$ $f(\theta) = (3-\sin(2 \theta))\sin\left( \theta - \frac{\pi}{4}\right) - \sin\left(3 \theta +\frac{\pi}{4}\right) \\$ $f(\theta) = (3-\sin(2 \theta)) \left(\dfrac{\sin(\theta)-\cos(\theta)}{\sqrt{2}}\right) - \left(\dfrac{\sin(3 \theta)+\cos(3 \theta)}{\sqrt{2}}\right) \\$ $\sqrt{2}f(\theta) = 3\sin(\theta) - 3\cos(\theta) -\sin(\theta)\sin(2 \theta) + \sin(2 \theta)\cos(\theta) - \sin(3 \theta) - \cos(3 \theta) \\$ $\sqrt{2}f(\theta) = 3(\sin(\theta) - \cos(\theta)) -\dfrac{1}{2}(\cos(\theta)-\cos(3 \theta)) +\dfrac{1}{2}(\sin(3 \theta)+\sin(\theta)) - \sin(3 \theta) - \cos(3 \theta) \\$ $\sqrt{2}f(\theta) = \dfrac{7}{2}\sin(\theta) - \dfrac{7}{2}\cos(\theta)-\dfrac{1}{2}\sin(3\theta) -\dfrac{1}{2}\cos(3\theta) \\$ $2 \sqrt{2}f(\theta) = 7(\sin(\theta) - \cos(\theta))- (4\cos^3(\theta) - 3\cos(\theta)) - (3\sin(\theta)- 4\sin^3(\theta)) \\$ $2 \sqrt{2}f(\theta) = 4\sin(\theta) - 4\cos(\theta) + 4\sin^3(\theta)- 4\cos^3(\theta) \qquad \color{#3D99F6}\text{ -A}\\$ $\dfrac{f(\theta)}{\sqrt{2}} = \sin(\theta)-\cos(\theta) - (\sin(\theta)-\cos(\theta))(1+\sin(\theta)\cos(\theta)) \\$ $f(\theta) = \sqrt{2}(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \\$

$\color{#D61F06} f(\theta) \geq 0 \implies (\sin(\theta)-\cos(\theta)) \geq0 \text{ because } (2+\frac{\sin(2\theta)}{2}) \text{ is always positive.} \\$

$f(\theta) \leq 2\sqrt2 \implies 2 \sqrt{2}f(\theta) \leq 8 \implies 4\sin(\theta) - 4\cos(\theta) + 4\sin^3(\theta)- 4\cos^3(\theta) \leq 8 \qquad \color{#3D99F6}\text{ -from A}\\$

$\implies 4\sin(\theta) - 4\cos(\theta) + 2\sin^2(\theta)\cos(\theta)- 2\sin(\theta)\cos^2(\theta) - 4 \leq 0 \qquad \color{#3D99F6}\text{ LHS}\\$

$\implies (\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \color{#3D99F6}\text{ RHS}\\$

$\color{#D61F06} \text{One can verify LHS = RHS. If anybody has a more intuitive solution, please provide it.}$

$\implies (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\$

$\color{#D61F06} f(\theta) \leq 2\sqrt2 \implies (\sin(\theta)-\cos(\theta)-1) \leq0 \text{ because } (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) \text{ is always positive.} \\$

$\text{Hence, } 0\leq f(\theta) \leq 2\sqrt2 \implies (\sin(\theta) - \cos(\theta) \geq 0) \text{ and } (\sin(\theta) - \cos(\theta) -1 \leq 0) \\$ $\implies 0 \leq \sin(\theta) - \cos(\theta) \leq 1 \implies 0 \leq \sqrt{2}\sin(\theta - \frac{\pi}{4}) \leq 1 \implies 0 \leq \sin(\theta - \frac{\pi}{4}) \leq \frac{1}{\sqrt{2}} \\$

$\color{#D61F06} \implies \theta \in [\frac{\pi}{4} + 2\pi n, \frac{\pi}{2} + 2\pi n] \cup [\pi + 2\pi n, \frac{5 \pi}{4} + 2\pi n] \\$

$\color{#D61F06} \implies x \in [\frac{1}{4} + 2 n, \frac{1}{2} + 2n] \cup [1 + 2n, \frac{5}{4} + 2n], n \in \mathbb{Z^+} \\$ when n = 50, it breaches the finish line. Hence n = 0, 1,2 ... 49. Distance ran = 50 (1/4) + 50 (1/4) = 25. Distance swam= 100-25 = 75.

$\dfrac{\text{Distance swam}}{\text{Distance ran}}$ = $\dfrac{75}{25} = 3$