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I posted this problem a few days back, but forgot to include a key information, which meant it was a meaningless question. I hope it makes sense now and someone could give me a hand with this.

It is given, that $\theta = \frac{\pi}{2N}$ for $N \geq 2$ a natural number, which also makes an oriented angle of $45^{\circ}$ with vertical axis and the height from the bottom of the triangle to the top is $\epsilon$. I need to estimate the length of the side $A$ in terms of $N$ and $\epsilon$, as far as I know the estimate $A \leq \frac{8\epsilon}{N} $ should hold but I am really having issues showing it.

In particular having the estimate involving $N$, when using simple trigonometric $\sin$/$\cos$/$\tan$ identities.

Any help would be greatly appreciated! Please see the figure below for an illustration.

enter image description here

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$$A=\epsilon-\epsilon\tan\left(\frac{\pi}{4}-\theta\right).$$ Thus, we need to prove that $$\epsilon-\epsilon\tan\left(\frac{\pi}{4}-\theta\right)\leq\frac{16\epsilon\theta}{\pi}$$ or $$1-\frac{1-\tan\theta}{1+\tan\theta}\leq\frac{16\theta}{\pi}$$ or $$\frac{\tan\theta}{1+\tan\theta}\leq\frac{8\theta}{\pi},$$ which is true because easy to show that even $$\frac{\tan\theta}{1+\tan\theta}\leq\theta$$ is true for all $0\leq\theta\leq\frac{\pi}{4}$.

Indeed, let $$f(\theta)=\theta-\frac{\tan\theta}{1+\tan\theta}.$$ Hence, $$f'(\theta)=1-\frac{1}{\cos^2\theta(1+\tan\theta)^2}=1-\frac{1}{(\sin\theta+\cos\theta)^2}=\frac{2\sin\theta\cos\theta}{(\sin\theta+\cos\theta)^2}\geq0.$$ Thus, $f(\theta)\geq f(0)=0$ and we are done!

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  • $\begingroup$ Thanks so much, somehow did not realize the horizontal distance to the edge is also $\epsilon$. Apologies for the wrong question the first time round! $\endgroup$ – John Smith Jul 29 '17 at 2:28
  • $\begingroup$ @John Smith You are welcome! $\endgroup$ – Michael Rozenberg Jul 29 '17 at 5:11

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