Bend

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This solution comes from Andrei Lazanu ,Tudor Vianu National College, Bucharest, Romania.

I calculate the length of the stick in terms of $a$, $b$ and $\theta$. From the figure I observe that the length of the stick could be seen as the sum of two hypotenuses of two right-angled triangles. Its length is: $$l(\theta )=\frac{a}{\sin \theta }+\frac{b}{\cos \theta }.$$ Now, I have to calculate the minimum of this expression, in order to make the stick pass through the corner. For this, I calculate the derivative of $l(\theta)$ and equate it to $0$. I must say from the beginning that derivatives are not so familiar to me. For a minimum value of the length: $$\frac{\mathrm{d}\mathrm{l}}{\mathrm{d}\theta }=\frac{-a\cos \theta }{{\sin }^2\theta }+\frac{b\sin \theta }{{\cos }^2\theta }=0.$$ So for a minimum value $a\cos^3 \theta = b\sin^3 \theta$ and $$\tan \theta ={\left(\frac{a}{b}\right)}^{1/3}.$$ Now, I have to calculate $\sin\theta$ and $\cos\theta$ as functions of $\tan\theta$. I know that: $$\cos x=\frac{1}{\sqrt{1+{\tan }^{2}x}}$$ and$$\sin x=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}}$$ In the case of the problem, I have: $$\frac{1}{\cos \theta }=\sqrt{1+{\left(\frac{a}{b}\right)}^{2/3}}$$ and $$\frac{1}{\sin \theta }={\left(\frac{b}{a}\right)}^{1/3}\sqrt{1+{\left(\frac{a}{b}\right)}^{2/3}}$$ So the minimum length is $$\begin{array}{rl}\frac{a}{\sin \theta }+\frac{b}{\cos \theta }& =(a^{2/3}{b}^{1/3}+b)\sqrt{\frac{a^{2/3}+b^{2/3}}{b^{2/3}}}\\ & ={(a^{2/3}+b^{2/3})}^{3/2}\end{array}.$$ The result is symmetric in $a$ and $b$.

If $a=65 \text{ cm}$ and $b=75 \text{ cm}$ then $65^{2/3}+ 75^{2/3}=16.16623563 + 17.78446652=33.95070215$ and $33.951^{3/2}=197.8213407$ so an object of about $197 \text{ cm}$ could be manoeuvred around the bend but it is not possible to manoeuvre a $200 \text {cm}$ object around this bend.