MHB Problem of the Week # 238 - Oct 20, 2016

[Ackbach]

Here is this week's POTW:

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Let $s$ be any arc of the unit circle lying entirely in the first quadrant. Let $A$ be the area of the region lying below $s$ and above the $x$-axis and let $B$ be the area of the region lying to the right of the $y$-axis and to the left of $s$. Prove that $A+B$ depends only on the arc length, and not on the position, of $s$.

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[Ackbach]

Re: Problem Of The Week # 238 - Oct 20, 2016

This was Problem A-2 in the 1998 William Lowell Putnam Mathematical Competition.

Congratulations to kiwi for his correct solution, which follows:

Spoiler

\(\displaystyle A=\int^{x=\cos(\phi_1)}_{x=\cos(\phi_2)}\sqrt{1-x^2}\,dx\) and \( \displaystyle B=\int^{y=\sin( \phi_2)}_{y=\sin(\phi_1)}\sqrt{1-y^2}\, dy\)

Now let \(x=\cos(\phi)\) and \(y=\sin(\phi)\); then \(dx=-\sin(\phi) \, d\phi\) and \(dy=\cos(\phi) \, d\phi.\)

Making the substitutions and simplifying yields:

\(\displaystyle A=\int^{\phi_1}_{\phi_2}-\sin^2(\phi) \, d\phi\) and \(\displaystyle B=\int^{\phi_2}_{\phi_1}\cos^2(\phi) \, d\phi.\)

So

\[A+B=\int^{\phi_2}_{\phi_1}\left[\cos^2(\phi) +\sin^2(\phi)\right] d\phi = \int^{\phi_2}_{\phi_1} d\phi=\phi_2-\phi_1.\]

So $A+B$ depends only on the arc length and not on the position of $s$.