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Problem Of The Week #471 June 8th 2021

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anemone

MHB POTW Director
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Feb 14, 2012
3,909
Here is this week's POTW:

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Find the area of the set of all points in the unit square, which are closer to the center of the square than to its sides.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,909
Congratulations to lfdahl for his correct solution(Cool), which you can find below:
By symmetry, it suffices to solve the problem in one half quadrant of the unit square.

The boundary of the solution set comprises of the points in the plane for which $|OA| = |AB| = \frac{1}{2}-y$.

From the figure, one immediately has

(a). $\cos \theta = \frac{y}{\frac{1}{2}-y}$

(b). $z^2 = \left ( \frac{1}{2} \right )^2+ x^2$

Since $\angle OAB = \pi - \theta$ the acute angles in the isosceles triangle $OAB$ are $\frac{\theta}{2}$, and it follows, that

(c). $z = 2 (\frac{1}{2}-y) (\cos \left ( \frac{\theta}{2} \right )$.

With the help of the half-angle formula: $\cos \theta = 2\cos^2\left ( \frac{\theta }{2} \right ) -1$ and combining (a), (b) and (c), we get the result:

\[y = \frac{1}{4}-x^2\]. The graph has endpoints in $\left ( 0,\frac{1}{4} \right )$ and $(r,r)$.

The right end point is determined by the condition: $r = \frac{1}{4}-r^2$, which has the (positive) solution:

$\frac{1}{2}\left ( \sqrt{2} -1\right )$.

The solution set for the 1st quadrant thus comprises the area under the red y-curve minus the area of the coloured triangle:

\[A = \int_{0}^{r}\left ( \frac{1}{4}-x^2 \right )dx - \frac{1}{2}r^2 = \left ( \frac{1}{4}-\frac{1}{2}r-\frac{1}{3}r^2\right )r\]

Thus, by symmetry, the solution is: $A_{sol} = 8 A = \frac{1}{3}\left ( 4\sqrt{2}-5 \right ) \approx 0.218951$
potw 471.png
 
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