Irrationally simple

Geometry Level 3

In the figure above, A B C D ABCD and A M P N AMPN are squares and B D BD and M N MN are arcs of circles with center A A .

What is the ratio of the area of the grey region and the square A B C D ABCD ?

π 4 \frac{\pi}{4} π 8 \frac{\pi}{8} π 5 \frac{\pi}{5} π 7 \frac{\pi}{7} π 6 \frac{\pi}{6}

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Victor Paes Plinio by Victor Paes Plinio , 21, Brazil

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1 solution

Let ∣ A B ∣ = r . |AB| = r. Then ∣ A P ∣ = ∣ A B ∣ = r , |AP| = |AB| = r, and so ∣ A M ∣ = r 2 . |AM| = \dfrac{r}{\sqrt{2}}.

The area S S of the grey region is then the area of a quarter-circle of radius r r minus the area of a quarter-circle of radius r 2 , \dfrac{r}{\sqrt{2}}, i.e., S = π 4 ( r 2 − r 2 2 ) = π × r 2 8 . S = \dfrac{\pi}{4}\left(r^{2} - \dfrac{r^{2}}{2}\right) = \dfrac{\pi \times r^{2}}{8}.

Since the area of the square A B C D ABCD is just r 2 , r^{2}, the desired ratio is S r 2 = π 8 . \dfrac{S}{r^{2}} = \boxed{\dfrac{\pi}{8}}.

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Moderator note:

Great job. Bonus question: What would the area of the grey region be if the circles and squares are nested indefinitely (as shown in the picture below)?

Note that the figure above only shown the first 3 largest grey regions, when in fact there should be infinite number of grey regions.

In response to Challenge Master Note: Let Side of Square A B C D ABCD be a a . Now the area of largest grey region is π a 2 8 \dfrac{\pi a^2}{8} .

Side length of square inscribing second grey region = a × 1 2 × 1 2 = a 2 a\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}} = \dfrac{a}{2} .

So area of second grey region = π ( a 2 ) 2 8 = π a 2 32 = area of preceding grey region 4 \dfrac{\pi (\frac{a}{2})^2}{8} = \dfrac{\pi a^2}{32} \\ = \dfrac{\text{area of preceding grey region}}{4} .

So total area of grey region = π a 2 8 + π a 2 32 + π a 2 128 + … = π a 2 6 \dfrac{\pi a^2}{8} + \dfrac{\pi a^2}{32} + \dfrac{\pi a^2}{128}+ \dots \\ = \dfrac{\pi a^2}{6} .

Purushottam Abhisheikh - 6 years, 1 month ago

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Good analysis, but I think that the sum at the end should come out to

π a 2 8 × 1 1 − 1 4 = π a 2 8 × 4 3 = π a 2 6 . \dfrac{\pi a^{2}}{8} \times \dfrac{1}{1 - \frac{1}{4}} = \dfrac{\pi a^{2}}{8} \times \dfrac{4}{3} = \dfrac{\pi a^{2}}{6}.

Brian Charlesworth - 6 years, 1 month ago

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There was slight error in my calculations. Thanks for pointing it out.

Purushottam Abhisheikh - 6 years, 1 month ago

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