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enter image description here

I guess I'm just too dumb to solve this PSAT problem -- so what I tried for this is to break down the shaded area into a difference of sectors and triangles. I want to utilize the area of the 20 degree sector, and I have a hunch that all of this can be achieved without trigonometry. I don't see a way to utilize that Q is the midpoint of PR. I've basically gone nowhere on this problem, and all help is appreciated.

Community
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Saketh Malyala
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  • Hint: area of the sector with $20^{\circ}$ should be equal to the area of the4 shaded region. – Anurag A Jun 24 '17 at 06:43
  • that was my guess, but why???? @AnuragA – Saketh Malyala Jun 24 '17 at 06:44
  • I was thinking: We have that the two intersections between both pairs of adjacent circles are equal. Just oriented differently. And we are subtracting an equal piece from each. So one happens to be a nice sector, and an ugly shaded area. So we have that the areas are equal because subtraction of equal quantities preserves equally. SO the area is $72 * 1/9 = 8$, is that right reasoning? – Saketh Malyala Jun 24 '17 at 07:04
  • but why would area of the 20º sector not be 72 * 1/9? @MichaelRozenberg – Saketh Malyala Jun 24 '17 at 07:06
  • yeah. so thats 1/8 of the semicircle . i just used directly from semicircle @MichaelRozenberg – Saketh Malyala Jun 24 '17 at 07:11

5 Answers5

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The area of the common wedge between the two horizontal semicircles = the area of the wedge between the left horizontal circle and the tilted circle (unfortunately I can't draw a diagram). Now delete the area common to both these regions.

Think: if you rotate back the tilted circle by 20∘, it will superimpose itself on the right horizontal circle.

Anurag A
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  • Is there some property that would let us guarantee that the two created regions have identical areas? – Saketh Malyala Jun 24 '17 at 06:58
  • @SakethMalyala this is where $Q$ being midpoint is helpful. If you look at the figure posted by CY Aries, the point $U$ is the center of the tilted circle. So by symmetry the two areas mentioned in my answer are equal. – Anurag A Jun 24 '17 at 07:00
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Let $A$ be the area of the region bounded by arc $QS$, arc $SU$ and line segment $QU$.

The shaded area is equal to $A$ minus the sum of the area of the minor segment with centre angle $60^\circ$ and the area of the sector with centre angle $40^\circ$.

But $A$ minus the sum of the area of the minor segment with centre angle $60^\circ$ is equal to the area of the sector with centre angle $60^\circ$.

So the shaded area is equal to the area of the sector with centre angle $20^\circ$, which is equal to $\displaystyle 72\times\frac{20}{180}=8$.

CY Aries
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For variety, we can apply a variation on Cavalieri's principle.

The slanted semicircle is obtained by rotating the other semicircle 20 degrees around $Q$. The variation we need is:

If all of the circular slices of two regions have the same length, then the regions have the same volume

By circular slice, I mean the intersections of the region with circles centered on $Q$.

Because of the rotation, it's easy to see slice of the shaded region has the same length as the corresponding slice of the $20^\circ$ wedge — each region is obtained by rotating a figure $20$ degrees — and thus they have the same area.

For the rigorously inclined, the aforementioned principle can be fairly easily proven via area integrals in polar coordinates.

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Let $S$ is an area of the common part of two semicircles (one of them with diameter PR).

Thus, the needed area is $$S+\frac{\pi R^2}{18}-S=8.$$

Michael Rozenberg
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A simple way to think through the answer:

  1. Think of rotating the semicircle in 20° increments to make a complete circle.
  2. This will take 18 steps for a full circle (20*18=360).
  3. Therefore, the small area is 1/18th of a circle or 1/9th of a semicircle.
  4. If the area of the whole semicircle is 72, then the area of the shaded part is 1/9*72=8
JOHN DUBOIS
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