At first I guessed that the sum of the two shaded areas would be half the area of circle they're inside of, since I figured for a problem like this the sum would be independent of the semicircles' sizes and you get half the big circle's area in the limiting cases when one of the semicircles vanishes.

I expected the proof to be something like the situation when a triangle is inscribed in a rectangle so that a side of the triangle coincides with one of the sides of the rectangle (the triangle's area will always be half the rectangle's area), but I couldn't see how to do something like that for this situation.

So I decided to cheat and grind out an algebraic calculation on paper. Letting the semicircles have radii r and R, the sum of their areas is (1/2)(pi)(r^2) + (1/2)(pi)(R^2), and half the circle's area is (1/2)(pi)(r + R)^2 . . . Hummm...

Setting these equal and canceling like terms gives (pi)rR = 0 (i.e. r = 0 or R = 0), so my initial guess is correct ONLY for the two limiting cases! Note that these limiting cases were the only two cases that I had actually originally verified my guess for. This shows why one needs to insist on proof in mathematics -- the results for a few simple test cases might not be what the results are for all cases.

In general, the difference between half the big circle's area and the sum of the two semicircle's areas is (pi * rR), and the ratio of this difference to the big circle's area is (pi * rR)/[pi * (r + R)^2]. Cancelling pi's and dividing numerator and denominator by R^2 gives

(r/R) / (r/R + 1)^2,

and this expression varies when the value of r/R varies, because the value of x/(x+1)^2 varies when x varies. Thus, the percentage of the big circle's area taken up by the two semicircles is a function of r/R. Not the function just above, since THAT function represents (C/2 - S)/C, where C is the big circle's area and S represents the sum of the semicircles' areas. But since (C/2 - S)/C = (1/2) - (S/C), we get

(r/R)/(r/R + 1)^2 = (1/2) - (S/C)

and hence the ratio of the sum of the semicircles' areas, which is S, to the area of the big circle, which is C, wind up being THIS function of r/R:

S/C = (1/2) - (r/R)/(r/R + 1)^2

Thus, if you specify the ratio of the radii of the semicircles, then the percentage of the big circle's area taken up by the two little semicircles will be determined, but this percentage will vary when you vary the ratio r/R.

At first I guessed that the sum of the two shaded areas would be half the area of circle they're inside of, since I figured for a problem like this the sum would be independent of the semicircles' sizes and you get half the big circle's area in the limiting cases when one of the semicircles vanishes.

ReplyDeleteI expected the proof to be something like the situation when a triangle is inscribed in a rectangle so that a side of the triangle coincides with one of the sides of the rectangle (the triangle's area will always be half the rectangle's area), but I couldn't see how to do something like that for this situation.

So I decided to cheat and grind out an algebraic calculation on paper. Letting the semicircles have radii r and R, the sum of their areas is (1/2)(pi)(r^2) + (1/2)(pi)(R^2), and half the circle's area is (1/2)(pi)(r + R)^2 . . . Hummm...

Setting these equal and canceling like terms gives (pi)rR = 0 (i.e. r = 0 or R = 0), so my initial guess is correct ONLY for the two limiting cases! Note that these limiting cases were the only two cases that I had actually originally verified my guess for. This shows why one needs to insist on proof in mathematics -- the results for a few simple test cases might not be what the results are for all cases.

In general, the difference between half the big circle's area and the sum of the two semicircle's areas is (pi * rR), and the ratio of this difference to the big circle's area is (pi * rR)/[pi * (r + R)^2]. Cancelling pi's and dividing numerator and denominator by R^2 gives

(r/R) / (r/R + 1)^2,

and this expression varies when the value of r/R varies, because the value of x/(x+1)^2 varies when x varies. Thus, the percentage of the big circle's area taken up by the two semicircles is a function of r/R. Not the function just above, since THAT function represents (C/2 - S)/C, where C is the big circle's area and S represents the sum of the semicircles' areas. But since (C/2 - S)/C = (1/2) - (S/C), we get

(r/R)/(r/R + 1)^2 = (1/2) - (S/C)

and hence the ratio of the sum of the semicircles' areas, which is S, to the area of the big circle, which is C, wind up being THIS function of r/R:

S/C = (1/2) - (r/R)/(r/R + 1)^2

Thus, if you specify the ratio of the radii of the semicircles, then the percentage of the big circle's area taken up by the two little semicircles will be determined, but this percentage will vary when you vary the ratio r/R.