| Theorem of Pappus: Continued |
Soroban noticed that the final answer to last week's problem, 9π2 + 32/3π2 + 32, is very close to 2, which is the length of the rectangle. (You may want to look over last week's solution.)
Is there some height h of a rectangle of length 2 such that a semi-circular "cap" will move its centroid exactly one unit to the right? If so, find it.
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Problem Moderated by: Sasha |
| Problem Solution |
Let h = 2R, where R = radius of semicircle.
Let C = centroid of entire region.
The area of the rectangle is 4R.
Its centroid is 1 unit to the left of C.
The area of the semicircle is 1/2πR2.
Its centroid is 4R/3π to the right of C.
Since the semicircle and rectangle "balance":
(4R)*(1) = 1/2πR2*4R/3π
Solving for R, we get: R = sqrt(6)
Therefore, h = 2*sqrt(6)
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I had expected an answer involving π...
wrong again. (Why am I always surprised?)
An earlier Surprise was the length of one
arch of a cycloid: 8R, where R = radius of
generating circle. We are rolling a circle
-- I expected π2 at the very least.
But noooo... it turned out to be the
perimeter of the circumscribed square.
Solution submitted by Soroban |
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