## May 2021

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In the Math Corner of my May Diary I included the following brainteaser.

I have parked two semicircles inside a circle, as shown. What looks to be the case, *is* the case: *AB* and *CD* are
parallel, *E* is a single point of contact.

Prove that, within the larger circle, the shaded area is equal to the non-shaded area; in other words, that the sum of the areas of the two semicircles is half the area of the big circle.

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**• Solution**

There are three circles in play here. There's the smallest one, of which *AEB* is a semicircle; then a slightly bigger one, of
which *CED* is a semicircle; and then the big enclosing circle. I'll call the radii of these three circles *p*, *q*, and
*r*. Just remember the order, smallest to largest: *p*, *q*, and *r*.

Applying the formula for the area of a circle, we have to prove that
π*p*²/2 + π*q*²/2 = π*r*²/2, which of course simplifies to
*p*² + *q*² = *r*². Aha! — Pythagoras' theorem. But where is the right-angled
triangle?

First construct a *diameter* of the big circle parallel to *AD*. I've drawn it in red here. Now imagine that diameter is a
mirror and *reflect* the line *CD* in the mirror.

*C* reflects to *C*′, which plainly is just the same as *B*. *D* reflects to some point on the big
circle's circumference a wee bit clockwise of *D* — call it *D*′; and it is easy to see that
*D*′*B* is perpendicular to *AB*, 45° plus 45°.

There's your right-angled triangle, *ABD*′. And guess what: *AD*′ = 2*r* (because any chord that
subtends a right angle at the circle's circumference must be a diameter), *BD*′ = 2*q* (because it's the reflection of
*CD*), and *AB* = 2*p* (duh). The result follows from Pythagoras' Theorem.

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** Added later**: A reader sent in a niftier solution. From the center of the big circle, draw raddi to

*A*and

*C*. We then have:

Now add first and last equations and square!