# »  Solutions to puzzles in my VDARE.com monthly Diary

## 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 πp²/2 + πq²/2 = πr²/2, which of course simplifies to p² + q² = r².  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 DB is perpendicular to AB, 45° plus 45°.

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