To try everything Brilliant has to offer for free for a full 30 days, visit brilliant.org/... . You'll also get 20% off an annual premium subscription.
You can also solve it by realizing that the size of the small circles isn't fixed, which means you can put the semicircles an arbitrarily short distance d apart. As d decreases, the areas of the small circles shrink proportionally to d squared and can be ignored. Meanwhile, the yellow area then approaches just a vertically sheared version of the black area, so they must be equal.
The text of the problem implies that the black area doesn’t depend on the radius of the small circles (as otherwise it would be impossible to calculate the result without additional data). In the limit where the circles are infinitesimally small compared to the semicircles, the two semicircles become identical, so the non overlapping regions on the top and on the bottom have the same area. In the same limit, the area of the circles is infinitesimal. So, if the problem can be solved, the black area must be the same as the yellow area.
Could you please share how you thought up using that triangle??? Everything before & after was computation, but that step specifically, how did you come up with it?
I did the same thing. Examining the black region, I could see that the bottom right corner is located on both semi-circles (for which I used radii r and R). So I knew that distance from this point to center of smaller half-circle was r, and that distance from this point to center of larger half-circle was R. Also, since the diameters of the two circles were parallel and diameter of smaller circle was a chord of larger circle, then the line joining the two centers would be perpendicular to both diameters, and therefore we get a right triangle Another thing to keep in mind: a lot of geometry problems involving circles are solved with right triangles. So you should always be on the lookout for right triangles when dealing with circles.
I got really deep into the trig of it based on the yellow area and never thought of making an equation for the black area which is literally the prompt of the question 🤦
Since the red around the small circles does not go outside the black region, then the red around these circles in combination with the white circles themselves form what becomes the blue circles (with radius x) in the solution shown.
There wasn't enough information to solve. There is no indication the blue circles are tangent to the top and/bottom of the black section. In fact, looking at how much black is above and below the blue circles, it's safer to assume that they aren't tangent.
