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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.
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.
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 🤦
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.
This problem was beatiful. Great resolution
That single right triangle brought everything together. How exciting, indeed!
how exciting
how exciting
how exciting
very :):):)
My son is currently working on the coding lessons in Brilliant. He loves it.
Wow that... was... mindblowingly easy for how intimidating it was. Amazing job!
What the hell!! That right angled triangle was beautiful
i came from that right triangle plot twist, jesus christ.
Absolute smut
Dawg what 😭🙏
@@thachookie6545 it was literally so godly
Aaaaand it’s all over the screen😅
Clean up on aisle MY PANTS 😂😂😂
German smiley
I was gonna say that
Man you always solve it so much simpler than I do, this one had me messing around with chords and areas of segments and all the rest of it.
Wow, it sure seemed like there wasn't enough information at first, but with those cancellations it worked out! Beautiful.
you’re the “kentucky ballistics” of math problems for me.
love watching your videos!
that was wonderful! thanks for showing this puzzle!
This was *delightfully* exciting!
I loved this problem! How exciting!
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.
Love the solution! Will definetly check out briliiant too! Had a quick question though. What software do you use?
This is the first math exercise I've seen that has a narrative arc.
How exciting...!
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.
That is brilliant!
I actually gasped when the terms cancelled out omg
Creepy smiley tbh, but the solution with radius and triangle was awesome
ngl this one was a bit difficult for me to follow and the use of traingles completely gave me a whiplash 😭
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.
You are brilliant sir 🙂
"smiley face"
meanwhile I see a criminal banana
How exciting! 🙂
Great video
Brilliant
What about the red around the small circles?
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.
@AndyMath - How do you create these spicy solutions in the form of Animation? What software/tool are you using? Can anyone else help me?
Further proof that everything is a triangle.
Yeeey! New vid!
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 🤦
so it doesn't matter how big the top semicircle is, right? it's freaky how that works.
✨Magic!✨
I wanted to solve this bc it's a smiley face but I figured it was like calc-level hard.
Your videos are like these '5 minutes craft' TikToks that take days to actually do them. How long does it take you to solve the problem?
Will the yellow area equal the black area for any scale of this problem?
Yes.
@@MarieAnne. Have you actually proven it to yourself or are you just guessing?
hi samuel
Its giving ninja turtle
i wonder how this guy can be himself.
Who else is he gonna be?
@@Qermaq i mean.
@@mekaindo Don't be mean. :D
😩
How in gods name…
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.
What I would call the “2024 Answer” to this question would Sadly be “The Question is Racist!” 🙄🤨😉
How exciting!