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Thanks Blake for the suggestion!
Square ABCD has a side length of 4. Construct a quarter circle with radius 4 centered at B and a semi-circle with diameter AD. What is the area of overlap between the quarter circle and semicircle?
This was said to be a primary school question in China and it went viral because many felt it was too difficult. The person who suggested it to me said many calculus students in America were finding it difficult. So what’s the answer?
As usual, watch the video for a solution.
Viral Primary School Question From China
Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To Viral Primary School Question From China
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
The area of overlap is equal to the areas of the two circular sectors minus the areas of the two right triangles.
Let’s calculate each area. The area of a single right triangle is:
0.5(2)(4) = 4
So two triangles have an area of 8.
The areas of the circular sectors are more tricky to calculate. A circular sector has an area of 0.5r2θ for a central angle θ.
Let t be half the semicircle central angle and s be half the quarter circle’s central angle. We can use trigonometry to calculate the angles.
Angle t is equal to arctan(4/2) = arctan 2. Angle s is complementary to t in the right triangle, so s = π/2 – t = π/2 – arctan(2). Thus we have:
area overlap
= sector(lower) + sector(upper) – 2(triangle)
= 0.5(22)(2t) + 0.5(42)(2t) – 2(4)
= 4t + 16s – 8
= 4 arctan 2 + 16(π/2 – arctan 2 – 8
= 4 arctan 2 + 8π – 16 arctan 2 – 8
= 8π – 12 arctan 2 – 8
≈ 3.847
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