Wie löst man diese chinesische Grundschule Mathematikaufgabe?

Question

DerRoll · Answer

Zunächst mal löst man das sohttps://www.youtube.com/watch?v=6YvlHt8dlHQund dann aus den Kommentaren:This is a classic trick question on social media that's faked as a "primary school question" because it looks simple at first. But no even grade 6 math competitions in China will not have arc tangent. Plus that using a calculator is not allowed for primary school exams anyway. However, apart from the calculating arctan(2) part, the rest is indeed primary school level, but as a bonus question that's only meant to be solved by top students....Since I have relatives from China, I asked them: They think it is highly unlikely that this is a questions asked in primary school. They confirmed that primary school is up to the 5th or 6th grade ... depending on where you go to primary school in china, but even for 6th-graders they said this question is to hard. Calculating with squares and circles to a certain degree yes, but not to this level of complexity. - They think it is more likely that the phrase "Primary school in China" was written beside it, to cater to prejudice that all Chinese people are top notch when it comes to math, which they are not....I’ve taught primary school in China. This is not the kind of question they have to deal with. Circles and geometry, yes; but not this....I was once a math competition student until middle school. No there's no arctan for primary school math. Unless it is somehow hinted in the question, like if a right triangle has sides equal to 1 and 2 the angle can be assumed to equal to blabla......etcetc. Also wie immer viel Lärm um Nichts, aber das Internet macht es möglich.

Rammstein53 · Answer

Alternative Lösung:

Kreisrand grün: x² + y² - 4 = 0
Kreisrand blau: (x+2)² + (y-4)² - 16 = 0
Schnittpunkt S der beiden Kreise:
x² + y² - 4 = (x+2)² + (y-4)² - 16
Die Lösungen liegen auf der Geraden y = x/2 + 1
Daraus folgt:
Sx = 6/5
Sy = 8/5
##
Für den Winkel γ an der Ecke C gilt:
AC*tan(γ/2) = AB
γ = 2*arctan(1/2)
Für den Winkel β an der Ecke B gilt (Winkel bei A und S betragen 90°):
β = pi - γ = π - 2*arctan(1/2)
Fläche des Kreissegments CAS (blau berandet):
r²*π * 2*arctan(1/2)/(2π) = 16*arctan(1/2)
Davon geht die Fläche des Dreiecks CAS ab: 1/2*AC*(2+Sx) = 6.4
Fläche des Kreissegments BAS (rot berandet):
r²*π * (π - 2*arctan(1/2)/(2π) ) = 2*π - 4*arctan(1/2)
Davon geht die Fläche des Dreiecks BAS ab: 1/2*AB*Sy = 1/2*2*8/5 = 1.6
In Summe:
16*arctan(1/2) - 6.4 + 2*π - 4*arctan(1/2) - 1.6
12*arctan(1/2) - 8 + 2*π ~ 3.84695661518926

Wie löst man diese chinesische Grundschule Mathematikaufgabe?

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