black area by 6 arcs of radius = circle radius (1995 Chile Level 2 P2)

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black area by 6 arcs of radius = circle radius (1995 Chile Level 2 P2)

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#1 h Aug 29, 2019, 10:03 PM • 1 Y

Y by Adventure10

In a circle of radius $1$ , six arcs of radius $1$ are drawn, which cut the circle as in the figure. Determine the black area.

https://cdn.artofproblemsolving.com/attachments/8/9/0323935be8406ea0c452b3c8417a8148c977e3.jpg

This post has been edited 1 time. Last edited by parmenides51, Nov 22, 2021, 12:56 PM

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#2 h Sep 2, 2019, 7:20 PM • 1 Y

Y by Adventure10

Consider $\frac{1}{6}$ of the black area.
Area of the sector of the circle $\frac{1}{6}\pi$ .
Area of the $\triangle OAB =\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$ .
Green area $=\frac{1}{6}\pi-\frac{\sqrt{3}}{4}$ .
Black area $=\frac{\sqrt{3}}{4}-(\frac{1}{6}\pi-\frac{\sqrt{3}}{4})=\frac{\sqrt{3}}{2}-\frac{1}{6}\pi$ .
Total black area $=6(\frac{\sqrt{3}}{2}-\frac{1}{6}\pi)=3\sqrt{3}-\pi$ .

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#3 h Sep 2, 2019, 11:36 PM • 1 Y

Y by Adventure10

or u could subtract 12 times the green area from the total circle area

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yshk

#4 h Sep 2, 2019, 11:59 PM • 2 Y

Y by Adventure10, Mango247

vanstraelen wrote:

Consider $\frac{1}{6}$ of the black area.
Area of the sector of the circle $\frac{1}{6}\pi$ .
Area of the $\triangle OAB =\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$ .
Green area $=\frac{1}{6}\pi-\frac{\sqrt{3}}{4}$ .
Black area $=\frac{\sqrt{3}}{4}-(\frac{1}{6}\pi-\frac{\sqrt{3}}{4})=\frac{\sqrt{3}}{2}-\frac{1}{6}\pi$ .
Total black area $=6(\frac{\sqrt{3}}{2}-\frac{1}{6}\pi)=3\sqrt{3}-\pi$ .

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#5 h Sep 3, 2019, 12:20 AM • 2 Y

Y by Adventure10, Mango247

Solution

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#6 h Sep 3, 2019, 4:44 AM • 1 Y

Y by Adventure10

or Integrals???

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