That looks like a moon!

Geometry Level 3

A semi circle of diameter 1 unit sits at the top of a semi circle of diameter 2 units. the shaded region inside the the semi circle but outside the larger semi circle is called a lune . If the area of the lune is $\alpha$ , then find the value of $\lfloor 1000\alpha\rfloor$

Note

Assume that the diameters are parallel to each other.

The answer is 302.

4 solutions

Nayanmoni Baishya
Nov 23, 2014

Area of lune =area of smaller semi circle - (1/3(area of bigger semi circle) - area of the equilateral triangle) = (Pi/8)-(pi/6-root3/4)=a,

A=0.3021

Can you explain why " - 1/3(area of greater semi circle)"? How do you know that the corresponding area should be 1/3?

Calvin Lin Staff - 6 years, 6 months ago

Extend two lines from the centre of the diameter of the bigger circle to the two ends of the chord of length 1,just opposite to it. We thus have an equilateral triangle of length 1,each. Angles for an equilateral triangle is pi/3. Hence the sector is 1/3 of the semi circle.

Nayanmoni Baishya - 6 years, 6 months ago

Right. So, as opposed to "1/3 (area of greater semi circle)", you actually mean " 1/3 area of greater semi circle - equilateral triangle of length 1".

I see that this is expressed in the 2nd equation in terms of "(pi/6 - root 3 / 4)"

If you agree with this, can you update the solution accordingly?

Calvin Lin Staff - 6 years, 6 months ago

@Calvin Lin – Area of lune =area of smaller semi circle - (1/3(area of bigger semi circle) - area of the equilateral triangle)

Nayanmoni Baishya - 6 years, 6 months ago

@Nayanmoni Baishya – Thanks. I have updated your solution accordingly. Note that you can edit your solution and comments by clicking on the "edit" icon at the bottom.

Calvin Lin Staff - 6 years, 6 months ago

A Former Brilliant Member
Oct 29, 2017

The area of the shaded part is area of the small semicircle minus the area of the segment of a circle(big). The area of the segment of the circle (big) is equal to the area of the sector of the circle minus area of the triangle.

$A_{segment}=\dfrac{60}{360}\pi (1^2)-\dfrac{1}{2}(1^2)(\sin 60) \approx 0.090586073$

$A_{shaded}=\dfrac{1}{2}\left(\dfrac{\pi}{4}\right)(1^2) -0.090586073 \approx 0.302113008$

The desired answer is $1000(0.302113008)=$ $\boxed{302}$

Josh Williams
Dec 12, 2014

So we all know that using a calculator for this would be fraudulent and not in the spirit of mathematics.

Nikola Djuric
Dec 3, 2014

Also can be solved using integrals,but it is harder...

That looks like a moon!

The answer is 302.

4 solutions

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