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# hard geometry question help (circles)

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A, B, and C are 3 points on a plane such that AB = BC = 1 and CA = $$\sqrt{3}$$. Draw three circles whose diameters are $$\overline{AB}$$$$\overline{BC}$$, and $$\overline {CA}$$ respectively. The area of the region included in all 3 circles (the grey region below) is $${a\over b}\pi + {\sqrt{c}\over d}$$ in simplest form, so a, b, c, d are positive integers, a and b are relatively prime, and c is not divisible by the square of a prime. Enter the values a, b, c, d.

Thanks for the help, also here is the diagram: Feb 8, 2022

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I got it! It is:

a = 2

b = 3

c = 3

d = 4

Here is how I got it:

First I calculated the area of the whole big circle, that is 3pi / 4.

Then I calculated the area of one of the circular segments for the small circles, the part that sticks out of the big circle like a crescent. That is a sector minus two 30-60-90 triangles, so the area of the sector is circle with diameter BC / 3. Since BC = 1, the area of the circle is pi / 4. Then the sector is pi / 12. The area of the circular segment is pi / 12 - the area of the triangle.

The area of the triangle is the triangle that connects point C to the midpoint of BC to the intersection of the small circle to the big circle. That is two 30 - 60 - 90 triangles. The area of the two triangles is (0.5 / 2)^2 * sqrt(3). The area of the two triangles is $$\sqrt{3}\over16$$. Thus the area of the circular segment for the small circle is pi / 12 - sqrt(3)/16.

Next we have to find the circular segment made by the big circle and the radii of it, and the intersection of the small circles with the big circle. Since that sector has a central angle of 60*, then the area of the sector is 1/6th of the big circle, which the sector area is pi / 8. The area of the equilateral triangle that makes up part of the sector is 3sqrt(3)/16. Thus, the circular segment made by the bigger circle is pi / 8 - 3sqrt(3)/16.

Now to get the crescent sticking out of the big circle, we subtract the circular segment of the big circle from the circular segment of the small circle. That is (pi / 12 - sqrt(3)/16 - pi / 8 + 3sqrt(3)/16) is the area of one of the crescents. Since we have two crescents in the total area and one big circle, the equation is:

$${{3\pi}\over4} + 2({\pi\over12} - {\sqrt{3}\over16} - {\pi\over8} + {3\sqrt{3}\over16})$$

Now we just have to simplify, and we get:

$${2\pi\over3} + {\sqrt{3}\over4}$$

Then as you can see, a = 2, b = c = 3, and d = 4. Feb 8, 2022