common area of intersecting circles (2002 UQ/QAMT PS Competition 8.5)

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common area of intersecting circles (2002 UQ/QAMT PS Competition 8.5)

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#1 h Feb 26, 2021, 5:03 PM

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The diagram below shows two circles, Circle $1$ and Circle $2$ . The centre $C_1$ of Circle $1$ lies on the circumference of Circle $2$ , and the centre $C_2$ of Circle 2 lies on the circumference of Circle $1$ . The two circles intersect at points $A$ and $B$ . The length of the line joining $C_1$ and $C_2$ is $2$ cm, rnd the length of the line joining $A$ and $B$ is $2\sqrt3$ cm. Find the area common to both circles (shaded).

https://cdn.artofproblemsolving.com/attachments/b/5/123980c2a9e57f7cd3127562e66c87aea62dd9.png

This post has been edited 2 times. Last edited by parmenides51, Feb 26, 2021, 11:37 PM

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#2 h Feb 26, 2021, 5:03 PM

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posted for the image link

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#3 h Feb 26, 2021, 8:16 PM

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Interesting how no one has tried this problem yet. Draw $AB$ , which splits the area into two regions. The area of one of those regions is $\frac 43 \pi - \sqrt 3$ . Thus, the area of both regions combined is $\frac 83 \pi - 2 \sqrt 3.$

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#4 h Feb 27, 2021, 2:38 AM • 1 Y

Y by Mango247

I'm getting the same result - just find that $C_1C_2=2$ via special triangles and then $4(\frac{2\pi}{3}-\sqrt3)+2\sqrt3 = \boxed{\frac{8\pi}{3}-2\sqrt3}$ .

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#5 h Feb 27, 2021, 2:56 AM

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@#4: Wait... $C_1C_2=2$ Already... D'ya mean $AC_2=2?$ Probably, since 30-60-90 is a special triangle

Simply note that $\triangle AC_1C_2$ is equilateral, and then $\angle AC_2B=120^\circ,$ from with you can find circular segment $AC_1B$ by taking Sector $\overarc{AB}$ with respect to $C_2$ minus $AC_2B.$ Then double this to obtain an answer of $\tfrac{8}{3}\pi-2\sqrt{3}.$

As a follow-up, try the intersection of three circles

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#6 h Feb 27, 2021, 3:07 AM

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Note that $AC_2 = C_1C_2$

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#7 h Feb 27, 2021, 3:09 AM • 3 Y

Y by Mango247, Mango247, Mango247

OlympusHero wrote:

Note that $AC_2 = C_1C_2$

LOL That was stupid on my part, but

OlympusHero wrote:

just find that $C_1C_2=2$ via special triangles

parmenides51 wrote:

The length of the line joining $C_1$ and $C_2$ is $2$

Instead, the fact that $AC_1C_2$ is equilateral gives us the fact that $\angle AC_2C_1=60^\circ,$ which gives us the area of the $120^\circ$ sector from which we can subtract $ABC_2,$ and we get half the area of the desired region

This post has been edited 1 time. Last edited by peace09, Feb 27, 2021, 3:10 AM

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#8 h Feb 27, 2021, 3:16 AM

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Oh oops I didn't realize they told us that. The problem has unnecessary information then.

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#9 h Feb 27, 2021, 3:26 AM • 2 Y

Y by peace09, Jndd

This problem is very very common, I've done lots of these, just different values for the radius/diameter given. Basically, the strat is to find the equilateral triangle by connecting the centers and the intersection points. Then, find the area of the sectors and subtract, for anyone who doesn't understand.

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#10 h Feb 27, 2021, 3:48 AM • 1 Y

Y by Mango247

Yes, the $2\sqrt 3$ condition is extraneous.

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