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Puzzle – Overlapping circles

January 13, 2007 am31 3:17 am

Maybe this one is easier than I will make it seem.

Given three circles, each of radius 3, drawn so that each circle passes through the center of the other 2 (in other words, the centers form a 3-3-3 equilateral triangle), find the area of the portion common to all 3.

(ripped off from an SAT II practice problem. All answers were in terms of pi and radicals, and there were five choices)

7 Comments leave one →
  1. January 13, 2007 am31 5:48 am 5:48 am

    Isn’t the area formula for a Reuleaux triangle one of your state standards?

    (Just kidding.)

  2. January 13, 2007 am31 6:07 am 6:07 am

    I actually gasped “WHHAATT?” out loud before I saw the parentheses. I was pretty pleased that it only took me a few minutes to figure this one out — especially since when I saw it I wasn’t sure I could do it (except at the SAT II level it was unlikely I could not).

    I was more pleased to be able to show the student how to slightly underestimate the area, get a numerical result, and convert the choices into numerical results as well. There was only one answer (the right one) that was close enough to be considered. Of course, it would be nice to just do the math…. But there is a point of pride in defeating, in outsmarting, a multiple choice problem.

  3. January 19, 2007 pm31 1:56 pm 1:56 pm

    I don’t have paper and pencil with me, but I don’t think I have even a clue how to solve this problem.

  4. January 19, 2007 pm31 2:09 pm 2:09 pm

    OK, armed with paper and pencil, it doesn’t seem so unsolvable! (unless I’m overlooking something…)

    So the shape is basically an equilateral triangle with three “crusts” (shape bounded by arc + chord), one on each side.

    First, the area of the equilateral triangle (there’s a formula for this but I don’t remember) is … 1/2 * 3 * 1.5r3 = 9r3/4.

    Next, take one triangle, and determine a 60º wedge: 1/6 pi r^2 = 3pi/2.

    Subtract wedge – triangle for the “crust”: 3pi/2 – 9r3/4. Multiply that by three and add triangle…let’s see:

    (using paper) 9/2 * (pi – r3). I don’t know where my TI-82 is anymore, and I’m really not sure how to check that, so could you tell me if I’m on the right track?

  5. January 20, 2007 pm31 6:41 pm 6:41 pm

    I guess we could generalize to [(r^2)/2][pi – sqr3], that would be the formula mrc was referring to.

    Had I given the choices from the multiple choice, it would have turned out that only was was a bit more than 9sqr3/4 (using the calculator to give us just under 4).

  6. January 21, 2007 am31 10:11 am 10:11 am

    Interesting—thanks!

  7. andrew permalink
    June 27, 2009 am30 8:28 am 8:28 am

    what you want to do here is find the area of one “wedge” and multiply that by 3.

    But that’ll be an overcount, since the area of the equilateral triangle will get counted two many times. Therefore, subtract the areas of 2 equilateral triangles.

    The area of a wedge (sector) is: (gotta know this one!)

    A = (angle in radians)/2*r^2.

    In this case, the area of our wedge is = (pi/3)/2*3^2.

    The area of any triangle is:

    A = a*b*sin(angle between a and b)/2

    In our problem, the area of the triangle is = 9/2 * sin(60) = 9sqrt[3]/4

    Finally, the area of the reauleaux triangle is:

    3*wedge area – 2*triangle area =

    3* (pi/3)/2*3^2 – 2*9sqrt[3]/4

    which simplifies to:

    9*pi/2 – 9*sqrt[3]/2

    phew!

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