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Monday, March 16, 2015

Problem of the Week

Happy Monday! Sorry we missed last week's problem of the week. To make up for it, we have a particularly challenging problem today- it can be done using calculus or geometry. I elected to solve it using calculus. We posted the following image on our social media pages:

And like usual, I've solved the problem below...


As I mentioned above, this problem can be solved using calculus or geometry, but I elected to take the calculus route. Click on any of the images below to make them larger!

Above we see my initial set up of the problem. I decided to take the first quadrant of the total image and find the shaded bit there. To do that, I had to use an entire unit circle.




After a lot of integration and parts labeling, I was able to determine the answer: 4(Pi)/3 - (3)^(.5).

*CORRECTION:* I made a weird mistake in the very last section of my answer. I incorrectly "simplified" by adding (3/3) to ((pi)/3) and managed to get (4(pi)/3). Thank you to Brian in the comments section for pointing it out! The first part of the line, before the boxed answer, is correct.


Did you solve this a different way? Do you have any questions about my solution? Please leave a comment! I'd love to see a geometric solution in the comments section.

4 comments:

  1. Hint: exploit the symmetry of the construction to reduce work.

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  2. In the last step, how does 1 + (pi)/3 = 3/3 +(pi)/3 = (3+(pi))/3 become 4(pi)/3 all of a sudden?

    That seems bigger than the 1 square unit of the original square ( 4(Pi)/3 - (3)^(.5) = 2.45673939722)

    ReplyDelete
    Replies
    1. Wow, thank you Brian. I just made a complete misstep in the easy section of the proof. I falsely attempted to simplify by adding (3/3) to ((pi)/3). I'm editing the post now to account for my mistake! The part right before the boxed answer is correct. -Tori

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  3. Also works out quite nicely by considering (as you did) quarter parts but then use sectors from the corner of the original square and add/subtract triangles.

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