AMC Math Problem: geometry (problem of the week 45)

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  • Опубліковано 6 січ 2025
  • Hey guys! I am back with another AMC problem of the week! This week's problem comes from the year 2003 10A problem 19. I hope you enjoy it!

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  • @codetaku
    @codetaku 7 днів тому +1

    This was always my favorite type of math problem, no huge shortcuts, just a lot of calculation to get a logical conclusion
    Worth noting that you didn't have to learn 30 60 90 triangles or do any trig at all for this problem--you already have the hypotenuse so you can just do the pythagorean theorem real fast in your head. c^2 - a^2 = b^2 so b = sqrt(3/4) or sqrt(3)/2. If you have your basic triangles super memorized already, cool, but if you don't, it's sometimes faster to just quickly calculate it out in the hyper-reliable way than to take even 5 seconds to try to verify in your head that a 30 60 90 triangle has certain ratios to their legs.
    Of course, if you weren't given the hypotenuse, it would indeed be critical to remember your triangles.

  • @muhammadseyan8361
    @muhammadseyan8361 5 днів тому

    i just integrated
    0.5 * pi * r^2 - 2 * Integration of sqrt(1-x^2) from sqrt(3)/2 to 1

  • @Anonymous-uu8fw
    @Anonymous-uu8fw 7 днів тому

    you should do some more programming videos

  • @longfds
    @longfds 5 днів тому

    Hey can I do math with you