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Astrophysics

  • one year ago

What is the most interesting math problem you've encountered that you could share? Post away!

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  1. amilapsn
    • one year ago
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    Do you know how to prove all triangles are isosceles?

  2. imqwerty
    • one year ago
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    - the sum of two opposite sides and a diagonal of a quadrilateral is 20cm. The area of quadrilateral is 50cm^2. Find the length of the other diagonal. this question is 100% correct and no information is missing :) :D

  3. Astrophysics
    • one year ago
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    Haha, that sounds surprisingly easy, but it's not..hmm

  4. imqwerty
    • one year ago
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    XD

  5. ganeshie8
    • one year ago
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    My all time favorite math problem is a proof: The quadratic reciprocity law. For distinct odd primes \(p,q\), show that \[\large (p/q)(q/p)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\] (Gauss got so much obsessed with this problem and called it law ) https://en.wikipedia.org/wiki/Quadratic_reciprocity

  6. hartnn
    • one year ago
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    *

  7. thomas5267
    • one year ago
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    @imqwerty Does the quadrilateral has to be convex?

  8. thomas5267
    • one year ago
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    Would this be a valid quadrilateral under the scope of that question? |dw:1439136713982:dw|

  9. imqwerty
    • one year ago
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    its not given in the question so u can take any quadrilateral.

  10. thomas5267
    • one year ago
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    "the sum of two opposite sides and a diagonal of a quadrilateral is 20cm." Which of the following is true? one side + opposite side + diagonal = 20 cm; or, one side + opposite side = diagonal = 20 cm

  11. imqwerty
    • one year ago
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    1st one

  12. mathmath333
    • one year ago
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    recently reasoning questions are fascinating

  13. nincompoop
    • one year ago
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    cool!

  14. nincompoop
    • one year ago
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    does it work with other integer or Z+ values?

  15. nincompoop
    • one year ago
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    the boxes for astro

  16. nincompoop
    • one year ago
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    nevermind, I just checked it myself

  17. ganeshie8
    • one year ago
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    @mukushla is the answer emptyset ?

  18. ganeshie8
    • one year ago
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    the given equation can be rearranged as a^3 - b^3 = 199*200*ab (a-b)(a^2+ab+b^2)=199*200*ab (a-b)(a/b+b/a+1) = 199*200 since the right hand side is integer, it must be the case that a/b+b/a is also an integer only integer value of a/b+b/a is 2 and this doesn't satisfy the equation, so there are no solutions

  19. dan815
    • one year ago
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    |dw:1439146432404:dw|

  20. ikram002p
    • one year ago
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    you might sound this crazy but my favorite is (The Alternate Angle Theorem ) Proof xD

  21. ikram002p
    • one year ago
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    i also liked 1+1 which Bertrand Russell stated in logic , i dont think that ur post wide enough to handle the rest of mine so i'll stay and see what others lovely problems that users have. thanks @Astrophysics :3

  22. Nnesha
    • one year ago
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    2+2=22

  23. amistre64
    • one year ago
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    abbot and costello: 7x13=28 https://www.youtube.com/watch?v=xkbQDEXJy2k

  24. Astrophysics
    • one year ago
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    Haha that was good amistre, thanks xD This is turning out pretty nice, thanks everyone for sharing! Nice posts @ikram002p hehe, I also enjoyed this post ganeshie made for a triple integral a while back http://openstudy.com/users/ganeshie8#/updates/54de8a16e4b0b0ad8854cba3

  25. Nnesha
    • one year ago
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    5+5+5=550

  26. ali2x2
    • one year ago
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    The most interesting I've found is 2+2 :) Its 22 :D

  27. Nnesha
    • one year ago
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    o^_^o

  28. zzr0ck3r
    • one year ago
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    Cantors argument that the real numbers are uncountable. It suffices to show that \((0,1)\) is uncountable. Suppose it was, then we can make a list as follows \(0.a_1a_2a_3a_4...\) \(0.b_1b_2b_3b_4...\) \(0.c_1c_2c_3c_4...\) \(0.d_1d_2d_3d_4...\) \(0.e_1e_2e_3e_4...\) \(0.f_1f_2f_3f_4...\) . . . Consider the number \(0.abcdef\) where \(a=1\) if \(a_1\ne a\) else \(a=0\) and \(b=1\) if \(b_2\ne b\) else \(b=0\) and \(c=1\) if \(c_3\ne c\) else \(c=0\). It is not on the list!

  29. zzr0ck3r
    • one year ago
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    There are little things I left out, but that is pretty much the thing :)

  30. Astrophysics
    • one year ago
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    Ahh yes, I've heard of this, thank you so much for bringing this up, this is great!

  31. zzr0ck3r
    • one year ago
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    |dw:1439174881229:dw| Area of entire square \((a+b)^2\) Area of entire square broken into chunks \(4*\frac{2}{1}a*b+c^2\) set them equal \(4*\frac{1}{2}a*b+c^2=(a+b)^2\\\cancel{2ab}+c^2=a^2+\cancel{2ab}+b^2\) \[a^2+b^2=c^2\]

  32. zzr0ck3r
    • one year ago
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    My drawing could be better lol :)

  33. Astrophysics
    • one year ago
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    Nice one, and don't worry about it haha

  34. zzr0ck3r
    • one year ago
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    That cantor argument should also say \(a=1\) if \(a_1\ne 1\) else \(a=0\) \(b=1\) if \(b_2\ne 1\) else \(b=0\) . . .

  35. Jhannybean
    • one year ago
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    *

  36. Jhannybean
    • one year ago
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    What about you,@Empty ?

  37. anonymous
    • one year ago
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    Do people encounter their own problem?

  38. ganeshie8
    • one year ago
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    ohh right, that is a blunder lol

  39. ikram002p
    • one year ago
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    lets try to solve it in separate question :)

  40. tkhunny
    • one year ago
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    One I missed on the Putnam, many years ago. 1) Given a rectangle with the center identified. 2) Shoot an infinitely elastic point in any direction. 3) Bounce off the infinitely elastic walls if necessary - and it usually will be. 4) The point escapes the rectangle ONLY at the corners. 5) What is the expected number of bounces for the point to exit? Obviously, there are 4 directions with 0 bounces. Obviously, there are 8 directions with 1 bounce. Now what?

  41. Jack1
    • one year ago
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    a girl on here asked me to calculate how much wood a woodchuck could chuck once... she was pretty so i tried...

  42. anonymous
    • one year ago
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    (9^62773 + 2)^83721 (9 to the 62773 power plus 2) to the 83721 power. It caught my eye

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