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

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

Mathematics
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Do you know how to prove all triangles are isosceles?
- 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
Haha, that sounds surprisingly easy, but it's not..hmm

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Other answers:

XD
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
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@imqwerty Does the quadrilateral has to be convex?
Would this be a valid quadrilateral under the scope of that question? |dw:1439136713982:dw|
its not given in the question so u can take any quadrilateral.
"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
1st one
recently reasoning questions are fascinating
cool!
does it work with other integer or Z+ values?
the boxes for astro
nevermind, I just checked it myself
@mukushla is the answer emptyset ?
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
|dw:1439146432404:dw|
you might sound this crazy but my favorite is (The Alternate Angle Theorem ) Proof xD
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
2+2=22
abbot and costello: 7x13=28 https://www.youtube.com/watch?v=xkbQDEXJy2k
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
5+5+5=550
The most interesting I've found is 2+2 :) Its 22 :D
o^_^o
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!
There are little things I left out, but that is pretty much the thing :)
Ahh yes, I've heard of this, thank you so much for bringing this up, this is great!
|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 \(4*\frac{1}{2}a*b+c^2=(a+b)^2\\\cancel{2ab}+c^2=a^2+\cancel{2ab}+b^2\) \[a^2+b^2=c^2\]
My drawing could be better lol :)
Nice one, and don't worry about it haha
That cantor argument should also say \(a=1\) if \(a_1\ne 1\) else \(a=0\) \(b=1\) if \(b_2\ne 1\) else \(b=0\) . . .
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What about you,@Empty ?
Do people encounter their own problem?
ohh right, that is a blunder lol
lets try to solve it in separate question :)
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?
a girl on here asked me to calculate how much wood a woodchuck could chuck once... she was pretty so i tried...
(9^62773 + 2)^83721 (9 to the 62773 power plus 2) to the 83721 power. It caught my eye

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