Can you find three integers a,b,c such that: Arctg(a)+Arctg(b)+Arctg(c)=pi?
104855
There is many answers\na=1 b=2 c=3 is true\nalso a=b=c=sqrt(3) is true but not integers\n\nHow we can get them? \na,b,c must satisfy abc=a+b+c\n\nWhy?\natan(a) + atan(b) + atan (c) =pi\nlet atan(b)+atan(c) = y     (eq1)\n\natan(a) + y =pi\ntaking tan to the 2 sides\n\ntan [atan(a) + y] =tan(pi) = 0\n[tan(atan(a)) +tan (y) ]/[1- tan(atan(a))*tan (y) ]=0\n\nso a + tan(y)=0                       (eqn2)\nbut tan(y) = tan (atan (b) +atan(c))    from eqn1\n=(b+c)/(1-bc)\n\nsubstitue in eqn2\na+(b+c)/(1-bc)=0\nmultiply by (1-bc)\na(1-bc)+b+c=0\na+b+c-abc=0\n***********\n*a+b+c=abc*\n***********\n\na=1 b=2 c=3 is true since 1+2+3 =1*2*3 =6\na=b=c=sqrt(3) is true since 3sqrt(3)=srqt(3)^3
208926
the distributive property says:\n\n a(b+c)=  ab + ac
329054
There is a formula to compute any of the desired digits of pi. If you want the eighth digit, this formula gives it to you. If you want the fifth digit, here it is.
589996
Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
832721
ab = a+b\na(b-1) = b\na = b/(b-1).\n\nSo any number of the form n/(n-1), where n is an integer.\nn = 2 and n = 3 gives you the two you wrote; n = 4, 5, 6, 7, 8 give you 5 more.
739170
Let \nA = The sum of the reciprocals of positive integers whose only prime factors are 2s and 3s\nS = Σ(1/3)^n; n=0 to ∞ (all powers of three only)\nT = Σ(1/2)^m; m=0 to ∞ (all powers of three only)\n\nAll factors of three and 1/2 = 1/2 S\nAll factors of three and (1/2)² = (1/2)² S\nAll factors of three and (1/2)³ = (1/2)³ S\n...\n\nA = Σ(1/2)^m S; m=0 to ∞\nTake the common factor out\nA = S Σ(1/2)^m ; m=0 to ∞\nA = S T\n\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - - - \n\n\nArea of a circle = πr² \nArea of a segment: A = r²/2 ( θ - sin θ )\nYou want a segment with an area equal to one thir of the circle.\nA = r²/2 ( θ - sin θ ) = 1/3πr²\n1/2 ( θ - sin θ ) = 1/3π\nθ - sin θ = 2/3π\n1 - (sin &theta) / θ = 2/3π/θ\n\nBy law of sines, (sin &theta) / &theta = 1/(2r)\n\n1 - 1/(2r) = 2/3 π/θ\nθ = 2/3 π / (1 - 1/(2r))\nθ = 4/3 πr / (2r - 1)\n\nr = h + d (the height of the segment + the distance from the center to the cut)\n\nd is the length of the adjacent segment of a triangle with angle θ/2 and hypotenuse r\n\nd/r = cos (θ/2)\nd = r cos (θ/2)\n\nd = r cos (2/3 πr / (2r - 1))\n\nCheck my work carefully.  I was making it up as I went.  I can't give you any diagrams except from the sources.  I used "r", not "R" as in the sources.
1130099
(√3)² + 1 ² = 2²\nsin α = √3/2 α = 60º;  ß = 90 - α;  ß = 30º\n\nAnd your second question is\naⁿ = c² - b².  I'm going to assume that you wanted to put in a, b and c are all integers and a > 1.  Because if not, your theorem is not true and cannot be proven.  With that stipulation, this theorem passes the "intuitive" test.  It makes sense.\nWe know for certain that, for all even n, it must be true by the Pythagorean theorem.  So only odd n are of consequence to this theorem.\n\nI'm going to take a moment here and say that if you're trying to use this theorem as a FLT lemma, you must understand that it won't fly.  The lemma must use the carry the same stipulations as the successor theorem.  Said the other way, the corollary is bound by the restrictions of the theorem.\n\na=2, n=3 gives us c=3 and b=1.  8=9-1.  I took some time trying to find a counter-example and could not.  I did find some neat patterns emerging.\n\nThe proof to your theorem is simple: aⁿ = a²[(½a+½)² - (½a-½)²] because the expression in brackets will always be equal to a (a is odd prime; composite a can be simplified to prime a).\n\nI also found that 2ⁿ= [2^(n-2)+1]² - [2^(n-2)-1]².  So, in addition to all the solutions from the above expression, there will be another set of solutions for 2ⁿ; n is odd integer > 5.
675696
I agree with changing Cot(Theta) into Cos(Theta)/Sin(Theta)\n\nsqrt(3)*Cot(Theta) = 2*Cos(Theta)\n\nsqrt(3)*Cos(Theta)/Sin(Theta) = 2 * Cos(Theta)\n\nsqrt(3)/2 = Sin(Theta)\n\nTheta = pi/3 and 2pi/3, or 60 degrees and 120 degrees.
775644
Duch, there's some confusion over what the problem is.  It could be a/(3+4b) - a/(3+b)  OR  a/3 +4b -a/3 -b\n\nIf the former, then it boils down to -(11ab)/(9 + 15b + 4b²)\n\nIf the later, 3b
870635
(2-3i)(4+6i) = 8 + 12i - 12i - 18i^2\n= 8 + 18\n26
481802
Let's see:\n\nI choose 3, then 3+3=6, 6+4=10, 10*1/2=5, 5-3=2!\n\ngood!
1178023
a*b is the same thing as adding a b's together. Here are a few examples:\n3*2 = 2+2+2 = 6\n2*3 = 3+3 = 6\n1*6 = 6\n6*1 = 1+1+1+1+1+1 = 6\n4*4 = 4+4+4+4 = 16\n9*9 = 9+9+9+9+9+9+9+9+9 = 81
910180
3ab(2ab^2)^4\n3ab(16a^4b^8)\n48a^5b^9\ntrust me!!!! i just got done doing this in class a week ago.
409648
Yes. Many. Here's a way to get more:\n\nPick and two numbers A and B with A more than B.\n\nLet x=A^2 -B^2\nLet y=2*A*B\nLet z=A^2 +B^2.\n\nThen it will turn out that x^2+y^2=z^2.\n\nAll primitive Pythagorean triples are of this form, by the way. Primitive means that there are no common divisors between the three numbers.\n\nFor example, if A=5 and B=2, then you get x=21, y=20 and z=29.
1181427
ABC+ABD+C'D\n=ABC+ABD(C+C')+C'D\n=ABC+ABCD+ABC'D+C'D\n=ABC(1+D)+C'D(AB+1)\n=1ABC+1C'D\n=ABC+C'D
821031
I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27