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Algebra Level 4

Given real numbers x , y , z x,y,z such that x + y + z = 0 x+y+z=0 , find the sum of the maximum and minimum values of x y + y z + x z x 2 + y 2 + z 2 \dfrac{xy+yz+xz}{x^2+y^2+z^2} .

If the sum of the maximum and minimum values of the above expression is α \alpha , submit your answer as the remainder when α 201 4 201 3 ⋯ 1 \huge \alpha^{2014^{2013^{\cdots^{1}}}} is divided by 3628800.


The answer is 1.

Manuel Kahayon by Manuel Kahayon , 21, Philippines

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2 solutions

Manuel Kahayon
Jul 5, 2016

( x + y + z ) 2 = 0 (x+y+z)^2=0

x 2 + y 2 + z 2 + 2 ( x y + y z + x z ) = 0 x^2+y^2+z^2+2(xy+yz+xz)=0

x 2 + y 2 + z 2 = − 2 ( x y + y z + x z ) x^2+y^2+z^2 = -2(xy+yz+xz)

x y + y z + x z x 2 + y 2 + z 2 = − 1 2 \frac{xy+yz+xz}{x^2+y^2+z^2} = -\frac{1}{2}

Therefore, our minimum is equal to our maximum, and the maximum is equal to the ONLY possible value of the expression, − 1 2 -\frac{1}{2} .

Therefore α = 2 − 1 2 = − 1 \alpha = 2\frac{-1}{2} = -1 .

But since ( − 1 ) 2 k = 1 (-1)^{2k} = 1 for all integral k k , our enormous power tower above simplifies to 1 1 , and it is a fun fact that 1 ≡ 1 ( m o d 3628800 ) 1 \equiv 1 \pmod{3628800} .

Therefore our answer is 1 \boxed{1} .

P.S. I named the problem as such since if you actually bought the problem with real money, you'd get scammed since it only looks difficult, but isn't. ;)

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Shree Ganesh
Jul 9, 2016

Actually I don't know the solution but once I saw saw that alpha to power smthing I surely knew that answer is 1 or 0

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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