USAMO 1978

Algebra Level 3

Real numbers a a , b b , c c , d d , and e e satisfy the system of equations below:

{ a + b + c + d + e = 8 a 2 + b 2 + c 2 + d 2 + e 2 = 16 \begin{cases} \begin{aligned} a + b + c + d + e & = 8 \\ a^2 + b^2 + c^2 + d^2 + e^2 & = 16 \end{aligned} \end{cases}

Find the sum of the maximum and minimum values of a a .


The answer is 3.2.

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1 solution

Sarthak Sahoo
Apr 3, 2020

b + c + d + e = ( 8 a ) b+c+d+e=(8-a)

The Artithematic mean is : x = ( 8 a ) 4 x=\frac{(8-a)}{4} Let b = x + b 1 , c = x + c 1 , d = x + d 1 b=x+b_1 , c=x+c_1 , d=x+d_1 and e = x + e 1 e=x+e_1

Then b 1 + c 1 + d 1 + e 1 = 0 b_1+c_1+d_1+e_1=0 and

16 = a 2 + 4 x 2 + b 1 2 + c 1 2 + d 1 2 + e 1 2 a 2 + 4 x 2 = a 2 + ( 8 a ) 2 4 16=a^2+4x^2+b_1^2+c_1^2+d_1^2+e_1^2 \ge a^2 +4x^2 =a^2 +\frac{(8-a)^2}{4}

or a ( 5 a 16 ) 0 a(5a-16)\le 0

And Hence the max and min of a a are 16 / 5 16/5 and 0 0 respectively

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