Let a , b , and c be real numbers such that a + b + c = 2 and a 2 + b 2 + c 2 = 1 2 .
If the difference between the maximum and minimum values of c can be expressed as q p , where p and q are coprime positive integers , find p + q .
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First solver XD
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Second solver!
Here comes the third.... I thought my method would be unique and I thought of posting it first but found the same method here XD
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I also apparently posted a very similar problem ages ago, too. What a coincidence!
Same solution
Same way... I seem to have written a solution to a similar problem before :-)
A little longer method but here it goes,
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + a c ) ⟹ 4 = 1 2 + 2 ( a b + b c + a c ) ⟹ a b + b c + a c = − 4
Now a + b + c = 2 ⟹ a + b = 2 − c
And a b + b c + a c = − 4 ⟹ a b = − 4 − c ( a + b ) = − 4 − c ( 2 − c ) = c 2 − 2 c − 4
So a and b are the real roots of the equation y = x 2 + ( c − 2 ) x + ( c 2 − 2 c − 4 )
So the discriminant of the last equation is greater than or equal to zero so ( c − 2 ) 2 − 4 ( c 2 − 2 c − 4 ) ≥ 0
⟹ − 3 c 2 + 4 c + 2 0 ≥ 0 ⟹ − 2 ≤ c ≤ 3 1 0
So q p = 3 1 0 − ( − 2 ) = 3 1 6 and p + q = 1 6 + 3 = 1 9
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From the given, we have:
a + b = 2 − c
a 2 + b 2 = 1 2 − c 2
By the Cauchy-Scwarz inequality, we have 2 ( a 2 + b 2 ) ≥ ( a + b ) 2 .
Substitution of the above results gives us 3 c 2 − 4 c − 2 0 = 0 , and we see that equality holds for c = − 2 and c = 3 1 0 .
The difference between these two values is 3 1 6 so the answer is 1 9 .