USAMO Problem 1
Let
a
,
b
,
c
,
d
be real numbers such that
b
−
d
≥
5
and all zeros
x
1
,
x
2
,
x
3
,
and
x
4
of the polynomial
P
(
x
)
=
x
4
+
a
x
3
+
b
x
2
+
c
x
+
d
are real. Find the smallest value the product
(
x
1
2
+
1
)
(
x
2
2
+
1
)
(
x
3
2
+
1
)
(
x
4
2
+
1
)
can take.This problem is part of
this set
.
The answer is 16.
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1 solution
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note that ( a + i ) ∗ ( a − i ) = a 2 + 1
So the product can be written as
( x 1 + i ) ( x 1 − i ) ( x 2 + i ) ( x 2 − i ) ( x 3 + i ) ( x 3 − i ) ( x 4 + i ) ( x 4 − i )
we know that
P ( x ) = ( x − x 1 ) ( x − x 2 ) ( x − x 3 ) ( x − x 4 )
and the above product is
P ( i ) ∗ P ( − i )
on calculating values of P ( i ) ∗ P ( − i ) ,
we get ( b − d − 1 ) 2 + ( a − c ) 2
b − d can have a minimum value of 5,and ( a − c ) 2 can have 0 thus required product has smallest value of 16.
By some algebraic manipulation it can be proven that the product is just
( b − d − 1 ) 2 + ( a − c ) 2 . (Hint: fit in x = i , − i )
This is at least ( 5 − 1 ) 2 + 0 = 1 6 .
Let a = 3 , b = 6 , c = 3 , d = 1 and we are done.