Unexpected Substitutions
Given x 2 + y 2 + z 2 + x y z = 4 where x , y , z > 0 , evaluate cos − 1 ( 2 x ) + cos − 1 ( 2 y ) + cos − 1 ( 2 z )
rounded to the nearest integer.
The answer is 3.
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1 solution
How do you know what substitution to use?
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Lemma For x , y , z > 0 , x 2 + y 2 + z 2 + x y z = 4 ⟺ ( x , y , z ) = ( 2 cos A , 2 cos B , 2 cos C ) if A , B , C are the angles of an acute triangle.
Proof Right to left is easy. Plug in the values and you get cos 2 A + cos 2 B + cos 2 C + 2 cos A cos B cos C = 1 , which is well-known.
For left to right, first observe that 0 < x , y , z < 2 . Thus you can find A , B ∈ ( 0 , 2 π ) such that x = 2 cos A , y = 2 cos B . Simply substituting this into the identity will get you z = 2 cos C . Thus we have proven the lemma.
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Interesting. I wonder what other inequality problems use trigonometry as a substitution.
THANKYOU
Challenge Master Note: Solve this problem via trigo substitution.
Recall the great substitution x = 2 cos A , y = 2 cos B , z = 2 cos C , where A , B , C are three angles of an acute triangle.
Therefore, the answer is A + B + C = π ≈ 3 .