For which real values of p does the equation
√(x2 - p) + 2 √(x2 - 1) = x have real roots? What are the roots?
Solution
I must admit to having formed rather a dislike for this type of question which came up in almost every one of the early IMOs. Its sole purpose seems to be to teach you to be careful with one-way implications: the fact that a2 = b2 does not imply a = b.
The lhs is non-negative, so x must be non-negative. Moreover 2√(x2 - 1) ≤ x, so x ≤ 2/√3. Also √(x2 - p) ≤ x, so p ≥ 0.
Squaring etc gives that any solution must satisfy x2 = (p - 4)2/(16 - 8p). We require x ≤ 2/√3 and hence (3p - 4)(p + 4) ≤ 0, so p ≤ 4/3.
Substituting back in the original equality we get |3p-4| + 2|p| = |p - 4|, which is indeed true for any p satisfying 0 ≤ p ≤ 4/3.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
21 Sep 1998
Last corrected/updated 24 Sep 2003