Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it United States

Problem:

Determine all pairs (a,b)(a, b) of real numbers such that 10,a,b,ab10, a, b, a b is an arithmetic progression.

Solution

Solution:

The answer is (4,2)(4, -2) and (52,5)\left(\frac{5}{2}, -5\right).

Since 10,a,b10, a, b is an arithmetic progression, we have
a=12(10+b). a = \frac{1}{2}(10 + b).
Also, we have a+ab=2ba + a b = 2 b, and so
a(1+b)=2b. a(1 + b) = 2b.
Substituting the expression for aa gives
(10+b)(1+b)=4b. (10 + b)(1 + b) = 4b.
Solving this quadratic equation gives the solutions b=2b = -2 and b=5b = -5. The corresponding values for aa can be found by a=12(10+b)a = \frac{1}{2}(10 + b).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.