Maths Olympiad Prep

Track / Stage 4 / 242 of 340 #502 of 1964

Problem 502

AMC 12 late, AIME early
Algebra Difficulty 4.8 Find the answer

(40 points) Find all pairs of rational numbers (a,b)(a, b), for which a+b=\sqrt{a}+\sqrt{b}= =2+3=\sqrt{2+\sqrt{3}}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Answer: (0.5;1.5)(0.5 ; 1.5) and (1.5;0.5)(1.5 ; 0.5).

Solution. Squaring the equality, we get a+b+2ab=2+3a+b+2 \sqrt{a b}=2+\sqrt{3}. Transferring a+ba+b to the right side and squaring again, we get 4ab=(2ab)2+2(2ab)3+4 a b=(2-a-b)^{2}+2(2-a-b) \sqrt{3}+ +3. In this expression, all terms except 2(2ab)32(2-a-b) \sqrt{3} are rational, and therefore this term must also be rational. This is possible only if 2ab=02-a-b=0. Then we have a+b=2,2ab=3a+b=2, 2 \sqrt{a b}=\sqrt{3}. Such a system is not difficult to solve: b=2ab=2-a, ab=a(2a)=34a b=a(2-a)=\frac{3}{4}. Solving the quadratic equation, we find both pairs of solutions.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.