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2021 AMC 12A Problem 2

Problem 2 of 25EasierAlgebra

Under what conditions is a2+b2=a+b\sqrt{a^2 + b^2} = a + b true, where aa and bb are real numbers?

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Solution

Because a2+b2\sqrt{a^2+b^2} is never negative, equality requires a+b0.a + b \ge 0. Squaring both sides gives a2+b2=(a+b)2a^2 + b^2 = (a+b)^2 =a2+2ab+b2,= a^2 + 2ab + b^2, which simplifies to 2ab=0,2ab = 0, i.e. ab=0.ab = 0. Conversely, if ab=0ab = 0 then a2+b2=(a+b)2,a^2 + b^2 = (a+b)^2, and if additionally a+b0a + b \ge 0 then a2+b2=a+b=a+b.\sqrt{a^2+b^2} = |a+b| = a+b. So both conditions together are exactly what is needed. Thus, the correct answer is D.

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Concepts: radical · algebraic manipulation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.