Maths Olympiad Prep

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Problem 1010

AMC 12 late, AIME early
Algebra Difficulty 4.9 Prove it NMO Selection Tests for JBMO · Romania

Let xx and yy be real nonzero numbers, such that x3+y3+3x2y2=x3y3x^3 + y^3 + 3x^2y^2 = x^3y^3. Determine the set of the possible values of E=1x+1yE = \frac{1}{x} + \frac{1}{y}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Rewrite the given condition successively (x+y)33xy(x+y)=x3y33x2y2(x + y)^3 - 3xy(x + y) = x^3y^3 - 3x^2y^2, i.e. (x+y)3(xy)3=3xy(x+y)3x2y2(x + y)^3 - (xy)^3 = 3xy(x + y) - 3x^2y^2, or (x+yxy)(x2+2xy+y2+x2y+xy2+x2y2)=3xy(x+yxy)(x + y - xy)(x^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2) = 3xy(x + y - xy). We either have x+y=xyx + y = xy, which leads to E=1E = 1 (obtained for x=y=2x = y = 2), or x2+2xy+y2+x2y+xy2+x2y2=3xyx^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2 = 3xy. The last equality, multiplied by 22, can be written equivalently x2(y+1)2+y2(x+1)2+(xy)2=0x^2(y + 1)^2 + y^2(x + 1)^2 + (x - y)^2 = 0, which is possible if and only if x=y=1x = y = -1 (x=y=0x = y = 0 is forbidden). In this case, E=2E = -2. In conclusion, the set of the possible values of EE is {2,1}\{-2, 1\}.

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