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HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

Let (x,y)(x, y) be a pair of real numbers satisfying 56x+33y=−yx2+y2, and 33x−56y=xx2+y256x+33y=\frac{-y}{x^{2}+y^{2}}, \quad \text { and } \quad 33x-56y=\frac{x}{x^{2}+y^{2}} Determine the value of ∣x∣+∣y∣|x|+|y|.

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

Solution

Observe that 1x+yi=x−yix2+y2=33x−56y+(56x+33y)i=(33+56i)(x+yi)\frac{1}{x+yi}=\frac{x-yi}{x^{2}+y^{2}}=33x-56y+(56x+33y)i=(33+56i)(x+yi) So (x+yi)2=133+56i=1(7+4i)2=(7−4i65)2(x+yi)^{2}=\frac{1}{33+56i}=\frac{1}{(7+4i)^{2}}=\left(\frac{7-4i}{65}\right)^{2} It follows that (x,y)=±(765,−465)(x, y)= \pm\left(\frac{7}{65},-\frac{4}{65}\right).

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