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Algebra Difficulty 4.3 AIME Find the answer United States

Problem:
Let (x,y)(x, y) be a pair of real numbers satisfying
56x+33y=−yx2+y2, and 33x−56y=xx2+y2 56 x + 33 y = \frac{-y}{x^{2} + y^{2}}, \quad \text{ and } \quad 33 x - 56 y = \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

Solution:
Answer: 1165\frac{11}{65} Observe that
1x+yi=x−yix2+y2=33x−56y+(56x+33y)i=(33+56i)(x+yi) \frac{1}{x + y i} = \frac{x - y i}{x^{2} + y^{2}} = 33 x - 56 y + (56 x + 33 y) i = (33 + 56 i)(x + y i)
So
(x+yi)2=133+56i=1(7+4i)2=(7−4i65)2 (x + y i)^{2} = \frac{1}{33 + 56 i} = \frac{1}{(7 + 4 i)^{2}} = \left(\frac{7 - 4 i}{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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.