Maths Olympiad Prep

Track / Stage 5 / 237 of 400 #837 of 1964

Problem 837

AIME late
Algebra Difficulty 5.4 Find the answer

Solve the equation:

3x4y=5x2+y2 3 x-4 y=5 \sqrt{x^{2}+y^{2}}

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

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Official solution

\triangle Let's introduce vectors uˉ\bar{u} and vˉ\bar{v}, and choose their coordinates in such a way that the left side of the equation expresses the dot product of the vectors in coordinates, while the right side expresses the product of the lengths of the vectors. Suitable vectors are uˉ(x;y)\bar{u}(x ; y) and vˉ(3;4)\bar{v}(3 ;-4). We apply inequality (4):

uˉvˉ=3x4yuˉvˉ=x2+y232+42=5x2+y2. \bar{u} \cdot \bar{v}=3 x-4 y \leq|\bar{u}| \cdot|\bar{v}|=\sqrt{x^{2}+y^{2}} \cdot \sqrt{3^{2}+4^{2}}=5 \sqrt{x^{2}+y^{2}} .

By the condition, the left and right sides of this inequality are equal. We use the proportion (3):

x3=y4,y=43x \frac{x}{3}=\frac{y}{-4}, \quad y=-\frac{4}{3} x

This, in fact, is the answer - with the addition of the restriction x0x \geq 0, since the ratio x3\frac{x}{3} must be non-negative - due to the fact that the proportionality coefficient in equality (3) when using inequality (4) is non-negative.

Answer: (x;4x/3)(x ;-4 x / 3), where xx is any non-negative number.

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