Maths Olympiad Prep

Track / Stage 5 / 181 of 400 #781 of 1964

Problem 781

AIME late
Algebra Difficulty 5.4 Find the answer

436. Solve the equation:

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

A number or a short expression. Spacing and $ signs are ignored.

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 and solution reproduced as published; topic, difficulty and ordering added by this site.