Maths Olympiad Prep

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, 2022

Geometry Difficulty 5.3 AIME, harder Prove it Bulgaria

Find all pairs of real numbers (x,y)(x, y) such that
x22x+y2+4y+5+x28x+y24y+20=516x2+9y2=68. \left| \begin{aligned} & \sqrt{x^2 - 2x + y^2 + 4y + 5} + \sqrt{x^2 - 8x + y^2 - 4y + 20} = 5 \\ & 16x^2 + 9y^2 = 68. \end{aligned} \right.

Solution

Consider orthogonal coordinate system and the points A(1,2)A(1, -2) and B(4,2)B(4, 2). Since
x22x+y2+4y+5=(x1)2+(y+2)2x28x+y24y+20=(x4)2+(y2)2, \begin{aligned} \sqrt{x^2 - 2x + y^2 + 4y + 5} &= \sqrt{(x-1)^2 + (y+2)^2} \\ \sqrt{x^2 - 8x + y^2 - 4y + 20} &= \sqrt{(x-4)^2 + (y-2)^2}, \end{aligned}
we conclude that the solutions of the first equation are all points C(x,y)C(x, y) such that CA+CB=5|CA| + |CB| = 5. But
AB=(41)2+(2(2))2=16+9=5=CA+CB, |AB| = \sqrt{(4-1)^2 + (2-(-2))^2} = \sqrt{16+9} = 5 = |CA| + |CB|,
and thus the point CC lies on the line segment ABAB. The equation of the line ABAB is 3y=4x103y = 4x - 10 so we want to find the solutions of the two equations satisfying x[1,4]x \in [1, 4]. Now the solution of the system is given by
32x280x+100=682x25x+2=0x1=12,x2=2. 32x^2 - 80x + 100 = 68 \Leftrightarrow 2x^2 - 5x + 2 = 0 \Leftrightarrow x_1 = \frac{1}{2}, x_2 = 2.
Only x2x_2 belongs to the interval [1,4][1, 4], thus the only solution is (2,23)(2, -\frac{2}{3}).

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