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Difference between revisions of "1970 Canadian MO Problems/Problem 8"

Latest revision as of 01:44, 28 November 2023

Problem

Consider all line segments of length 4 with one end-point on the line y=x and the other end-point on the line y=2x. Find the equation of the locus of the midpoints of these line segments.

Solution

Point on $y=x$ line: $(a,a)$

Point on $y=2x$ line: $(b,2b)$

Then,

$(b-a)^2+(2b-a)^2=4^2$

$5b^2-6ab+2a^2-16=0$

Using the quadratic equation,

$b=\frac{3a \pm \sqrt{80 - a^2}}{5}$

Midpoint $x$ and $y$ as follows:

$x=\frac{b+a}{2}=\frac{8a \pm \sqrt{80 - a^2}}{10}$

$y=\frac{2b+a}{2}=\frac{11a \pm 2\sqrt{80 - a^2}}{10}$

Solving for $\sqrt{80 - a^2}$ we have:

$\pm \sqrt{80 - a^2}=10x-8a=\frac{10y-11a}{2}$

Solving for $a$ we get:

$a=4x-2y$ which we put into one of the equations for x as:

$x=\frac{8(4x-2y) \pm \sqrt{80 - (4x-2y)^2}}{10}$

$(10x-8(4x-2y))^2=80-(4x-2y)^2$

$100x^2-160x(4x-2y)+65(4x-2y)^2-80=0$

$20x^2-32x(4x-2y)+13(4x-2y)^2-16=0$

$20x^2-128x^2+64xy+208x^2-208xy+52y^2-16=0$

$100x^2-144xy+52y^2-16=0$

which simplifies to the equation of the an ellipse:

$25x^2-36xy+13y^2-4=0$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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