Difference between revisions of "1970 Canadian MO Problems/Problem 8"
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<math>b=\frac{3a \pm \sqrt{80 - a^2}}{5}</math> | <math>b=\frac{3a \pm \sqrt{80 - a^2}}{5}</math> | ||
| + | |||
| + | Midpoint <math>x</math> and <math>y</math> as follows: | ||
<math>x=\frac{b+a}{2}=\frac{8a \pm \sqrt{80 - a^2}}{10}</math> | <math>x=\frac{b+a}{2}=\frac{8a \pm \sqrt{80 - a^2}}{10}</math> | ||
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<math>x=\frac{8(4x-2y) \pm \sqrt{80 - (4x-2y)^2}}{10}</math> | <math>x=\frac{8(4x-2y) \pm \sqrt{80 - (4x-2y)^2}}{10}</math> | ||
| + | |||
| + | <math>(10x-8(4x-2y))^2=80-(4x-2y)^2</math> | ||
| + | |||
| + | <math>100x^2-160x(4x-2y)+65(4x-2y)^2-80=0</math> | ||
| + | |||
| + | <math>20x^2-32x(4x-2y)+13(4x-2y)^2-16=0</math> | ||
| + | |||
| + | <math>20x^2-128x^2+64xy+208x^2-208xy+52y^2-16=0</math> | ||
| + | |||
| + | <math>100x^2-144xy+52y^2-16=0</math> | ||
| + | |||
| + | which simplifies to the equation of the an ellipse: | ||
| + | |||
| + | <math>25x^2-36xy+13y^2-4=0</math> | ||
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 
line: 
Point on 
line: 
Then,
Using the quadratic equation,
Midpoint 
and 
as follows:
Solving for 
we have:
Solving for 
we get:
which we put into one of the equations for x as:
which simplifies to the equation of the an ellipse:
~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.
