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

Revision as of 01:22, 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,

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

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

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

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:

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

~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.

@@ Line 26: / Line 26: @@
 <math>\pm \sqrt{80 - a^2}=10x-8a=\frac{10y-11a}{2}</math>
+Solving for <math>a</math> we get:
+<math>a=4x-2y</math> which we put into one of the equations for x as:
+<math>x=\frac{8(4x-2y) \pm \sqrt{80 - (4x-2y)^2}}{10}</math>

Page

Toolbox

Search

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

Revision as of 01:22, 28 November 2023

Problem

Solution