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

Revision as of 01:18, 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,

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

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

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

~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 4: / Line 4: @@
 == Solution ==
-{{Solution}}
+Point on <math>y=x</math> line: <math>(a,a)</math>
+Point on <math>y=2x</math> line: <math>(b,2b)</math>
+Then,
+<math>(b-a)^2+(2b-a)^2=4^2</math>
+<math>5b^2-6ab+2a^2-16=0</math>
+Using the quadratic equation,
+<math>b=\frac{3a \pm \sqrt{80 - a^2}}{5}</math>
+<math>x=\frac{b+a}{2}=\frac{8a \pm \sqrt{80 - a^2}}{10}</math>
+<math>y=\frac{2b+a}{2}=\frac{11a \pm 2\sqrt{80 - a^2}}{10}</math>
+~Tomas Diaz. orders@tomasdiaz.com
+{{alternate solutions}}

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

Revision as of 01:18, 28 November 2023

Problem

Solution