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Locus of ratios

A triangle $$ABC$$ has a fixed base $$BC$$. If $$AB:AC= 1:2$$, then prove that the locus of the vertex $$A$$ is a circle whose centre is on the side $$BC$$ but not the midpoint of $$BC$$.

Note by Paramjit Singh
4 years, 1 month ago

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See http://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml

- 4 years, 1 month ago

Thanks!

- 4 years, 1 month ago

Assume the side BC lies on X axis with the mid point as Origin (0,0) and B (-a,0) C (a,0). Let the locus point A be (x,y)

Now since AB: AC = 1:2

\frac { \sqrt { { (x+a) }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { (x-a) }^{ 2 }+{ y }^{ 2 } } } =\frac { 1 }{ 2 }

Solving this you'll get a circle equation proving that the locus of A is a circle with centre on BC i.e on X axis

- 4 years, 1 month ago

Nice...

- 4 years, 1 month ago