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 $BC$ .

#Locus #Goldbach'sConjurersGroup

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Xuming Liang - 7 years, 3 months ago

Thanks!

A Brilliant Member - 7 years, 3 months 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

Sai Prashanth D - 7 years, 3 months ago

Nice...

A Brilliant Member - 7 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$