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If a triangle ABC remains always similar to a given triangle,and if the point A is fixed andB always move along a given straight line, find the locus of C.

I have been able to solve it for a right angled triangle but not a normal one.How to do this?Solve using coordinate geometry.i have already done with geometrical proof.

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  • $\begingroup$ Missing assumptions? Please see my Comment on Beni's Answer. $\endgroup$ – hardmath Oct 12 '13 at 12:41
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To get from B to C you always have to make a rotation of angle $\pm\theta=\angle ABC$ around $A$ and a homothety of ratio $AC/AB$ of center $A$. Both these transformations map lines to lines, so if $B$ lies on a line, the locus of $C$ is also a line, which you can find by performing the above transformations.

If you don't want, you do not need to use geometric transformations. Consider $B, B'$ two positions for $B$ and the corresponding $C,C'$ Then you can prove that the triangles $ABB'$ and $ACC'$ are similar, and from there it is quite straight forward.

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    $\begingroup$ Perhaps you are assuming the point $A$ remains fixed? And that the location of $C$ varies continuously with the location of $B$? Such assumptions suffice to support your approach, but I didn't read these into the original Question. $\endgroup$ – hardmath Oct 12 '13 at 12:40
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    $\begingroup$ @hardmath: If you don't fix $A$ then the problem is void. If $B$ varies on a line, $A$ can be everywhere, then $C$ can be everywhere. I'm sure that in the original question $A$ should be fixed. $\endgroup$ – Beni Bogosel Oct 12 '13 at 12:43
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    $\begingroup$ My critique is more about the problem formulation than about your solution, but note continuity is also an issue (since we can at any time reflect $ABC$ across $AB$ and get another similar triangle). $\endgroup$ – hardmath Oct 12 '13 at 12:47
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    $\begingroup$ @hardmath: Yeah, that's true. $\endgroup$ – Beni Bogosel Oct 12 '13 at 12:52
  • $\begingroup$ ya A is fixed.. $\endgroup$ – maths lover Oct 12 '13 at 13:30
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Sounds to me like $C$ can be anywhere in the plane. For every possible $B$ on the line and every $C$ anywhere, you can always draw the line $BC$ and find a place to put $A$ such that $ABC$ is similar to the given triangle.

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    $\begingroup$ Note: this answer was written before the condition that $A$ remains fixed was added to the question. $\endgroup$ – Henning Makholm Oct 12 '13 at 14:01

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