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On the sides $BC,CA,AB$ of a given triangle are taken points $P,Q,R$ such that the triangle $PQR$ is of given species. Prove that the locus of the circumcentre of triangle $PQR$ is a straight line.

Here i am stuck at "species" of triangle. What does it actually mean, and how to approach it? Please help.

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  • $\begingroup$ jstor.org/stable/3603970?seq=1 $\endgroup$ Dec 6, 2020 at 6:09
  • $\begingroup$ Alas, the solution that @strawberry-sunshine links to (from which your problem is taken word-for-word) assumes you already know what "of given species" means and various theorems about it, such as that the circles $AQR, BPR, CPQ$ intersect in a common point $O$. This is certainly not true of generally picked $P, Q, R$. $\endgroup$ Dec 6, 2020 at 18:01
  • $\begingroup$ @Paul Sinclair. Sorry sir but i am a student preparing for an Indian exam (JEE) and the above problem is from my coaching material. I didn't even knew it was picked from some book. $\endgroup$ Dec 7, 2020 at 3:37
  • $\begingroup$ I looked around but the most I could find was a book defining two angles to be "of the same species" if they were both acute, both right, or both obtuse. I don't know if you can get past the paywall on strawberry-sunshine's link, but you don't actually need to. The solution it gives is entirely on the free Preview page. Unfortunately, It doesn't give much in the way of hints as to the meaning of this phrase. $\endgroup$ Dec 7, 2020 at 17:32
  • $\begingroup$ @Paul Sinclair. Thanks for your effort and time. $\endgroup$ Dec 8, 2020 at 6:34

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'of given species' means if triangle $ABC$ is equilateral then triangle $PQR$ would be equilateral too. If triangle$ABC$ is scalene, so would be triangle $PQR$, that too proportionally. i.e. triangle$ABC$ would be similar to triangle $PQR$. Now, you can find many such triangles inside the triangle $ABC$. And if you join the circumcenters of those triangles, you would get a straight line.

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