No X can be any interior point. PX QX and RX need to be lifted out of the diagram to form another triangle.I think it's because X must only be in the center for a triangle to form?
Must the angles be expressed only in terms of and ?No X can be any interior point. PX QX and RX need to be lifted out of the diagram to form another triangle.
Given similarity, the solution cannot depend on the side length of the original triangle. Answer is disturbingly simple and the proof is stunning.Must the angles be expressed only in terms of and ?
The best I could do is
where is the side length of the equilateral triangle and are the angles of the newly formed triangle.
(Hopefully that's correct...)
Both proofs a little clumsyRandom point inside an equilateral triangle
Take any equilateral triangle and pick a random point inside the triangle. Draw from each vertex a line to the random point. Two of the three angles at the point are known let's say $x$,$y$. If themath.stackexchange.com
The quantityGiven similarity, the solution cannot depend on the side length of the original triangle. Answer is disturbingly simple and the proof is stunning.
I think the first answer given is essentially the same as the one you posted, just a bit more direct.Both proofs a little clumsy
The use of congruence makes it less direct.The quantity
can probably be simplified so as to remove .
The answer I gave holds numerically for all such equilateral and isosceles triangles formed and most likely for the rest of them too.
The other solutions are definitely much nicer though.
I think the first answer given is essentially the same as the one you posted, just a bit more direct.