Hardest geometry question in history answered by student trivially............ How? (1 Viewer)
- Thread starter no_arg
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No X can be any interior point. PX QX and RX need to be lifted out of the diagram to form another triangle.
Must the angles be expressed only in terms of
The best I could do is
where
(Hopefully that's correct...)
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 ofand
?
The best I could do is
whereis the side length of the equilateral triangle and
are the angles of the newly formed triangle.
(Hopefully that's correct...)
Random 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 the
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 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.
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.
Suppose that PQR is a 50-60-70 (respectively) triangle with an interior point X. Let angle PXR = s and angle QXR = t.
Find (in terms of s and t) the three angles of any triangle with side lengths equal to PX, QX and RX.
Find (in terms of s and t) the three angles of any triangle with side lengths equal to PX, QX and RX.
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Can someone please solve this lol.
Velocifire
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hit up jesus and tell him to solve the variant question.
Velocifire
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