I'm surprised nobody else has posted this. Yes, there are plenty of discussions in Art of Problem Solving, but why not here? And of course there are more problems.
Triangle has a right angle at . Let be the point on line such that and lies between and . Point is chosen so that and is the bisector of . Point is chosen so that and is the bisector of . Let be the midpoint of . Let be the point such that is a parallelogram. Prove that , , and are concurrent.
Find all positive integers for which each cell of an table can be filled with one of the letters I, M, and O in such a way that:
Note. The rows and columns of an table are each labelled to in a natural order. Thus each cell corresponds to a pair of positive integer with . For , the table has diagonals of two types. A diagonal of first type consists all cells for which is a constant, and the diagonal of this second type consists all cells for which is constant.
Let be a convex polygon in the plane. The vertices have integral coordinates and lie on a circle. Let be the area of . An odd positive integer is given such that the squares of the side lengths of are integers divisible by . Prove that is an integer divisible by .
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I have an extension to Q1. (Almost got a complete solution for it, but i observed this, which follows):
Let this intersection point be labelled O. Prove O is the circumcentre of ΔXBE.
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Nice observation. If you complex trig bash it, the solution to your extension becomes very apparent. I'm currently typing up my solution to the original problem and can include your extension at the end.
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Nice, I have my own solution for it as well. Th essence for this extension is to just prove OE=OX since OX=OB is quite apparent.
Here's my synthetic solution. Let XF intersect BD at O and EM at O′. We need to show that O coincides with O′. A clear observation shows that B,C,F,D,X are concyclic because ∠{DCF}=∠{DBF}. Moreover D is the circumcenter of triangle CBA. A clear angle chase shows that ∠{DXF}=∠{DBF} showing O as the circumcenter of triangle OBE thus EO′=BO Thus O coincides with O′.proved Your extension can be proved easily if you watch F as the incenter of triangle DAB and watching XFAD and CDEM as the parallelogram. Thanks proving O as the circumcenter was really very helpful.
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Oh about it, u were talking to me.. OK I am also trying to solve it with another way...
Provide a solution using complex numbers.
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Well ur solution was really nice, I am also seeing it and trying to find a mistake..... 😂 😂😂😂😂😂😂😂😇😇😇😇😇😝😝😝😝😝😋