I almost forgot how fun challenging math was until my plane ride last week where I solved IMO problem 1. Then I fell right back in love with math again. Anyway, let's begin. (Please excuse my notations and terminology, I'm a bit rusty on that).
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.
We are given
Now let angle and the rest are labeled in the picture for you. (IGNORE POINT G). Also, define .
Now define as the -coordinate of point and define similarly. Segments and are both length . Thus and .
Some arithmetic yields
Next, it is obvious that since , segment .
And it follows
Similar triangles tells us that and point tells us that .
Thus the picture shows us that
Amazingly, this means that point is directly over point with coordinates
Next, it is apparent that .
Pythagorean theorem tells us that . Unbelievably, this is pretty much the only root we have to deal with besides in all of this problem. And this root goes away immediately. Props to the maker of this beautiful problem.
From we know what is and thus .
Not surprisingly, everything cancels once again and we are left with
We now have
I didn't take the time to actually work out the lines in variable form. But plugging in values for a and b show that the three lines are concurrent.
I unfortunately lost the paper that I had the coordinates of intersection on. But the coordinates are sooo pretty, no radicals, just simple, elegant fractions.