2005 USAMO Problems/Problem 3
(Zuming Feng) Let be an acute-angled triangle, and let and be two points on side . Construct point in such a way that convex quadrilateral is cyclic, , and and lie on opposite sides of line . Construct point in such a way that convex quadrilateral is cyclic, , and and lie on opposite sides of line . Prove that points , and lie on a circle.
Let be the second intersection of the line with the circumcircle of , and let be the second intersection of the circumcircle of and line . It is enough to show that and . All our angles will be directed, and measured mod .
Since points are concyclic and points are collinear, it follows that But since points are concyclic, It follows that lines and are parallel. If we exchange with and with in this argument, we see that lines and are likewise parallel.
It follows that is the intersection of and the line parallel to and passing through . Hence . Then is the second intersection of the circumcircle of and the line parallel to passing through . Hence , as desired.
Motivation: One can notice that if you take such a then is cyclic, and that similarly is also cyclic. One gets the intuition that only one such circle should exist where the other chord passes through A, and so sets up a ghost point for , which works. ~cocohearts
Lemma. are collinear.
Suppose they are not collinear. Let line intersect circle (i.e. the circumcircle of ) again at distinct from . Because , we have that is cyclic. Hence , so . Then must be the other intersection of the parallel to through with circle . Then is on segment , so is contained in triangle . However, line intersects this triangle only at point because lies on arc not containing of circle , a contradiction. Hence, are collinear, as desired.
As a result, we have , so is cyclic, as desired.
Due to the parallel lines, Therefore, it suffices to prove that Note that by the cyclic quadrilaterals. Now, the condition simplifies to proving collinear.
Use barycentric coordinates. Let By the parallel lines, for some As is a vertex of the reference triangle, we must prove that
Now we find the circumcircle of Let its equation be Substituting in gives Substituting in yields Substituting in yields
Similarly, the circumcircle of is Now, we substitute into the equation
Let and note that both are negative. By the quadratic formula, we have Multiplying these two, we have as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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