2011 USAMO Problems/Problem 3
In hexagon , which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy , , and . Furthermore , , and . Prove that diagonals , , and are concurrent.
Let , , and , , , . Define the vectors: Clearly, .
Let intersect at . Note that . Define the points and on lines and respectively so that and . Then . As is isosceles with , the base angles are both . Thus, . Similarly, and .
Next we will find the angles between , , and . As , the angle between the vectors and is . Similarly, the angle between and is . Since the angle between and is , the angle between and is . Similarly, the angle between and is , and the angle between and is .
And since , we can arrange the three vectors to form a triangle, so the triangle with sides of lengths , , and has opposite angles of , , and , respectively. So by the law of sines: and the triangle with sides of length , , and has corresponding angles of , , and . It follows by SAS congruency that this triangle is congruent to , , and , so , , and , and , , and are the reflections of the vertices of triangle about the sides. So , , and concur at the orthocenter of triangle .
We work in the complex plane, where lowercase letters denote their corresponding point's poition. Let denote hexagon . Since , the condition is equivalent to .
Construct a "phantom hexagon" as follows: let be a triangle with , , and (this is possible since by the angle conditions), and reflect over its sides to get points , respectively. By rotation and reflection if necessary, we assume and have the same orientation (clockwise or counterclockwise), i.e. . It's easy to verify that for and opposite sides of have equal lengths. As the corresponding sides of and must then be parallel, there exist positive reals such that , , and . But then , etc., so the non-parallel condition "transfers" directly from to and If , then must be similar to and the conclusion is obvious.
Otherwise, since and , we must have and . Now let , , be the feet of the altitudes in ; by the non-parallel condition in , are pairwise distinct. But , whence are three distinct collinear points, which is clearly impossible. (The points can only be collinear when is a right triangle, but in this case two of must coincide.)
Alternatively (for the previous paragraph), WLOG assume that is the unit circle, and use the fact that , etc. to get simple expressions for and .
We work in the complex plane to give (essentially) a complete characterization when the parallel condition is relaxed.
WLOG assume are on the unit circle. It suffices to show that uniquely determine , since we know that if we let be the reflection of over , be the reflection of over , and be the reflection of over , then satisfies the problem conditions. (*)
It's easy to see with the given conditions that Note that so plugging into the third equation we have Simplifying, this becomes Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if then whence If , then eliminating , we get The first case corresponds to (*) (since uniquely determine and ), the second corresponds to (or equivalently, since , ), and by symmetry, the third corresponds to .
Otherwise, if , then we easily find from the first of the two equations in (we actually don't need this, but it tells us that the locus of working is a line through the origin). It's easy to compute and , so , and we're done.
Comment. It appears that taking the unit circle is nicer than, say or the unit circle (which may not even be reasonably tractable).
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