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.
Solutions
Solution 1
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 .
Solution 2
We work in the complex plane, where lowercase letters denote point affixes. 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 .
Solution 3
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).
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2011 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.