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.
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Solution 1
Let , , and , , , , intersect at , intersect at , and intersect at . Define the vectors: Clearly, .
Note that . By sliding the vectors and to the vectors and respectively, 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 , and the angle between and is . Thus, the angle between and is , or just in the other direction if we take it modulo . 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 corrosponding angles of , , and . But then triangles , , 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 If , then must be similar to and the conclusion is obvious. Otherwise, since and , we must have and . Construct parallelograms and ; if is the reflection of over and is the reflection of over , then by simple angle chasing we can show that and . But means and must be linearly dependent (note that and ), so we must have . But then , which is impossible, so we're done.
Alternatively, WLOG assume the is the unit circle, and compute , etc.
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 |