# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Guess What? It’s Parallel! Topic: Vector Geometry. Level: Olympiad. February 17th, 2006

Problem: (Arbelos – Volume 5, Chapter 4 Cover Problem) Let be a triangle and let be a point on the circumcircle of . Denote the orthocenter of and the orthocenter of . Prove that is a parallelogram.

Solution: Let be the circumcenter of the two triangles and set it to be the origin. Let denote the vectors to each of the points, respectively. Note that .

We claim that and (as vectors).

To prove this we only need to show that is perpendicular to , and by symmetry everything else follows.

Assume . Then . Consider the dot product . But the dot product of two vectors can be zero iff they are perpendicular, and the result follows.

Now back to the problem. Since and are both perpendicular to , they are clearly parallel.

Consider writing as the vector and as the vector , which by our claim above is equivalent to . Since and are the same vector, they are going in the same direction and thus parallel. Then is a parallelogram, as desired. QED.

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Comment: A very powerful use of vector geometry, as it simplifies the problem immensely, especially with the common formula for the orthocenter.

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Practice Problem: Let be the centroid of and be the centroid of . Prove that is parallel to .

### One Response to “Guess What? It’s Parallel! Topic: Vector Geometry. Level: Olympiad.”

1. Mathematical Food for Thought Says:

[...] We know and from here, so we’ll use these results. Also, by the definition of a midpoint. Using these three equations, we can solve for . [...]