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    Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.
  
Guess What? It’s Parallel! Topic: Vector Geometry. Level: Olympiad. February 17th, 2006

Problem: (Arbelos – Volume 5, Chapter 4 Cover Problem) Let ABC be a triangle and let A^{\prime} be a point on the circumcircle of ABC . Denote H the orthocenter of \triangle ABC and H^{\prime} the orthocenter of \triangle A^{\prime}BC . Prove that AHH^{\prime}A^{\prime} is a parallelogram.

Arbelos-Vol5 Ch4 Cover

Solution: Let O be the circumcenter of the two triangles and set it to be the origin. Let A, A^{\prime}, B, C denote the vectors to each of the points, respectively. Note that |A| = |A^{\prime}| = |B| = |C| .

We claim that H = A+B+C and H^{\prime} = A^{\prime}+B+C (as vectors).

To prove this we only need to show that H-A is perpendicular to B-C , and by symmetry everything else follows.

Assume H = A+B+C . Then H-A = B+C . Consider the dot product (B+C) \cdot (B-C) = B \cdot B – C \cdot C = |B|^2-|C|^2 = 0 . But the dot product of two vectors can be zero iff they are perpendicular, and the result follows.

Now back to the problem. Since AH and A^{\prime}H^{\prime} are both perpendicular to BC , they are clearly parallel.

Consider writing AA^{\prime} as the vector A^{\prime}-A and HH^{\prime} as the vector H^{\prime}-H , which by our claim above is equivalent to (A^{\prime}+B+C) – (A+B+C) = A^{\prime}-A . Since AA^{\prime} and HH^{\prime} are the same vector, they are going in the same direction and thus parallel. Then AHH^{\prime}A^{\prime} 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 G be the centroid of \triangle ABC and G^{\prime} be the centroid of \triangle A^{\prime}BC . Prove that GG^{\prime} is parallel to AA^{\prime} .

Posted in Geometry, Olympiad || 1 Comment »

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

  1. Mathematical Food for Thought Says:
    January 27th, 2007 at 12:50 pm

    [...] 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 . [...]

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