## Thursday, February 16, 2006

### Guess What? It's Parallel! Topic: Vector Geometry. Level: Olympiad.

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.

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 $AA^{\prime}$.

#### 1 comment:

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