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Given $\triangle ABC$. On the side AB externally is constructed square ABPQ. On the side AC internally is constructed square ACMN. AH is the altitude. If $O_1$ and$O_2$ are the centers of the two squares, prove that $O_1, O_2 $ and $H$ are collinear.

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    From the question, $AH$ is an altitude, so $H$ lies on $BC$.2012-06-04

1 Answers 1

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Let $A'$ be the intersection of the altitude through $A$ with $BC$ (the point called $H$ by the OP).

Let $X$ be the point on $BC$ such that $AA'X$ forms an isosceles rectangular triangle. (The vector $BC$ points in the same direction as the vector $A'X$.)

Then, a clockwise rotation by $45^\circ$ around $A$ followed by a scaling by $\frac1 {\sqrt 2}$ moves the points $X,C,B$ to the points $A',O_2,O_1$.

Clearly, a line remains a line after rotating and scaling.

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    Beautiful solution indeed. Do you have any idea how can I avoid using the rotation?2012-06-04