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orl wrote: A line $l$ is drawn through the intersection point $H$ of altitudes of acute-angle triangles. Prove that symmetric images $l_a, l_b, l_c$ of $l$ with respect to the sides $BC,CA,AB$ have one point in common, which lies on the circumcircle of $ABC.$ See the note "Anti-Steiner points with respect to a triangle" on my website http://de.geocities.com/darij_grinberg/ . Darij

Hey, the link is broken !!

vickymano wrote: Hey, the link is broken !! I guess I can rectify that situation Here's my solution: Let $AD, BE, CF$ be altitudes of $\triangle ABC$. Also, let $l$ meet sides $BC, CA, AB$ of $\triangle ABC$ at points $K, L, M$ respectively. WLOG we may assume that $K$ lies on the extension of $CB$ beyond $B$, and points $L$ and $M$ lie on the boundary of the triangle. It suffices to show that $l_b, l_c$ concur on the circumcircle of $\triangle ABC$. Then, by analogous arguments $l_a, l_b$ also concur on the circumcircle of $\triangle ABC$, hence $l_a, l_b, l_c$ concur on the circumcircle of $\triangle ABC$. Now, it is well-known that the reflections of $H$ with respect to the sides of $\triangle ABC$ all lie on the circumcircle of $\triangle ABC$. So let $Q, R$ be the reflections of $H$ in sides $AC, AB$ respectively. Also, note that $RM \equiv l_c$ and $QL \equiv l_b$. So, our problem reduces to proving that $RM$ and $QL$ concur on the circumcircle of $\triangle ABC$, which is also the circumcircle of $\triangle RAQ$. Also let $RM$ intersect $QL$ at $O$. We have: $\angle ARM+\angle AQL=\angle AHM+\angle AHL=180^0$ Also, the angle sum in pentagon $RAQLM$ is $540^0$, hence we have that: $\angle RAQ+\angle RML+\angle QLM=360^0 \leftrightarrow \angle RAQ+\angle ROQ=180^0$ Hence $R, A, Q, O$ are concyclic, and therefore $O$ does indeed lie on the circumcircle of $\triangle ABC$, as desired.

Dear Mathlinkers, see also http://jl.ayme.pagesperso-orange.fr/Docs/Droite%20de%20Simson%20de%20Fe.pdf p. 7-12. Sincerely Jean-Louis