Competition problem — Geometry — Maths Olympiad Prep

Library / /199 of 520

Geometry Difficulty 6.3 National olympiad Prove it

15. (ICE 1) Let $A B C$ be an acute-angled triangle. Three lines $L_{A}, L_{B}$ , and $L_{C}$ are constructed through the vertices $A, B$ , and $C$ respectively according to the following prescription: Let $H$ be the foot of the altitude drawn from the vertex $A$ to the side $B C$ ; let $S_{A}$ be the circle with diameter $A H$ ; let $S_{A}$ meet the sides $A B$ and $A C$ at $M$ and $N$ respectively, where $M$ and $N$ are distinct from $A$ ; then $L_{A}$ is the line through $A$ perpendicular to $M N$ . The lines $L_{B}$ and $L_{C}$ are constructed similarly. Prove that $L_{A}$ , $L_{B}$ , and $L_{C}$ are concurrent.

Solution

15. Referring to the description of $L_{A}$ , we have $\angle A M N=\angle A H N=90^{\circ}-$ $\angle H A C=\angle C$ , and similarly $\angle A N M=\angle B$ . Since the triangle $A B C$ is acute-angled, the line $L_{A}$ lies inside the angle $A$ . Hence if $P=L_{A} \cap B C$ and $Q=L_{B} \cap A C$ , we get $\angle B A P=90^{\circ}-\angle C$ ; hence $A P$ passes through the circumcenter $O$ of $\triangle A B C$ . Similarly we prove that $L_{B}$ and $L_{C}$ contains the circumcenter $O$ also. It follows that $L_{A}, L_{B}$ and $L_{C}$ intersect at the point $O$ . Remark. Without identifying the point of intersection, one can prove the concurrence of the three lines using Ceva's theorem, in usual or trigonometric form.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.