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Concursul "Viitorii matematicieni" Problema 4
Posted: Sat Jan 09, 2010 1:06 pm
by salazar
4. Pe planul triunghiului ascutitunghic \( OBC \) se duce perpendiculara\( AO \). Fie \( H,H^\prime \) ortocentrele triunghiurilor \( ABC \) respectiv \( OBC \). Sa se arate ca dreapta \( HH^\prime \) este perpendiculara pe planul \( (ABC) \).
Posted: Fri Feb 26, 2010 6:03 pm
by Andi Brojbeanu
Fie \( OA^{\prime}\perp BC \), \( BB^{\prime}\perp OC \) si \( BB^{\prime \prime }\perp AC \).
Cum \( OA^{\prime}\perp BC, OA\perp (OBC) \) si \( OA^{\prime}, BC\in (OBC) \), din teorema celor trei perpendiculare rezulta ca \( AA^{\prime}\perp BC \), deci \( HH^{\prime}\in (OAA^{\prime}) \).
\( OA\perp (OBC)\Rightarrow OA\perp BC\Rightarrow BC\perp OA \).
Din \( BC\perp OA \) si \( BC\perp OA^{\prime}\Rightarrow BC\perp (OAA^{\prime})\Rightarrow BC\perp HH^{\prime}\Rightarrow HH^{\prime}\perp BC \).
\( OA\perp (OBC)\Rightarrow OA\perp BB^{\prime}\Rightarrow BB^{\prime}\perp OA \).
Din \( BB^{\prime}\perp OA \) si \( BB^{\prime}\perp OC \Rightarrow BB^{\prime}\perp (OAC)\Rightarrow BB^{\prime}\perp AC\Rightarrow AC\perp BB^{\prime} \).
Din \( AC\perp BB^{\prime} \) si \( AC\perp BB^{\prime \prime}\Rightarrow AC\perp (BB^{\prime}B^{\prime \prime})\Rightarrow AC\perp HH^{\prime}\Rightarrow HH^{\prime}\perp AC \).
Din \( HH^{\prime}\perp BC \) si \( HH^{\prime}\perp AC \Rightarrow HH^{\prime}\perp (ABC) \).