Fie \( ABC \) un triunghi isoscel cu \( AB=AC \) si fie \( n \) un numar natural, \( n>1 \). Pe latura \( AB \) consideram punctul \( M \) astfel incat \( n\cdot AM=AB \). Pe latura \( BC \) consideram punctele \( P_1, P_2, ....., P_{n-1} \) astfel incat
\( BP_1=P_1P_2=....=P_{n-1}C=\frac{1}{n}BC \).
Sa se arate ca:
\( \angle{MP_1A}+\angle{MP_2A}+....+\angle{MP_{n-1}A}=\frac{1}{2}\angle{BAC} \).
Severius Moldoveanu
JBTST I 2010, Problema 3
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Andi Brojbeanu
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JBTST I 2010, Problema 3
Last edited by Andi Brojbeanu on Mon Apr 26, 2010 3:56 pm, edited 1 time in total.
Andi Brojbeanu
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca
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Destul de interesanta problema...
Considera \( MN \| BC,\ N \in AC \). Atunci dac\u a not\u am cu \( S \) suma unghiurilor din enunt, vom obtine din simetrie ca \( 2S =\angle MP_1 N +...+\angle MP_{n-1}N \). Deoarece patrulaterele \( P_kP_{k+1}NM \) sunt paralelograme, obtinem ca \( \angle MP_k N=\angle P_kNP_{k+1} \).
Prin urmare suma cautata este \( 2S=\angle P_1NC=\angle A \). De aici se deduce usor concluzia.
Considera \( MN \| BC,\ N \in AC \). Atunci dac\u a not\u am cu \( S \) suma unghiurilor din enunt, vom obtine din simetrie ca \( 2S =\angle MP_1 N +...+\angle MP_{n-1}N \). Deoarece patrulaterele \( P_kP_{k+1}NM \) sunt paralelograme, obtinem ca \( \angle MP_k N=\angle P_kNP_{k+1} \).
Prin urmare suma cautata este \( 2S=\angle P_1NC=\angle A \). De aici se deduce usor concluzia.
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Tomorow is a mistery,
But today is a gift.
That's why it's called present.
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