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  • 1. XXIX OPM - 10.11.2010 - Pre-Olimp´adas - 5o ´ ı Questao 1: ˜ cada opcao correcta: 4 pontos ¸˜Sugestoes para a resolucao dos problemas ˜ ¸˜ cada opcao errada: -1 ponto ¸˜ Questoes 2, 3: 10 pontos cada ˜ S1. (a) Opcao C. ¸˜ (b) Opcao D. ¸˜ (c) Opcao C. ¸˜ E (d) Opcao D. ¸˜ (e) Opcao C. ¸˜ ¸˜2. Solucao 1: A area da regiao assinalada e igual a area do rectˆ ngulo ´ ˜ ´ ` ´ a [ADHG] menos a area do rectˆ ngulo ´ a [BCF E]. G A CO ¸ ˜ E B F C H D Os lados do rectˆ ngulo [ADHG] medem 40 m e 17 m, logo a area de [ADHG] mede 40 × 17 a ´ = 680 m2 . Os lados do rectˆ ngulo [BCF E] medem 40 − 2 × 11 a = 18 m e 17 − 11 = 6 m, logo a area de [BCF E] mede ´ 18 × 6 = 108 m2 . Portanto, a area da regiao assinalada mede 680 − 108 ´ ˜ = 572 m2 . Solucao 2: A area da regiao assinalada e igual a soma das areas dos rectˆ ngulo [ABEI], [CDJF ] e [IJHG]. ¸˜ ´ ˜ ´ ` ´ a G H LU E F I J A B C D a = 6 m, logo a area de [ABEI] mede 11 × 6 = 66 m2 . Os lados do rectˆ ngulo [ABEI] medem 11 m e 17 − 11 ´ Da mesma forma se conclui que a area de [CDJF ] mede 66 m 2. ´ Os lados do rectˆ ngulo [IJHG] medem 40 m e 11 m, logo a area de [IJHG] mede 40 × 11 a ´ = 440 m2 . Portanto, a area da regiao assinalada mede 66 + 66 + 440 ´ ˜ = 572 m2 . SO3. Existem cinco tipos de setas e cada tipo repete-se de cinco em cinco setas. Ora 1001 = 200 × 5 + 1, logo a seta na ultima posicao e do mesmo tipo da seta na primeira posicao. Portanto, a sequˆ ncia termina com a ´ ¸˜ ´ ¸˜ e seta . spm