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Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)

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Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)

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Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)

1. 1. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 1 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” PEMBAHASAN OSN MATEMATIKA SMP 2015 TINGKAT KABUPATEN BAGIAN B : ISIAN SINGKAT BAGIAN B : ISIAN SINGKAT 1. Jawaban : 𝑥 = −4, −1 Pembahasan : 𝑥 𝑎𝑑𝑎𝑙𝑎𝑕 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑥2 + 5𝑥 + 6 𝑎𝑑𝑎𝑙𝑎𝑕 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . (𝑥 + 3) 𝐾𝑎𝑟𝑒𝑛𝑎 𝑥 + 2 < 𝑥 + 3 𝑚𝑎𝑘𝑎 𝑢𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 2, 3, 5, 7, 11, … 𝑏𝑒𝑟𝑙𝑎𝑘𝑢 ∶ 𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 2 𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 2 = 1 .2 = −2 . −1 𝐼𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶ 𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .2 𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 2 𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 2 − 3 𝑥 = −1 𝑑𝑎𝑛 𝑥 = −1 (𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑠𝑎𝑚𝑎) 𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = −2 . −1 𝑥 + 2 = −2 𝑑𝑎𝑛 𝑥 + 3 = −1 𝑥 = −2 − 2 𝑑𝑎𝑛 𝑥 = −1 − 3 𝑥 = −4 𝑑𝑎𝑛 𝑥 = −4 (𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑠𝑎𝑚𝑎) 𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 3 𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 3 = 1 .3 = −3 . −1 𝐼𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶ 𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .3 𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 3 𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 3 − 3 𝑥 = −1 𝑑𝑎𝑛 𝑥 = 0 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎) 𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = −3 . −1 𝑥 + 2 = −3 𝑑𝑎𝑛 𝑥 + 3 = −1 𝑥 = −3 − 2 𝑑𝑎𝑛 𝑥 = −1 − 3 𝑥 = −5 𝑑𝑎𝑛 𝑥 = −4 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎) 𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 5 𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 5 = 1 .5 = −5 . −1 𝐼𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶ 𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .5
2. 2. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 2 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 5 𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 5 − 3 𝑥 = −1 𝑑𝑎𝑛 𝑥 = 2 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎) 𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = −5 . −1 𝑥 + 2 = −5 𝑑𝑎𝑛 𝑥 + 3 = −1 𝑥 = −5 − 2 𝑑𝑎𝑛 𝑥 = −1 − 3 𝑥 = −7 𝑑𝑎𝑛 𝑥 = −4 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎) 𝐷𝑎𝑟𝑖 𝑝𝑒𝑟𝑕𝑖𝑡𝑢𝑛𝑔𝑎𝑛 𝑑𝑖𝑎𝑡𝑎𝑠 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑝𝑜𝑙𝑎 𝑏𝑎𝑕𝑤𝑎 𝑢𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 ≥ 3 𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑑𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑦𝑎𝑛𝑔 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎 𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑥 = −4, −1 2. Jawaban : 12 Pembahasan : 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑚𝑒𝑙𝑎𝑙𝑢𝑖 𝑡𝑖𝑡𝑖𝑘 (−2, 6) 𝑠𝑢𝑚𝑏𝑢 𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑛𝑦𝑎 𝑥 = −1 𝑎, 𝑏, 𝑑𝑎𝑛 𝑐 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑒𝑛𝑎𝑝 𝑝𝑜𝑠𝑖𝑡𝑖𝑓 𝑏𝑒𝑟𝑢𝑟𝑢𝑡𝑎𝑛 −2 𝑥 , 6 𝑦 → 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 6 = 𝑎 . −2 2 + 𝑏 . −2 + 𝑐 6 = 4𝑎 − 2𝑏 + 𝑐 4𝑎 − 2𝑏 + 𝑐 = 6 … 1 𝑆𝑢𝑚𝑏𝑢 𝑠𝑖𝑚𝑒𝑡𝑟𝑖 𝑥 = −1 → 𝑥 = − 𝑏 2𝑎 −1 = − 𝑏 2𝑎 −2𝑎 = −𝑏 𝑏 = 2𝑎 … 2 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑠𝑖𝑘𝑎𝑛 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 2 𝑘𝑒 1 ∶ 4𝑎 − 2𝑏 + 𝑐 = 6 4𝑎 − 2 . 2𝑎 + 𝑐 = 6 4𝑎 − 4𝑎 + 𝑐 = 6 𝑐 = 6 𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑎𝑕 𝑐 = 6 , 𝑏 = 2𝑎 𝑑𝑎𝑛 𝑘𝑎𝑟𝑒𝑛𝑎 𝑎, 𝑏, 𝑑𝑎𝑛 𝑐 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑒𝑛𝑎𝑝 𝑝𝑜𝑠𝑖𝑡𝑖𝑓 𝑏𝑒𝑟𝑢𝑟𝑢𝑡𝑎𝑛 𝑚𝑎𝑘𝑎 𝑏𝑒𝑟𝑙𝑎𝑘𝑢 ∶ 2 𝑎 , 4 𝑏=2𝑎 , 6 𝐽𝑎𝑑𝑖 𝑎 + 𝑏 + 𝑐 = 2 + 4 + 6 = 12
3. 3. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 3 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 3. Jawaban : 7 3 + 12 𝑐𝑚2 Pembahasan : 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝑃, 𝑄 𝑑𝑎𝑛 𝑅 𝑎𝑑𝑎𝑙𝑎𝑕 𝑡𝑖𝑡𝑖𝑘 𝑠𝑖𝑛𝑔𝑔𝑢𝑛𝑔 𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝑝𝑎𝑑𝑎 𝑠𝑖𝑠𝑖 − 𝑠𝑖𝑠𝑖 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷, 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 ∶ ∠𝐴𝑃𝐷 = ∠𝐶𝑃𝐷 = ∠𝐷𝑅𝑆 = ∠𝐶𝑅𝑆 = ∠𝐴𝑄𝑆 = ∠𝐷𝑄𝑆 = 90 𝑜 𝑆𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 → 𝐴𝐵 = 𝐴𝐶 𝑆𝑅 = 𝑆𝑄 = 𝑃𝑆 = 1 𝑅𝐷 = 3 3 𝑐𝑚 ∠𝑆𝐷𝑅 = 60 𝑜 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 ∶ ∠𝐷𝑆𝑅 = 180 𝑜 − 90 𝑜 − 60 𝑜 = 30 𝑜 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑄𝑆 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶ ∠𝑄𝐷𝑆 = ∠𝑆𝐷𝑅 = 60 𝑜 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶ ∠𝐷𝐶𝑃 = 180 𝑜 − 90 𝑜 − 60 𝑜 = 30 𝑜
4. 4. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 4 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 ∶ ∠𝐷𝐴𝑃 = 180 𝑜 − 90 𝑜 − 60 𝑜 = 30 𝑜 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶ ∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵 = 30 𝑜 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐷 ∶ ∠𝐴𝐷𝐵 = 180 𝑜 − 60 𝑜 − 60 𝑜 = 60 𝑜 ∠𝐵𝐴𝐷 = 180 𝑜 − 60 𝑜 − 30 𝑜 = 90 𝑜 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 ∶ 𝐷𝑆 = 𝑅𝐷2 + 𝑆𝑅2 = 3 3 2 + 12 = 3 9 + 1 = 3 9 + 9 9 = 12 9 = 12 9 = 4 .3 9 = 2 3 3 𝐷𝑃 = 𝑃𝑆 + 𝐷𝑆 = 1 + 2 3 3 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶ 𝐶𝑃 𝑆𝑅 = 𝐷𝑃 𝑅𝐷 𝐶𝑃 1 = 1+ 2 3 3 3 3 𝐶𝑃 = 1 + 2 3 3 . 3 3 𝐶𝑃 = 3 3 + 2 𝐶𝑃 = 3 3 . 3 3 + 2 𝐶𝑃 = 3 + 2 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶ 𝐴𝑃 = 𝐶𝑃 = 3 + 2 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶ 𝐴𝐵 = 𝐴𝐶 = 𝐴𝑃 + 𝐶𝑃 = 3 + 2 + 3 + 2 = 2 3 + 4 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶ 𝐴𝐷 𝑅𝐷 = 𝐴𝐵 𝑆𝑅 𝐴𝐷 3 3 = 2 3+4 1 𝐴𝐷 = 2 3 + 4 . 3 3 𝐴𝐷 = 2 + 4 3 3
5. 5. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 5 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 = 1 2 . 𝐷𝑃 . 𝐶𝑃 = 1 2 . 1 + 2 3 3 . 3 + 2 = 1 2 . 3 + 2 + 2 + 4 3 3 = 1 2 . 3 3 3 + 4 + 4 3 3 = 1 2 . 7 3 3 + 4 = 7 3 6 + 2 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 = 7 3 6 + 2 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 = 1 2 . 𝐴𝐵 . 𝐴𝐷 = 1 2 . 2 3 + 4 . 2 + 4 3 3 = 1 2 . 4 3 + 8 + 8 + 16 3 3 = 1 2 . 12 3 3 + 16 + 16 3 3 = 1 2 . 28 3 3 + 16 = 14 3 3 + 8 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 + 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 + 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 = 7 3 6 + 2 + 7 3 6 + 2 + 14 3 3 + 8 = 14 3 6 + 12 + 14 3 3 = 7 3 3 + 12 + 14 3 3 = 21 3 3 + 12 = 7 3 + 12 𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 7 3 + 12 𝑐𝑚2 4. Jawaban : 55 ∶ 153 Pembahasan : 𝐵𝑜𝑡𝑜𝑙 𝐼 → 𝐺𝐼 ∶ 𝐴𝐼 = 2 ∶ 11 𝐵𝑜𝑡𝑜𝑙 𝐼𝐼 → 𝐺𝐼𝐼 ∶ 𝐴𝐼𝐼 = 3 ∶ 5 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝑉𝑏𝑜𝑡𝑜𝑙 = 𝑉
6. 6. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 6 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝑉𝑜𝑙𝑢𝑚𝑒 𝐺𝑢𝑙𝑎 𝑑𝑎𝑛 𝐴𝑖𝑟 𝑑𝑎𝑙𝑎𝑚 𝑏𝑜𝑡𝑜𝑙 𝐼 𝑑𝑎𝑛 𝑏𝑜𝑡𝑜𝑙 𝐼𝐼 𝑠𝑒𝑏𝑒𝑙𝑢𝑚 𝑑𝑖𝑐𝑎𝑚𝑝𝑢𝑟 ∶ 𝐺𝐼 = 2 2+11 . 𝑉 = 2𝑉 13 𝐴𝐼 = 11 2+11 . 𝑉 = 11𝑉 13 𝐺𝐼𝐼 = 3 3+5 . 𝑉 = 3𝑉 8 𝐴𝐼𝐼 = 5 3+5 . 𝑉 = 5𝑉 8 𝑉𝑜𝑙𝑢𝑚𝑒 𝐺𝑢𝑙𝑎 𝑑𝑎𝑛 𝐴𝑖𝑟 𝑠𝑒𝑡𝑒𝑙𝑎𝑕 𝑏𝑜𝑡𝑜𝑙 𝐼 𝑑𝑎𝑛 𝑏𝑜𝑡𝑜𝑙 𝐼𝐼 𝑑𝑖𝑐𝑎𝑚𝑝𝑢𝑟 ∶ 𝐺𝐼+𝐼𝐼 = 𝐺𝐼 + 𝐺𝐼𝐼 = 2𝑉 13 + 3𝑉 8 = 16𝑉 104 + 39𝑉 104 = 55𝑉 104 𝐴𝐼+𝐼𝐼 = 𝐴𝐼 + 𝐴𝐼𝐼 = 11𝑉 13 + 5𝑉 8 = 88𝑉 104 + 65𝑉 104 = 153𝑉 104 𝐺𝐼+𝐼𝐼 ∶ 𝐴𝐼+𝐼𝐼 = 𝐺𝐼+𝐼𝐼 𝐴 𝐼+𝐼𝐼 = 55𝑉 104 153 𝑉 104 = 55𝑉 104 . 104 153𝑉 = 55 153 = 55 ∶ 153 𝐽𝑎𝑑𝑖 𝑟𝑎𝑠𝑖𝑜 𝑘𝑎𝑛𝑑𝑢𝑛𝑔𝑎𝑛 𝑔𝑢𝑙𝑎 𝑑𝑎𝑛 𝑎𝑖𝑟 𝑕𝑎𝑠𝑖𝑙 𝑐𝑎𝑚𝑝𝑢𝑟𝑎𝑛𝑛𝑦𝑎 𝑎𝑑𝑎𝑙𝑎𝑕 55 ∶ 153 5. Jawaban : 19 Pembahasan : 𝑓 𝑥 = 209 − 𝑥2 𝑓 𝑎𝑏 = 𝑓 𝑎 + 2𝑏 − 𝑓 𝑎 − 2𝑏 𝑑𝑖𝑚𝑎𝑛𝑎 ∶ 𝑎 , 𝑏 𝑎𝑑𝑎𝑙𝑎𝑕 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑝𝑜𝑠𝑖𝑡𝑖𝑓 𝑑𝑎𝑛 𝑎 < 𝑏 𝑓 𝑎𝑏 = 𝑓 𝑎 + 2𝑏 − 𝑓 𝑎 − 2𝑏 209 − 𝑎𝑏 2 = 209 − 𝑎 + 2𝑏 2 − 209 − 𝑎 − 2𝑏 2 209 − 𝑎2 𝑏2 = 209 − 𝑎2 + 4𝑎𝑏 + 4𝑏2 − 209 − 𝑎2 − 4𝑎𝑏 + 4𝑏2 209 − 𝑎2 𝑏2 = 209 − 𝑎2 − 4𝑎𝑏 − 4𝑏2 − 209 − 𝑎2 + 4𝑎𝑏 − 4𝑏2 209 − 𝑎2 𝑏2 = 209 − 𝑎2 − 4𝑎𝑏 − 4𝑏2 − 209 + 𝑎2 − 4𝑎𝑏 + 4𝑏2 209 − 𝑎2 𝑏2 = −8𝑎𝑏 209 = 𝑎2 𝑏2 − 8𝑎𝑏 19 .11 = 𝑎𝑏 . 𝑎𝑏 − 8 → 𝑎𝑏 = 19 𝑎𝑏 − 8 = 11 𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑎𝑕 𝑎𝑏 = 19 𝑦𝑎𝑛𝑔 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 , 𝑑𝑎𝑛 𝑘𝑎𝑟𝑒𝑛𝑎 𝑎 < 𝑏 𝑚𝑎𝑘𝑎 𝑏𝑒𝑟𝑙𝑎𝑘𝑢 ∶ 𝑎𝑏 = 19 𝑎𝑏 = 1 . 19 → 𝑎 = 1 𝑏 = 19 𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑏 𝑎 = 19 1 = 19
7. 7. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 7 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 6. Jawaban : 10080 Pembahasan : 𝑈1 + 𝑈2 + 𝑈3 + 𝑈4 = 70 → 𝑆4 = 70 𝑈5 + 𝑈6 + ⋯ + 𝑈16 = 690 𝑈1 + 𝑈2 + ⋯ + 𝑈16 = 760 𝑆16 = 760 𝑆 𝑛 = 𝑛 2 . 2𝑎 + 𝑛 − 1 . 𝑏 𝑆4 = 4 2 . 2𝑎 + 4 − 1 . 𝑏 = 70 2 . 2𝑎 + 3𝑏 = 70 2𝑎 + 3𝑏 = 70 2 2𝑎 + 3𝑏 = 35 … (1) 𝑆 𝑛 = 𝑛 2 . 2𝑎 + 𝑛 − 1 . 𝑏 𝑆16 = 16 2 . 2𝑎 + 16 − 1 . 𝑏 = 760 8 . 2𝑎 + 15𝑏 = 760 2𝑎 + 15𝑏 = 760 8 2𝑎 + 15𝑏 = 95 … (2) 𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 2 𝑑𝑎𝑛 (1): 2𝑎 + 15𝑏 = 95 2𝑎 + 3𝑏 = 35 12𝑏 = 60 𝑏 = 60 12 𝑏 = 5 → 1 : 2𝑎 + 3𝑏 = 35 2𝑎 + 3 . 5 = 35 2𝑎 + 15 = 35 2𝑎 = 35 − 15 2𝑎 = 20 𝑎 = 20 2 𝑎 = 10 𝑈𝑛 = 𝑎 + 𝑛 − 1 . 𝑏 𝑈2015 = 10 + 2015 − 1 .5 = 10 + 2014 .5 = 10 + 10070 = 10080 𝐽𝑎𝑑𝑖 𝑠𝑢𝑘𝑢 𝑘𝑒 − 2015 𝑏𝑎𝑟𝑖𝑠𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 10080
8. 8. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 8 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 7. Jawaban : 1 ∶ 2 Pembahasan : 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐴𝐵 𝑠𝑒𝑗𝑎𝑗𝑎𝑟 𝐸𝐹 𝐴𝐸 = 𝐵𝐹 𝐴𝐵 = 2 . 𝐸𝐹 𝐴𝑃 = 𝑃𝐵 = 𝐷𝑄 = 𝑄𝐶 = 𝐸𝐹 𝐴𝐷 ⊥ 𝐴𝐵 𝑑𝑎𝑛 𝐸𝐻 ⊥ 𝐸𝐹 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝑡𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 𝐴𝐵𝐹𝐸 = 𝑡 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝑃𝐸 = 𝑡𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑃𝐵𝐹𝐸 = 𝐸𝑅 𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸.𝐷𝑄𝐻 ∶ 𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸.𝑄𝐶𝐺𝐻 = 𝑉 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻 𝑉 𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸 .𝑄𝐶𝐺𝐻 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝑃𝐸 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻 𝐿 𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑃𝐵𝐹𝐸 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸 .𝑄𝐶𝐺𝐻 = 1 2 . 𝐴𝑃 . 𝐸𝑅 . 𝐸𝐻 𝑃𝐵 . 𝐸𝑅 . 𝐸𝐻 = 1 2 . 𝐸𝐹 . 𝐸𝑅 . 𝐸𝐻 𝐸𝐹 . 𝐸𝑅 . 𝐸𝐻 = 1 2 = 1 ∶ 2 𝐽𝑎𝑑𝑖 𝑝𝑒𝑟𝑏𝑎𝑛𝑑𝑖𝑛𝑔𝑎𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸. 𝐷𝑄𝐻 𝑑𝑎𝑛 𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸. 𝑄𝐶𝐺𝐻 𝑎𝑑𝑎𝑙𝑎𝑕 1 ∶ 2 8. Jawaban : 28 Pembahasan : 𝑀𝑎𝑡𝑒𝑚𝑎𝑡𝑖𝑘𝑎 → 𝐴 , 𝐵 , 𝐶 , 𝐷 𝐼𝑃𝐴 → 𝐴 , 𝐵 , 𝐶 , 𝐸 𝐼𝑃𝑆 → 𝐴 , 𝐷 , 𝐸 , 𝐹 𝐴 𝑑𝑎𝑛 𝐵 𝑏𝑒𝑟𝑠𝑎𝑢𝑑𝑎𝑟𝑎 , 𝑗𝑎𝑑𝑖 𝑗𝑖𝑘𝑎 𝐴 𝑡𝑒𝑟𝑝𝑖𝑙𝑖𝑕 𝑚𝑎𝑘𝑎 𝐵 𝑡𝑖𝑑𝑎𝑘 𝑡𝑒𝑟𝑝𝑖𝑙𝑖𝑕, 𝑏𝑒𝑔𝑖𝑡𝑢 𝑝𝑢𝑙𝑎 𝑠𝑒𝑏𝑎𝑙𝑖𝑘𝑛𝑦𝑎 𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑝𝑒𝑚𝑖𝑙𝑖𝑕𝑎𝑛 𝐵𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎 𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛𝑀𝑎𝑡𝑒𝑚𝑎𝑡𝑖𝑘𝑎 𝐴 , 𝐵 , 𝐶 , 𝐷 𝐼𝑃𝐴 𝐴 , 𝐵 , 𝐶 , 𝐸 𝐼𝑃𝑆 𝐴 , 𝐷 , 𝐸 , 𝐹 𝐴 𝐶 𝐷 , 𝐸 , 𝐹 3 𝐸 𝐷 , 𝐹 2
9. 9. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 9 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝐵 𝐶 𝐷 , 𝐸 , 𝐹 3 𝐸 𝐷 , 𝐹 2 𝐶 𝐴 𝐷 , 𝐸 , 𝐹 3 𝐵 𝐷 , 𝐸 , 𝐹 3 𝐸 𝐴 , 𝐷 , 𝐹 3 𝐷 𝐴 𝐸 , 𝐹 2 𝐵 𝐸 , 𝐹 2 𝐶 𝐴 , 𝐸 , 𝐹 3 𝐸 𝐴 , 𝐹 2 𝑇𝑜𝑡𝑎𝑙 𝑏𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎 𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛 28 𝐽𝑎𝑑𝑖 𝑐𝑎𝑟𝑎 𝑦𝑎𝑛𝑔 𝑚𝑢𝑛𝑔𝑘𝑖𝑛 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑚𝑖𝑙𝑖𝑕 𝑤𝑎𝑘𝑖𝑙 𝑠𝑒𝑘𝑜𝑙𝑎𝑕 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑘𝑒 𝑂𝑆𝑁 𝑆𝑀𝑃 𝑡𝑎𝑕𝑢𝑛 𝑖𝑛𝑖 𝑎𝑑𝑎 𝑠𝑒𝑏𝑎𝑛𝑦𝑎𝑘 28 9. Jawaban : 𝐴 −8, 6 , 𝐵 −8, 10 , 𝐶 −4, 6 Pembahasan : ∆𝐴𝐵𝐶 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑠𝑢𝑚𝑏𝑢 𝑌, 𝑘𝑒𝑚𝑢𝑑𝑖𝑎𝑛 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑙𝑎𝑔𝑖 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑔𝑎𝑟𝑖𝑠 𝑦 = 3 , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 𝑕𝑎𝑠𝑖𝑙 𝑝𝑒𝑛𝑐𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑛𝑦𝑎 𝑎𝑑𝑎𝑙𝑎𝑕 ∆𝐴′ 𝐵′ 𝐶′ , 𝑦𝑎𝑖𝑡𝑢 𝐴′ 8, 0 , 𝐵′ 8, −4 , 𝐶′ 4, 0 𝐷𝑒𝑛𝑔𝑎𝑛 𝑑𝑒𝑚𝑖𝑘𝑖𝑎𝑛 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑛𝑑𝑎𝑝𝑎𝑡𝑘𝑎𝑛 𝑘𝑒𝑚𝑏𝑎𝑙𝑖 ∆𝐴𝐵𝐶, 𝑙𝑎𝑘𝑢𝑘𝑎𝑛 𝑙𝑎𝑛𝑔𝑘𝑎𝑕 𝑚𝑢𝑛𝑑𝑢𝑟 𝑦𝑎𝑖𝑡𝑢 ∶ ∆𝐴′ 𝐵′ 𝐶′ 𝑕𝑎𝑟𝑢𝑠 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑔𝑎𝑟𝑖𝑠 𝑦 = 3 , 𝑘𝑒𝑚𝑢𝑑𝑖𝑎𝑛 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑠𝑢𝑚𝑏𝑢 𝑌 𝐽𝑎𝑑𝑖 𝑘𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡 𝑡𝑖𝑡𝑖𝑘 − 𝑡𝑖𝑡𝑖𝑘 ∆𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝐴 −8, 6 , 𝐵 −8, 10 , 𝐶 −4, 6
10. 10. www.siap-osn.blogspot.com @ Maret 2015 Sosuke D. Aizen 2 SMPN 1 Tambelangan Pembahasan OSN Matematika SMP 2015 / Page 10 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 10. Jawaban : 61600 Pembahasan : 𝑃𝑢𝑡𝑖𝑕 = 3 𝑀𝑒𝑟𝑎𝑕 = 3 𝐾𝑢𝑛𝑖𝑛𝑔 = 3 𝐻𝑖𝑗𝑎𝑢 = 3 𝐵𝑖𝑟𝑢 = 3 𝐽𝑢𝑚𝑙𝑎𝑕 = 12 𝑀𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑝𝑎𝑑𝑎 𝑔𝑒𝑙𝑎𝑛𝑔 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑎𝑡𝑢𝑟𝑎𝑛 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 2 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑝𝑢𝑡𝑖𝑕 𝑠𝑒𝑙𝑎𝑙𝑢 𝑡𝑒𝑟𝑑𝑎𝑝𝑎𝑡 4 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑒𝑙𝑎𝑖𝑛 𝑝𝑢𝑡𝑖𝑕 𝐾𝑎𝑟𝑒𝑛𝑎 𝑀𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑎𝑡𝑢𝑟𝑎𝑛 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 2 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑝𝑢𝑡𝑖𝑕 𝑠𝑒𝑙𝑎𝑙𝑢 𝑡𝑒𝑟𝑑𝑎𝑝𝑎𝑡 4 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑒𝑙𝑎𝑖𝑛 𝑝𝑢𝑡𝑖𝑕, 𝑚𝑎𝑘𝑎 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑏𝑒𝑟𝑝𝑜𝑙𝑎 ∶ 𝐷𝑒𝑛𝑔𝑎𝑛 𝑑𝑒𝑚𝑖𝑘𝑖𝑎𝑛 𝑝𝑜𝑠𝑖𝑠𝑖 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑝𝑢𝑡𝑖𝑕 𝑡𝑒𝑡𝑎𝑝, 𝑡𝑒𝑡𝑎𝑝𝑖 𝑢𝑛𝑡𝑢𝑘 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑚𝑒𝑟𝑎𝑕, 𝑘𝑢𝑛𝑖𝑛𝑔, 𝑕𝑖𝑗𝑎𝑢 𝑑𝑎𝑛 𝑏𝑖𝑟𝑢 𝑑𝑎𝑝𝑎𝑡 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑠𝑖 𝑑𝑎𝑟𝑖 𝑢𝑛𝑠𝑢𝑟 𝑦𝑎𝑛𝑔 𝑠𝑎𝑚𝑎 ∶ 𝐵𝑎𝑛𝑦𝑎𝑘 𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑚𝑒𝑟𝑎𝑕, 𝑏𝑖𝑟𝑢, 𝑘𝑢𝑛𝑖𝑛𝑔 𝑑𝑎𝑛 𝑕𝑖𝑗𝑎𝑢 = 12! 3! .3! .3! .3! 𝑇𝑒𝑡𝑎𝑝𝑖 , 𝑠𝑒𝑏𝑢𝑎𝑕 𝑔𝑒𝑙𝑎𝑛𝑔 𝑏𝑖𝑠𝑎 𝑑𝑖 𝑝𝑢𝑡𝑎𝑟 𝑑𝑎𝑛 𝑑𝑖𝑏𝑎𝑙𝑖𝑘 𝑘𝑎𝑛𝑎𝑛 𝑘𝑖𝑟𝑖𝑛𝑦𝑎 (𝑦𝑎𝑛𝑔 𝑘𝑎𝑛𝑎𝑛 𝑏𝑖𝑠𝑎 𝑗𝑎𝑑𝑖 𝑘𝑖𝑟𝑖, 𝑏𝑒𝑔𝑖𝑡𝑢 𝑗𝑢𝑔𝑎 𝑠𝑒𝑏𝑎𝑙𝑖𝑘𝑛𝑦𝑎) , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑚𝑝𝑒𝑟𝑚𝑢𝑑𝑎𝑕 𝑝𝑒𝑟𝑕𝑖𝑡𝑢𝑛𝑔𝑎𝑛, 𝑔𝑒𝑙𝑎𝑛𝑔 𝑑𝑖𝑘𝑒𝑙𝑜𝑚𝑝𝑜𝑘𝑘𝑎𝑛 𝑗𝑎𝑑𝑖 3 𝑏𝑎𝑔𝑖𝑎𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑎𝑐𝑢𝑎𝑛 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑝𝑢𝑡𝑖𝑕, 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐵𝑎𝑛𝑦𝑎𝑘 𝑘𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑔𝑒𝑙𝑎𝑛𝑔 𝑑𝑖𝑝𝑢𝑡𝑎𝑟 𝑎𝑡𝑎𝑢 𝑑𝑖𝑏𝑜𝑙𝑎𝑘 − 𝑏𝑎𝑙𝑖𝑘 = 3! 𝐴𝑔𝑎𝑟 𝑔𝑒𝑙𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑡𝑖𝑑𝑎𝑘 𝑎𝑑𝑎 𝑦𝑎𝑛𝑔 𝑙𝑒𝑏𝑖𝑕 𝑑𝑎𝑟𝑖 𝑠𝑎𝑡𝑢 𝑝𝑜𝑙𝑎 𝑔𝑒𝑙𝑎𝑛𝑔 (𝑎𝑘𝑖𝑏𝑎𝑡 𝑝𝑒𝑚𝑢𝑡𝑎𝑟𝑎𝑛 𝑎𝑡𝑎𝑢 𝑑𝑖𝑏𝑜𝑙𝑎𝑘 − 𝑏𝑎𝑙𝑖𝑘) 𝑚𝑎𝑘𝑎 𝑕𝑎𝑟𝑢𝑠 𝑑𝑖𝑙𝑎𝑘𝑢𝑘𝑎𝑛 𝑝𝑒𝑟𝑕𝑖𝑡𝑢𝑛𝑔𝑎𝑛 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐵𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑠𝑢𝑠𝑢𝑛𝑎𝑛 𝑔𝑒𝑙𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑚𝑢𝑛𝑔𝑘𝑖𝑛 𝑑𝑖𝑏𝑢𝑎𝑡 = 12! 3! .3! .3! .3! . 3! 𝑏𝑎𝑛𝑦𝑎𝑘 𝑝𝑒𝑚𝑢𝑡𝑎𝑟𝑎𝑛 / 𝑏𝑎𝑙𝑖𝑘𝑎𝑛 = 61600