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Osn texas Document Transcript

  • 1. Problem 1 HOME EXITProblem 2Problem 3Problem 4Problem 5 BC EXAM TEXAS A&M HIGH SCHOOL MATH CONTESTProblem 6Problem 7Problem 8 NOVEMBER 12, 2011Problem 9Problem 10 Editing by - Muhammad YusufProblem 11Problem 12 GO Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 2. Problem 1 HOME EXITProblem 2Problem 3 Orang yang gagal selalu mencari jalan untukProblem 4Problem 5Problem 6 menghindari kesulitan,Problem 7Problem 8 sementara orang yang suksesProblem 9 selalu menerjang kesulitanProblem 10 untuk menggapai kesuksesan.Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 3. Problem 1 Problem 1 Solusi HOME EXITProblem 2Problem 3 JikaProblem 4Problem 5Problem 6Problem 7 Maka x = ...Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 4. Problem 1 Problem 2 Solusi HOME EXITProblem 2Problem 3Problem 4Problem 5 Berapakah jumlah semua pembagi dari 36,Problem 6 termasuk 1 dan dirinya sendiriProblem 7Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 5. Problem 1 Problem 3 Solusi HOME EXITProblem 2Problem 3 Keliling sebuah persegi panjang adalah 28 m.Problem 4 Sebuah persegi panjang yang lain denganProblem 5 panjang 3 kali panjang persegi panjangProblem 6 pertama dan lebar 2 kali lebar persegi panjangProblem 7 pertama memiliki luas 72 m. Berapakah luasProblem 8 persegi panjang pertama?Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 6. Problem 1 Problem 4 Solusi HOME EXITProblem 2Problem 3Problem 4 Sebuah operasi “ * “ didefinisikan dengan :Problem 5 a * b = a2 + 3bProblem 6 Tentukan nilai dari (2 * 0) * (0 * 1)Problem 7Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 7. Problem 1 Problem 5 Solusi HOME EXITProblem 2Problem 3Problem 4 Segitiga ABC dengan panjang AB = 25Problem 5 meter, AC = 24 meter, dan BC = 23 meter.Problem 6 Suatu titik D yang merupakan titik padaProblem 7 AC sehingga BD tegak lurus dengan AC.Problem 8 Berapakah AD − DC?Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 8. Problem 1 Problem 6 Solusi HOME EXITProblem 2Problem 3 JikaProblem 4Problem 5Problem 6Problem 7 Maka x = ...Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 9. Problem 1 Problem 7 Solusi HOME EXITProblem 2Problem 3Problem 4 Tentukan penyelesaian dari :Problem 5Problem 6Problem 7 Untuk 0 ≤ x ≤ 2Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 10. Problem 1 Problem 8 Solusi HOME EXITProblem 2Problem 3Problem 4 What is the last digit of the sumProblem 5 1! + 2! + 3! + · · · + 2010! + 2011! ?Problem 6Problem 7Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 11. Problem 1 Problem 9 Solusi HOME EXITProblem 2Problem 3Problem 4 Suppose that f(x) is a function such thatProblem 5 for every real number x :Problem 6Problem 7 i) f(x) + f(1 − x) = 11Problem 8 ii) f(1 + x) = 3 + f(x)Problem 9 Then f(x) + f(−x) =Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 12. Problem 1 Problem 10 Solusi HOME EXITProblem 2Problem 3Problem 4 Let a, b and c be the three roots ofProblem 5 x3 − 64x− 14 . What is the value ofProblem 6Problem 7 a3 + b3 + c3 ?Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 13. Problem 1 Problem 11 Solusi HOME EXITProblem 2Problem 3Problem 4 Let ABC be an isosceles right triangleProblem 5 with right angle at C. Let P be a pointProblem 6 inside the triangle such that AP = 3, BP =Problem 7 5, and CP = 2√2. What is the area of theProblem 8 triangle ABC?Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 14. Problem 1 Problem 12 Solusi HOME EXITProblem 2Problem 3Problem 4 In how many distinct ways can one writeProblem 5 1,000,000 as the product of three positiveProblem 6Problem 7 integers? Treat all orderings of the sameProblem 8 set of factors as one wayProblem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 15. Problem 1 Problem 13 Solusi HOME EXITProblem 2Problem 3Problem 4 A cube is inscribed in a ball. What is theProblem 5 ratio of the volume of the cube to theProblem 6Problem 7 volume of the ball?Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 16. Problem 1 HOME EXITProblem 2Problem 3Problem 4Problem 5Problem 6Problem 7Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 17. Problem 1 HOME EXITProblem 2Problem 3Problem 4 Pembagi dari 36 :Problem 5 1, 2, 3, 4, 6, 9, 12, 18, 36Problem 6Problem 7 Jumlahnya :Problem 8 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36Problem 9Problem 10 = 91Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 18. Problem 1 HOME EXITProblem 2 Persegi panjang A, dengan panjang = p dan lebar = lProblem 3Problem 4 Kelilingnya = 28, sehingga 2p + 2l = 28Problem 5 Persegi panjang B, dengan panjang = 3pProblem 6 dan lebar = 2lProblem 7 Kelilingnya = 72, sehingga 2(3p + 2l) = 72Problem 8 Disederhanakan menjadi 3p + 2l = 36 Persamaan 2 3p + 2l = 36Problem 9 p = 8 m dan l = 6 mProblem 10 Persamaan 1 2p + 2l = 28Problem 11 Sehingga luas segitiga pertama = 8 × 6 = 48 m2Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 19. Problem 1 HOME EXITProblem 2Problem 3Problem 4 a * b = a2 + 3 bProblem 5Problem 6 (2 * 0) = 22 + 30 = 4 + 1 = 5Problem 7 (0 * 1) = 02 + 31 = 0 + 3 = 3Problem 8 (2 * 0) * (0 * 1) = (5 * 3) = 52 + 33 = 25 + 27 = 52Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 20. Problem 1 HOME EXITProblem 2Problem 3 C BD2 = BC2 – CD2 BD2 = 242 – CD2Problem 4 D Disamping itu, 23 meter 24 meterProblem 5 BD2 = AB2 – AD2 BD2 = 252 – AD2Problem 6 BD2 = 625 – AD2Problem 7 A 25 meter B SehinggaProblem 8 576 – CD2 = 625 – AD2Problem 9 Sehingga AD2 – CD2 = 49Problem 10 49 (AD – CD)(AD + CD) = 49 AD – CD =Problem 11 23 (AD – CD) × 23 = 49Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 21. Problem 1 HOME EXITProblem 2Problem 3Problem 4Problem 5Problem 6Problem 7Problem 8Problem 9 Pangkat disamakan,Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 22. Problem 1 HOME EXITProblem 2Problem 3Problem 4 Kedua ruas dikuadratkan,Problem 5Problem 6Problem 7Problem 8Problem 9Problem 10Problem 11 Penyelesaian persamaan tersebutProblem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 23. Problem 1 HOME EXITProblem 2Problem 3Problem 4Problem 5 Sehingga nilai xProblem 6 adalahProblem 7Problem 8Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 24. Problem 1 HOME EXITProblem 2Problem 3 Mulai dari 5! Dan seterusnya, angkaProblem 4 satuanya adalah nol.Problem 5Problem 6 Sehingga yang perlu diperhatikan adalahProblem 7 jumlah dariProblem 8 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33Problem 9Problem 10 Sehingga angka satuannya adalah 3Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 25. Problem 1 HOME EXITProblem 2Problem 3 f(x) + f(1 − x) = 11 f(-x) + f(1 + x) = 11Problem 4Problem 5 f(1 + x) = 3 + f(x) f(1 - x) = 3 + f(-x)Problem 6 Sehingga :Problem 7 f(x) + 3 + f(-x) = 11 f(x) + f(1 − x) = 11Problem 8 atau f(x) + f(-x) = 8Problem 9Problem 10 f(-x) + f(1 + x) = 11 f(-x) + 3 + f(x) = 11Problem 11 f(-x) + f(x) = 8Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 26. Problem 1 HOME EXITProblem 2Problem 3 x3 − 64x− 14Problem 4 = (x − a)(x − b)(x − c)Problem 5 = x3 − (a + b + c)x2 + (ab + bc + ac)x − abcProblem 6 Dengan (a + b + c) = 0, makaProblem 7 a3 + b3 + c3 = (64a + 14) + (64b + 14) + (64c + 14)Problem 8 = 64(a + b + c) + 42Problem 9 = 64(0) + 42Problem 10 = 42Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 27. Problem 1 HOME EXITProblem 2 A Misalkan AC = CB = lProblem 3 PD2 = 9 – (l – DC)2Problem 4 P PD2 = 8 – CD2Problem 5 D 9 – (l – DC)2 = 8 – DC2Problem 6 l2 - 2DC = 1Problem 7 C E B Karena DC = CE , makaProblem 8 PE2 = 25 – (l – CE)2 2l2 - 4DC = 18Problem 9 PE2 = 8 – CE2 l2 - 2DC = 1Problem 10 25 – (l – CE)2 = 8 – CE2Problem 11 l2 - 2CE = 17Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 28. Problem 1 HOME EXITProblem 2 A l2 - 2DC = 1 dan l2 - 2CE = 17Problem 3 DC = ½ (l2 – 1) CE = ½ (l2 – 17)Problem 4Problem 5 D P Karena DC2 + CE2 = 8l , makaProblem 6 ¼(l2 – 1)2 + ¼(l2 – 17)2 = 8lProblem 7 C E B (l2 – 1)2 + (l2 – 17)2 = 32l2Problem 8 l4 – 2l2 + 1+ l4 – 34l2 + 289 = 32l2Problem 9 l4 – 34l2 + 145 = 0 l2 = 5 atau 29Problem 10 Yang memenuhi hanya l2 = 29Problem 11 Sehingga luasnya = 29/2Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO
  • 29. Problem 1 HOME EXITProblem 2 Let (x, y) be the coordinates of P . Then x2 + y2 = 8 , (s −Problem 3 x)2 + y2 = 9 and (s − y)2 + x2 = 25 . We obtain s2 − 2sx = 1Problem 4 and s2 − 2sy = 17 . Solve for x and y in terms of s , andProblem 5 place in the first equation, obtaining (s2 − 1)2 + (s2 − 17)2Problem 6 = 32s2 . This simplifes to s4 − 34s2 + 145 = 0 , hence s2 = 5Problem 7 or 29 . Only s2 = 29 is consistent with the given data (notice that if s = √5 , then √10 = √2s = AB > BP = 5, aProblem 8 contradiction). Thus, the area is 29/2Problem 9Problem 10Problem 11Problem 12 Muhammad Yusuf, S.Pd.Problem 13 SMP NEGERI 1 BOLO