SEAMO PAPER E 2019 [PROBLEM And SOLUTION] - BorneoMath

18. The length of the 3 sides of a triangle are \(a, b\) and \(c\), with opposite angles A, B, and C, respectively.
Given that \(a^2 = 1\) and \(b^2 + c^2 = 2019\) ,
find the value of \(\frac{\cot A}{\cot B+\cot C}\).

(A) 1009
(B) 1010
(C) 2019
(D) 2020
(E) None of the abov

19. For \(i = 1, 2, 3,…, 100\) , \(x_i\) can be either \(\sqrt 3 − \sqrt 2\) or \(\sqrt 3 + \sqrt 2\).
Denoting S as \(x_1 x_2 +x_3 x_4 +x_5 x_6 +…+x_{99} x_{100}\) ,
how many possible different positive integer values of S are there?
(A) 25
(B) 50
(C) 75
(D) 100
(E) None of the above

\(𝐴 = (\sqrt 3 − \sqrt 2)(\sqrt 3 + \sqrt 2) = 3 − 2 = 1\)
\(𝐵 = (\sqrt 3 − \sqrt 2)(\sqrt 3 – \sqrt 2) = 5 − 2\sqrt 6\)
\(𝐶 = (\sqrt 3 + \sqrt 2)(\sqrt 3 + \sqrt 2) = 5 + 2\sqrt 6\)
\(𝐵 + 𝐶 = 10\)

Diperoleh \(A\) bilangan bulat dan \(B+C\) bilangan bulat
karna banyaknya ada \(50\), maka kemungkinan banyak pilihan untuk \(A\) harus genap agar dipastikan \(B\) dan \(C\) berpasangan. Banyak kemungkinan bilangan bulat positif dari \(S\) sama saja dengan banyak cara memilih nilai \(A\). Banyak cara memilih nilai \(A : \{0, 2, 4,…,50\}\) banyaknya ada \(26\) cara

20. 13 identical circles with radius 1 are inscribed within a larger circle as shown in the figure below.
Given that all the intersections occur at points of tangency, find the shaded area.

(A) \(4π\sqrt 3\)
(B) \(7π\)
(C) \(π(3\sqrt 3 +2)\)
(D) \(π(\sqrt 3 +6)\)
(E) None of the above

\(ΔABO≅ΔABC\)
Jari-jari lingkaran besar \((R)\) diperoleh dua kali tinggi \((t)\) segitiga \(ΔABO\) ditambah jari-jari lingkaran kecil \((r)\).
Tinggi \(Δ𝐴𝑂𝐵 = 𝑡 = \sqrt {2^2 − 1^2} = \sqrt 3\)
\(𝑅 = 2𝑡 + 𝑟 = 2\sqrt 3 + 1\)
L arsiran= Luas lingkaran besar – 13 × luas lingkaran kecil
\(= 𝜋𝑅^2 − 13𝜋𝑟^2\)
\(= 𝜋(2\sqrt 3 + 1)^2 − 13𝜋\)
\(= 𝜋(12 + 1 + 4\sqrt 3) − 13𝜋\)
\(= 13𝜋 + 4\sqrt 3 𝜋 − 13𝜋 = 4\sqrt 3𝜋\)

21. Find the smallest positive integer k such that

\(\frac{1}{\log_{2^k} 2019!} +\frac{1}{\log_{3^k} 2019!} +\frac{1}{\log_{4^k} 2019!} +…+\frac{1}{\log_{2019^k} 2019!} >\sqrt{2019}\)

22. Let \(\{a_n\}\) and \(\{b_n\}\) be two arithmetic sequences such that \(a_1=2019\) and \(b_1=1\)

Suppose \(a_{100}+b_{100}=1010\), evaluate
\(a_1+a_2 +…+a_{20} +b_1+b_2+…+b_{20}\)

\(𝑎_{100} + 𝑏_{100} = 1010\)
\(𝑎_1 + 99𝑝 + 𝑏_1 + 99𝑞 = 1010\)
\(2019 + 1 + 99(𝑝 + 𝑞) = 1010\)
\(𝑝 + 𝑞 =−\frac{1010}{99}\)
Selanjutnya
\(𝑎_1 + 𝑎_2 + ⋯ + 𝑎_{20} + 𝑏_1 + 𝑏_2 + ⋯ + 𝑏_{20}\)
\(=\frac{20}{2}(𝑎_1 + 𝑎_{20}) +\frac{20}{2}(𝑏_1 + 𝑏_{20})\)
\(= 10(2019 + 2019 + 19𝑝) + 10(1 + 1 + 19𝑞)\)
\(= 10(4038 + 19𝑝) + 10(2 + 19𝑞)\)
\(= (40380 + 190𝑝) + (20 + 190𝑞)\)
\(= 40400 + 190(𝑝 + 𝑞)\)
\(= 40400 − 190 (\frac{1010}{99}) =\frac{3807700}{99}\)

23. Shawn starts a number sequence with a positive integer \(n\). Each successive number is obtained by subtracting the largest possible square number less than or equal to the current number till he obtains 0.
For example, if \(n=60\), then

\(60 − 7^2 = 11\)
\(11 − 3^2 = 2\)
\(2 − 1^2 = 1\)
\(1 − 1^2 = 0\)

His sequence will be 60, 11, 2, 1, 0.
Let N be the smallest number for which Shawn’s sequence contains 8
numbers. What is the sum of digits of N?

24. Find the maximum value of

\(Q=\frac{\sin x + \sin 2x + \sin 3x}{\cos x+\cos 2x+\cos 3x}\)

for \(0≤x≤\frac{π}{8}\).

25. Let \(⌊x⌋\) denote the largest integer less than or equal to \(x\).

For example, \(⌊5⌋=5\) and \(⌊7.2⌋=7\)

Evaluate

\(S=⌊\sqrt 1⌋+⌊\sqrt 2⌋+⌊\sqrt 3⌋+…+⌊\sqrt{2019}⌋\)

\(⌊\sqrt1⌋ + ⌊\sqrt2⌋ + ⌊\sqrt3⌋ = 3 × 1\)
\(⌊\sqrt4⌋ + ⌊\sqrt5⌋ + ⋯ + ⌊\sqrt8⌋ = 5 × 2\)
\(⌊\sqrt9⌋ + ⌊\sqrt{10}⌋ + ⋯ + ⌊\sqrt{15}⌋ = 7 × 3\)
\(⌊\sqrt{16}⌋ + ⌊\sqrt{17}⌋ + ⋯ + ⌊\sqrt{24}⌋ = 9 × 4\)
\(⌊\sqrt{25}⌋ + ⌊\sqrt{26}⌋ + ⋯ + ⌊\sqrt{35}⌋ = 11 × 5\)
…
\(⌊\sqrt{442}⌋ + ⌊\sqrt{1936}⌋ + ⋯ + ⌊\sqrt{2024}⌋ = 89 × 44\)

Bentuk umumnya \(𝑛(2𝑛 + 1)=2n^2+n\), untuk \(𝑛\) bilangan asli

\(𝑆 = ⌊\sqrt1⌋ + ⌊\sqrt2⌋ + ⌊\sqrt3⌋ + ⋯ + ⌊\sqrt{2019}⌋\)

\(= \sum_{n=1}^{44}(2𝑛^2 + 𝑛) − (⌊√2020⌋ + ⌊√2021⌋ + ⋯ + ⌊√2024⌋)\)

\(= 2\sum_{n=1}^{44} 𝑛^2 +\sum_{n=1}^{44} n − (44 × 5)\)

\(= 2(\frac{44(44+1)(88+1)}{6})+ (\frac{44(44+1)}{2}) − 220\)

\(= 2(29370) + (990) − 220\)

\(= 58.740 + 990 − 220\)

\(= 59.510\)

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