Asian Science And Math Olympiad (ASMO) 2018 For Grade 9 - BorneoMath

11. A semicircle overlaps with a circle segment as shown in figure above. Find the area of shaded region. Express your answer in π.

Perhatikan \(Δ𝑂𝐷𝐶\), karena \(∠𝑂𝐶𝐷 = 60°\) maka jika \(𝑂𝐶 = 4\) maka \(𝐶𝐷 = 2\).

\([𝐴𝐵𝐶] =\frac{1}{2}𝐴𝐵. 𝐶𝐷 =\frac{1}{2}(8)(2) = 8\)

\(\begin{align}
Luas\; tembereng\; 𝑋 &= 𝐿𝑢𝑎𝑠\; 𝐽𝑢𝑟𝑖𝑛𝑔\; 𝑂𝐶𝐵 − [𝑂𝐶𝐵]\\
&=\frac{30}{360}𝜋(4)2 −\frac{1}{2}𝑂𝐵. 𝐶𝐷\\
&=\frac{16}{12}𝜋 −\frac{1}{2}(4)(2)\\
&=\frac{4}{3}𝜋 − 4\\
\end{align}\)

𝐿𝑢𝑎𝑠 daerah yang tidak diarsir adalah \([𝐴𝐵𝐶] + 𝑋 = 8 +
\frac{4}{3}𝜋 − 4 =\frac{4}{3}𝜋 + 4\)

\(\begin{align}
Luas\; 𝑌&= 𝐿𝑢𝑎𝑠\; 𝑠𝑒𝑡𝑒𝑛𝑔𝑎ℎ\; 𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛\; − 𝑙𝑢𝑎𝑠\; 𝑑𝑎𝑒𝑟𝑎ℎ\; 𝑡𝑖𝑑𝑎𝑘\; 𝑑𝑖𝑎𝑟𝑠𝑖𝑟\\
&=\frac{1}{2}𝜋(8)^2 − (\frac{4}{3}𝜋 + 4)\\
&= 32𝜋 −\frac{4}{3}𝜋 − 4\\
&=\frac{92}{3}𝜋 − 4\\
\end{align}\)

\(\begin{align}
Luas\; 𝑍 &= 𝐿𝑢𝑎𝑠\; 𝑗𝑢𝑟𝑖𝑛𝑔 − 𝑙𝑢𝑎𝑠\; 𝑑𝑎𝑒𝑟𝑎ℎ\; 𝑡𝑖𝑑𝑎𝑘\; 𝑑𝑖𝑎𝑟𝑠𝑖𝑟\\
&=\frac{15}{360}𝜋(8)2 − (\frac{4}{3}𝜋 + 4)\\
&=\frac{8}{3}𝜋 −\frac{4}{3}𝜋 − 4\\
&=\frac{4}{3}𝜋 − 4\\
\end{align}\)

Jadi jumlah daerah yang diarsir \(𝑌 + 𝑍 =\frac{92}{3}𝜋 − 4 +\frac{4}{3}𝜋 − 4 = 32𝜋 − 8\)

12. Ali tells Siti that the product of 3 positive integers is 36. Ali tells Siti what the sum of the 3 numbers is, but Siti still does not know what the 3 numbers are. What is the sum of the three numbers?

not yet available

13. An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

banyaknya bilangan dari \(1000\) sampai dengan \(9999 = 9999 – 1000 + 1 = 9000\).
Banyaknya bilangan \(4\) digit ganjil yang angkanya berbeda semua

Banyaknya ada \(8 × 8 × 7 × 5\)
Jadi peluangnya adalah \(\frac{8×8×7×5}{90×10×10}=\frac{56}{225}\)

14.

Figure above shows a triangle \(ABC\) with angle \(∠BAC=40°, PQ=BQ\) and \(QR=QC\). Find the angle \(∠PQR\).

Berdasarkan jumlah sudut pada segitiga

\(40 + 𝑎 + 𝑏 = 180\)
\(⇒𝑎 + 𝑏 = 140\)

Berdasarkan jumlah sudut pada segiempat

\(40 + 180 − 𝑎 + 180 − 𝑏 + 𝑐 = 360\)
\(⇒40 + 360 − (𝑎 + 𝑏) + 𝑐 = 360\)
\(⇒40 − 140 + 𝑐 = 360 − 360 = 0\)
\(⇒𝑐 = 100\)

Jadi besar \(∠𝑃𝑄𝑅 = 100°\)

15. Nineteen children, aged 1 to 19 respectively, are standing in a circle. The difference between ages of each pair of adjacent children is recorded. What is the maximum value of the sum of these 19 positive integers?

Jadi total selisih maksimum adalah \((1 + 2 + 3 + ⋯ + 18) + 9 = 180\)

16. What is the tens digit of \(7^{2018}\)?

\(7^1\; mod\; 100 ≡ 7\)
\(7^2 mod\; 100 ≡ 49\)
\(7^3 mod\; 100 ≡ 43\)
\(7^4 mod\; 100 ≡ 01\)
Polanya berulang sebanyak \(4\) kali
Jadi \(7^{2018} mod\; 100 ≡ 7^{2018\;𝑚𝑜𝑑\; 4} mod\; 100 ≡ 7^2\; mod\; 100 ≡ 49\)

17. In how many ways can the six letters in the word MOUSEY be arranged in a row without containing either the word YOU or ME? For example, the word YOMUSE is such an arrangement.

Banyak permutasi seluruhnya adalah 6!
Banyak permutasi YOU berdekatan ada 4!
Banyak permutasi ME berdekatan ada 5!
Banyak permutasi YOU dan ME ada 3!
Jadi banyak susunan adalah 6! – 4! – 5! + 3! = 582

18. The total number of mushroom gathered by \(11\) boys and \(𝑛\) girls is \(𝑛^2 + 9𝑛 − 2\) , which each gathering exactly the same number. Determine the positive integer \(𝑛\).

misalkan masing dari mereka mengumpulkan \(m\) jamur dimana \(m\) merupakan bulat positif, maka

\(11𝑚 + 𝑚𝑛 = 𝑛^2 + 9𝑛 − 2\)
\(⇒ 𝑚(11 + 𝑛) = 𝑛^2 + 9𝑛 − 2\)
\(⇒ 𝑚 =\frac{𝑛^2 + 9𝑛 − 2}{𝑛 + 11}\)
\(=\frac{n(n+11)}{n+11}-\frac{2n+2}{n+11}\)
\(= 𝑛 −\frac{2(𝑛 + 11)}{𝑛 + 11}+\frac{20}{𝑛 + 11}\)
\(= 𝑛 − 2 +\frac{20}{𝑛 + 11}\)
Supaya m bilangan bulat positif maka \(\frac{20}{𝑛+11}\)
bulat positif, hanya satu nilai \(n\) yang memenuhi yaitu \(𝑛 = 9\)

19. Suppose that \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\), where is the value of \(\frac{xyz(b+c)(c+a)(a+b)}{abc(y+z)(z+x)(x+y)}\)?

Misalkan \(\frac{𝑥}{𝑎}=\frac{𝑦}{𝑏}=\frac{𝑧}{𝑐}= 𝑘\)
\(𝑥 = 𝑎𝑘, 𝑦 = 𝑏𝑘, 𝑧 = 𝑐𝑘\) subtitusi ke

\(\frac{𝑥𝑦𝑧(𝑏 + 𝑐)(𝑐 + 𝑎)(𝑎 + 𝑏)}{𝑎𝑏𝑐(𝑦 + 𝑧)(𝑧 + 𝑥)(𝑥 + 𝑦)}\)
\(=\frac{(𝑎𝑘)(𝑏𝑘)(𝑐𝑘)(𝑏 + 𝑐)(𝑐 + 𝑎)(𝑎 + 𝑏)}{𝑎𝑏𝑐(𝑏𝑘 + 𝑐𝑘)(𝑐𝑘 + 𝑎𝑘)(𝑎𝑘 + 𝑏𝑘)}\)
\(=\frac{(𝑎𝑘)(𝑏𝑘)(𝑐𝑘)(𝑏 + 𝑐)(𝑐 + 𝑎)(𝑎 + 𝑏)}{𝑎𝑏𝑐𝑘(𝑏 + 𝑐)𝑘(𝑐 + 𝑎)𝑘(𝑎 + 𝑏)}\)
\(= 1\)

20. A circle of radius 3 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

Luas lingkaran = \(\frac{22}{7}(9) =\frac{198}{7}\)
Luas bintang = \(36 -\frac{22}{7}(9) =\frac{54}{7}\)
Jadi perbandingan luas bintang dan lingkaran adalah \(\frac{54}{7} ∶ \frac{198}{7}= 3 ∶ 11\)

21. Find all integers \(n\) such that \(1+2+3+…+ n\) is equal to 3-digit number with identical digits.

\(1 + 2 + 3 + ⋯+ 𝑛 = \overline{aaa}\)
\(\frac{(𝑛 + 1)𝑛}{2}= 𝑎(111)\)
\((𝑛 + 1)𝑛 = 2𝑎(3)(37) = 36(37) = 37(38)\)

Yang mememenuhi adalah ketika nilai \(𝑎 = 6, (𝑛 + 1)𝑛 = 36(37)\). Jadi nilai \(n\) yang memenuhi adalah \(36\)

22. Simplify the following expression into a single numerical value

\(\sqrt{12 − \sqrt{35} + \sqrt{60} − \sqrt{84}} + \sqrt{12 + \sqrt{35} + \sqrt{60} + \sqrt{84}}\)

Gunakan rumus :
\((𝑥 + 𝑦 + 𝑧)^2 = 𝑥^2 + 𝑦^2 + 𝑧^2 + 2(𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧)\)
\((\sqrt{𝑥} + \sqrt{𝑦} + \sqrt{𝑧})^2 = 𝑥 + 𝑦 + 𝑧 + 2(\sqrt{𝑥𝑦} + \sqrt{𝑥𝑧} + \sqrt{𝑦𝑧})\)

\(\sqrt{12 + \sqrt{35} + \sqrt{60} + \sqrt{84}}\)
\(=\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)^2\)\(=\frac{7}{2}+\frac{5}{2} + 6 + 2\left(\sqrt{\frac{7}{2}·\frac{5}{2}}+\sqrt{\frac{7}{2}· 6}+\sqrt{\frac{5}{2}·6}\right)\)
\(=\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)^2=12 + \sqrt{35} + \sqrt{60} + \sqrt{84}\)

Diperoleh \(\sqrt{12 + \sqrt{35} + \sqrt{60} + \sqrt{84}}=\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)\)

\(\sqrt{12 – \sqrt{35} + \sqrt{60} – \sqrt{84}}\)
\(=\left(-\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)^2\)\(=\frac{7}{2}+\frac{5}{2} + 6 + 2\left(-\sqrt{\frac{7}{2}·\frac{5}{2}}-\sqrt{\frac{7}{2}· 6}+\sqrt{\frac{5}{2}·6}\right)\)
\(=\left(-\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)^2=12 – \sqrt{35} + \sqrt{60} – \sqrt{84}\)

Diperoleh \(\sqrt{12 – \sqrt{35} + \sqrt{60} – \sqrt{84}}=\left(-\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)\)

Jadi

\(\sqrt{12 − \sqrt{35} + \sqrt{60} − \sqrt{84}} + \sqrt{12 + \sqrt{35} + \sqrt{60} + \sqrt{84}}\)
\( =\left(-\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)+\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}+\sqrt 6\right)\)
\(=2\sqrt{\frac{5}{2}}+2\sqrt 6=\sqrt{10}+2\sqrt 6\)

23. \(APQ\) is a right isosceles triangle inscribed in a rectangle \(ABCD\), with the vertex \(P\) of the right angle on \(BC\) and \(Q\) on \(CD\). If \(BP=1 cm\) and \(∠APB = 60°\) , what is the area, in cm² , of the triangle \(ADQ?\)

Sisi-sisi di atas diperoleh dari perbandingan sisi pada segitiga siku-siku yang salah satu sudutnya 60°

Luas \(ADQ=\frac{1}{2}(\sqrt 3 − 1)(\sqrt 3 + 1) =\frac{1}{2}(3 − 1) = 1\;𝑐𝑚^2\)

24. \(P\) is a point inside a rectangle \(ABCD\). If \(PA=4cm, PB = 6cm\) and \(PD = 9cm\), find the length, in cm, of \(PC\).

Gunakan dalil Britis Flag Theorem

\(𝐴𝑃^2 + 𝑃𝐶^2 = 𝐷𝑃^2 + 𝑃𝐵^2\)
\(16 + 𝑃𝐶^2 = 81 + 36\)
\(𝑃𝐶^2 = 117 − 16\)
\(𝑃𝐶 = \sqrt {101}\)

25. \(ab+bc +ca + 2(a +b+c) = 8045\) and \(abc -a -b-c = -2\) where \(a,b\) and \(c\) are positive integers. Find the value of \(a+b+c\).

Jumlahkan kedua persamaan
\(𝑎𝑏𝑐 + 𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 + 𝑎 + 𝑏 + 𝑐 = 8043\)
\(𝑎𝑏𝑐 + 𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 + 𝑎 + 𝑏 + 𝑐 + 1 = 8043 + 1\)
\((𝑎 + 1)(𝑏 + 1)(𝑐 + 1) = 8044\)
\((𝑎 + 1)(𝑏 + 1)(𝑐 + 1) = 2 × 2 × 2011\)
Diperoleh \(𝑎 = 1, 𝑏 = 1\) dan \(𝑐 = 2010\)
Jadi nilai \(𝑎 + 𝑏 + 𝑐 = 1 + 1 + 2010 = 2012\)

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