# WMI Preliminary Round 2021 [Grade 10A]

World Mathematics Invitational (WMI) is the first international competition founded by Taiwan. It gathers institutes and organizations worldwide that make efforts in promoting and popularizing mathematics. Through interacting with other math-loving students that represent their countries, students can expand their worldview, experience different cultures, and thus their horizon as well as their future will be broaden. (sc : http://www.wminv.org/)

Berikut ini soal dan solusi WMI grade 10A tahun 2021

1.$$\sqrt{19 + 5\sqrt{19}}$$ is between two successive positive integers. What is the smaller of the two positive integers?

(A) 5
(B) 6
(C) 7
(D) 8

Karena $$\sqrt{16} < \sqrt{19}$$, maka

$$19 + 5\sqrt{16} < 19 + 5\sqrt{19}$$
$$39 < 19 + 5\sqrt{19}$$

Karena $$\sqrt{19} < \sqrt{25}$$, maka

$$19 + 5\sqrt{19} < 19 + 5\sqrt{25}$$
$$19 + 5\sqrt{19}<44$$

Selanjutnya

$$39 < 19 + 5\sqrt{19} < 44$$
$$36 < 39 < 19 + 5\sqrt{19} < 44 < 49$$

Sehingga diperoleh

$$6 < \sqrt{19 + 5\sqrt{19}} < 7$$

Jadi bilangan terkecilnya adalah $$6$$

2. The regular heptagon on the right is formed by AB, BC, CD, DE, EF, and FG . If DE is perpendicular to y-axis, which of the seven lines has the smallest slope?

(A) CD
(B) EF
(C) FG
(D) GA

Gradien $$=\frac{Ξπ¦}{Ξπ₯}$$
, dari gambar garis yang memiliki gradient negative adalah garis $$FG, GA$$ dan $$DC$$. Dengan melakukan obeservasi, pada garis $$FG$$ nilai $$Ξπ¦ > Ξπ₯$$, karenaΒ  $$\frac{Ξπ¦}{Ξπ₯}<0$$Β  maka nilai gradien garis $$FG$$ adalah yang paling kecil

3. If $$π(π₯) = 2π₯^3 β 3π₯^2 β 8π₯ + 5 = π(π₯ β 1)(π₯ β 2)(π₯ β 3)οΌπ(π₯ β 2)(π₯ β 3)οΌπ(π₯ β 3) + π$$, find $$4π + 3π + 2π + π$$.

(A) 69
(B) 75
(C) 67
(D) 73

Untuk $$π₯ = 0$$

$$π(0) = π(β1)(β2)(β3) + π(β2)(β3) + π(β3) + π = 5$$
$$β β6π + 6π β 3π + π = 5 β¦ (1)$$

Untuk $$π₯ = 1$$

$$π(1) = π(β1)(β2) + π(β2) + π = 2 β 3 β 8 + 5 = β4$$
$$β 2π β 2π + π = β4 β¦ (2)$$

Untuk $$π₯ = 2$$

$$π(2) = π(β1) + π = 16 β 12 β 16 + 5 = β7$$
$$β βπ + π = β7 β¦ (3)$$

Untuk $$π₯ = 3$$

$$π(3) = π = 54 β 27 β 24 + 5 = 8$$
$$β π = 8$$

Subtitusi nilai $$π$$ ke persamaan (3), dan lanjutkan ke persamaan (2) dan (1) diperoleh nilai $$a, b, c$$ berturut turut adalah $$2, 9$$ dan $$15$$.
Jadi nilai dari $$4π + 3π + 2π + π = 4(2) + 3(9) + 2(15) + 8 = 73$$

4.Β  Set $$a$$ to be a positive real number. If the graph of the function $$f(x)=a^x$$ passes through four points $$(Ξ±, 1), (-2, Ξ²), (\frac{4}{3}, Ξ³)$$, and $$(\frac{1}{3}, \frac{1}{2})$$, find $$Ξ±+Ξ²Ξ³?$$

(A) 4
(B) 5
(C) 8
(D) 9

5. $$\frac{8}{4Γ7}+\frac{8}{7Γ10}+\frac{8}{10Γ13}+ β― +\frac{8}{61Γ64}= β―$$?

(A) $$\frac{15}{64}$$
(B) $$\frac{7}{32}$$
(C) $$\frac{3}{16}$$
(D) $$\frac{5}{8}$$

\begin{align} \frac{8}{4Γ7}+\frac{8}{7Γ10}+\frac{8}{10Γ13}+ β― +\frac{8}{61Γ64}&=8(\frac{1}{4 Γ 7}+\frac{1}{7 Γ 10}+\frac{1}{10 Γ 13}+ β― +\frac{1}{61 Γ 64})\\ &=\frac{8}{3}\left(\frac{1}{4}β\frac{1}{7}+\frac{1}{7}β\frac{1}{10}+\frac{1}{10}β\frac{1}{13}+ β― +\frac{1}{61}β\frac{1}{64}\right)\\ &=\frac{8}{3}\left(\frac{1}{4}β\frac{1}{64}\right)\\ &=\frac{8}{3}\left(\frac{16 β 1}{64}\right)\\ &=\frac{8}{3}\left(\frac{15}{64}\right)\\ &=\frac{5}{8}\\ \end{align}

6. Set $$π(π) = \cos^π π + \sin^π π$$, and $$π(1) = β1$$. Find $$π(2) β π(3) β π(5) + π(10) + π(18)$$.

(A) 3
(B) 5
(C) οΌ1
(D) 1

Rumus identitas trigonometri $$\cos^2 π + \sin^2 π = 1$$

$$(\cos π + \sin π)^2 = \cos^2 π + 2\cos π \sin π + \sin^2 π = 1$$
$$2 \cos π \sin π + 1 = 1$$
$$\cos π \sin π = 0$$

$$π(1) = (\cos π + \sin π) = β1$$
$$π(2) = \cos^2 π + \sin^2 π = 1$$
$$π(3) = \cos^3 π + \sin^3 π = (\cos π + \sin π)(\cos^2 π β \cos π \sin π + \sin^2 π) = (β1)(1 β 0) = β1$$
$$π(4) = \cos^4 π + \sin^4 π = (\cos^2 π + \sin^2 π)^2 β 2\cos^2 π \sin^2 π = 1 β 0 = 1$$
β¦
diperoleh

$$π(π) = \begin{cases}1, & \text{π πππππ}\\β1, & \text{π ππππππ}\end{cases}$$

$$π(2) β π(3) β π(5) + π(10) + π(18) = 1 β (β1) β (β1) + 1 + 1 = 5$$

7. $$(\log_9 2 + \log_3 \sqrt{128})(\log_{16} 27 β \log_2 \frac{1}{9})= β―$$

(A) 11
(B) 10
(C) 13
(D) 12

8. $${ π_π}$$ is an arithmetic sequence, and $${ π_π}$$ is a geometric sequence. If $$π_4 = π_4 = 9, π_7 = π_7 = 72$$, find the value of $$π_9 + π_9$$.

(A) 384
(B) 396
(C) 402
(D) 414

Misalkan suku pertama pada barisan aritmatika dan geometri adalah $$π$$ πππ $$π$$, beda dan rasio pada kedua barisan adalah $$π$$ dan $$π$$, selanjutnya
untuk barisan aritmetika

$$π_4 = π + 3π = 9$$
$$π_7 = π + 6π = 72$$

Eliminasi keduanya diperoleh $$π = β54$$ dan $$π = 21$$

Untuk barisan geometri

$$π_4 = ππ^3 = 9$$
$$π_7 = ππ^6 = 72$$

Bagi kedua persamaan diperoleh $$π = 2$$ dan $$π =\frac{9}{8}$$

Jadi nilai $$π_9 + π_9 = π + 8π + ππ^8 = β54 + 8(21) + \frac{9}{8}(2)^8 = 114 + 288 = 402$$

9. As in the picture, $$β B=90Β°, \overline{AB}:\overline{BC} : \overline{CA} = 15 : 8 : 17, \overline{CA}=\overline{CD}$$, and $$B, C$$ and $$D$$ are collinear points. Make $$β ACB=ΞΈ$$, what is the value of $$\tan\frac{ΞΈ}{2}$$?

(A) $$\frac{3}{4}$$
(B) $$\frac{5}{6}$$
(C) $$\frac{2}{3}$$
(D) $$\frac{3}{5}$$

10. Set $$m$$ and $$n$$ to be integers. If the equation $$x^3+mx^2+nx+3=0$$ has three different rational roots, find $$3mοΌ4n$$.

(A) οΌ6
(B) οΌ5
(C) οΌ4
(D) οΌ3

karena konstantanya 3, maka kemungkinan Akar-akarnya adalah 1, 3 dan -1

$$π₯^3 + ππ₯^2 + ππ₯ + 3 = (π₯ β 1)(π₯ + 1)(π₯ β 3) = 0$$
$$(π₯ β 1)(π₯ + 1)(π₯ β 3) = 0$$
$$(π₯^2 β 1)(π₯ β 3) = 0$$
$$π₯^3 β 3π₯^2 β π₯ + 3 = 0$$

Samakan dengan persamaan $$π₯^3 + ππ₯^2 + ππ₯ + 3 = 0$$, diperoleh $$π = β3$$ dan $$π = β1$$.

jadi nilai dari $$3π β 4π$$ adalah $$3(β3) β 4(β1) = β9 + 4 = β5$$

11. Set $$n$$ to be a positive integer. If $$π_π$$ means to βconvert $$\frac{π}{7}$$ into a decimal and find the $$π^{th}$$ number after the decimal pointβ, find $$π_{2021}$$

(A) 8
(B) 7
(C) 5
(D) 4

$$π_{2021} =\frac{2021}{7}= 288,714285714285 β¦$$

Angka setelah koma membentuk pengulangan $$714285714285β¦.$$, berulang setiap 6 angka, maka untuk angka ke$$-2021$$

$$2021\; πππ\; 6 β‘ 5$$

Angka ke$$-2021$$ sama saja dengan angka ke$$-5$$ pada pola bilangan di atas yaitu angka $$8$$

12. Given that $$n$$ is a positive integer, and the sequence $${π_π}$$ satisfies $$π_1 οΌ5$$ and $$π_{π+1} = π_π + (3π + 2)$$,Β  for $$π > 1$$. Find $$π_{20}$$ .

(A) 670
(B) 675
(C) 608
(D) 613

$$π_{π+1} = π_π + (3π + 2)$$
$$π_{π+1} β π_π = (3π + 2)$$

Selanjutnya

$$π_2 β π_1 = 3(1) + 2$$
$$π_3 β π_2 = 3(2) + 2$$
$$π_4 β π_3 = 3(3) + 2$$
$$π_5 β π_4 = 3(4) + 2$$
$$β¦.$$
$$π_{20} β π_{19} = 3(19) + 2$$

Jumlahkan persamaan di atas diperoleh

$$π_{20} β π_1 = 3(1 + 2 + 3 + β― + 19) + 2(19) = 570 + 38$$
$$π_{20} β 5 = 570 + 38$$
$$π_{20} = 570 + 38 + 5 = 613$$

13. Pick 4 different numbers at will from 13 successive odd numbers 1, 3, 5, β¦β¦, 25. If any two numbers from these 4 numbers are not successive odd numbers, how many methods are there to pick them?

(A) 126
(B) 162
(C) 210
(D) 252

14. The streets are shown below, and each grid is a square. Today, Anna goes from P to Q, and Blake goes from Q to P. Each of them starts off at the same time and goes the shortest path in the same speed. Suppose at each crossroads, the probability of choosing any direction to go forward is the same, and that the probability for Anna and Blake to meet on the way is $$\frac{p}{q} [p, qβN$$, and $$(p, q)οΌ1]$$, find $$pοΌq$$.

(A) 209
(B) 227
(C) 29
(D) 47

15. Put 6 ββs and 9 ββs in a row. Suppose the pattern of the shapes has to change 6 times, and one of the methods is shown below, how many methods are there to arrange the shapes?

(A) 560
(B) 630
(C) 840
(D) 910