# WMI Preliminary Round 2021 [Grade 10B]

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 10B tahun 2021

1) Which option below can represent the value of

$$\frac{1}{7}οΌ\frac{1}{9}$$?

(A) $$0.\overline{253968}$$
(B) $$0.\overline{242857}$$
(C) $$0.\overline{253963}$$
(D) $$0.\overline{253968}$$

2. How many integer solutions are there for equation $$(π₯^2 β π₯ β 1)^{5π₯^2+9π₯β2} = 1$$?

• Syarat 1
$$π₯^2 β π₯ β 1 = 1$$
$$π₯^2 β π₯ β 2 = 0$$
$$(π₯ β 2)(π₯ + 1) = 0$$
$$π₯ = 2 β¨ π₯ = β1$$
• Syarat 2
$$5π₯^2 + 9π₯ β 2 = 0$$
$$(5π₯ β 1)(π₯ + 2) = 0$$
$$π₯ =\frac{1}{5}(ππ) β¨ π₯ = β2$$
Nilai $$π₯ = β2$$ memenuhi karena $$π₯^2 β π₯ β 1 β 0$$
• Syarat 3
$$π₯^2 β π₯ β 1 = β1$$
$$π₯^2 β π₯ = 0$$
$$π₯(π₯ β 1) = 0$$
$$π₯ = 0 β¨ π₯ = 1$$
nilai $$π₯ = 0 β¨ π₯ = 1$$ memenuhi karena $$5π₯^2 + 9π₯ β 2$$ bernilai genap.

Jadi semua nilai $$x$$ yang memenuhi adalah $$\{-2, -1, 0, 1, 2\}$$ ada 5 bilangan

3) Given a regular sequence $$4, 5, 11, 25, 50, x, 145, β¦.$$ What should $$x$$ be?
(A) 121
(B) 97
|(C) 89
(D) 76

4) Set $$a, b β C, \overline{a}+2\overline{b} οΌi,$$ and $$\overline{a}Β·\overline{b} οΌ-5-i$$. Find the possible value of $$|a|^2$$ .
(A) 7
(B) 9
(C) 11
(D) 13

5. Look at the picture. $$π$$ is a point in $$Ξπ΄π΅πΆ$$. Draw $$3$$ lines which pass through $$P$$ and are parallel to each side of $$Ξπ΄π΅πΆ$$. If the areas of the small triangles $$π‘_1, π‘_2$$ and $$π‘_3$$ are $$4, 9$$, and $$49$$, respectively, find the area of $$Ξπ΄π΅πΆ$$.

(A) 121
(B) 144
(C) 169
(D) 172

Perhatikan kesebangunan $$Ξπ·ππΈ$$ dan
$$ΞπΌππΉ$$

$$(\frac{π}{π})^2=\frac{4}{49}β\frac{π}{π}=\frac{2}{7}$$

Perhatikan kesebangunan $$ΞπΌπ·πΆ$$ dan $$ΞπΌππΉ$$

$$(\frac{π}{π + π})^2=\frac{4}{49 + 4 + π}$$

$$β(\frac{2}{9})^2=\frac{4}{53 + π}$$

$$β 81 = 53 + π β π = 28$$

Selanjutnya, Perhatikan kesebangunan $$Ξπ·ππΈ$$ dan $$Ξπ»πΊπ$$

$$(\frac{π}{π})^2=\frac{9}{49}β\frac{π}{π}=\frac{3}{7}$$

Perhatikan kesebangunan $$Ξπ»π΅πΈ$$ dan $$Ξπ»πΊπ$$

$$(\frac{π}{π + π})^2=\frac{9}{49 + 9 + π}$$

$$β (\frac{3}{10})^2=\frac{9}{58 + π}$$

$$β 100 = 58 + π β π = 42$$

Perhatikan kesebangunan $$ΞπΊπ»π$$ dan $$ΞπΌππΉ$$

$$(\frac{π}{π})^2=\frac{4}{9}β\frac{π}{π}=\frac{2}{3}$$

Perhatikan kesebangunan $$Ξπ΄πΊπΉ$$ dan $$ΞπΌππΉ$$

$$(\frac{π}{π + π})^2=\frac{4}{9 + 4 + π}$$

$$β (\frac{2}{5})^2=\frac{4}{13 + π}$$

$$β 25 = 13 + π β π = 12$$

Jadi $$[ABC] = 4 + 9 + 49 + 28 + 42 + 12 = 144$$

6. Given that
$$\frac{1}{\log_x 3}+\frac{1}{\log_y 3} β₯8$$. If the smallest value of $$3^x + 3^y$$ is $$aΓ3^b$$ , and $$HCF(a, 3)οΌ1$$, find $$3aοΌb$$.

(A) 87
(B) 75
(C) 67
(D) 33

7. If $$k$$ is a positive integer and

$$\frac{1}{10!0!}+\frac{1}{8!2!}+\frac{1}{6!4!}+\frac{1}{4!6!}+\frac{1}{2!8!}+\frac{1}{0!10!}=\frac{π}{10!}$$,

find $$π$$.

$$\frac{1}{10!0!}+\frac{1}{8!2!}+\frac{1}{6!4!}+\frac{1}{4!6!}+\frac{1}{2!8!}+\frac{1}{0!10!}=\frac{π}{10!}$$

$$β\frac{10!}{10!0!}+\frac{10!}{8!2!}+\frac{10!}{6!4!}+\frac{10!}{4!6!}+\frac{10!}{2!8!}+\frac{10!} {0!10!}=k$$

$$β1 + 45 + 210 + 210 + 45 + 1 = π$$
$$βπ = 512$$

8. A sequence $${π_π}$$ satisfies $$π_1 = π_2 = 1$$ and $$π_π = π_{πβ1} β π_{πβ2} + π$$ (when $$π β₯ 3$$). Find $$π_{2020}$$

(A) 2020
(B) 2021
(C) 2022
(D) 2024

$$π_1 = 1$$
$$π_2 = 1$$
$$π_3 = π_2 β π_1 + 3 = 1 β 1 + 3 = 3$$
$$π_4 = π_3 β π_2 + 4 = 3 β 1 + 4 = 6$$
$$π_5 = π_4 β π_3 + 5 = 6 β 3 + 5 = 8$$
$$π_6 = π_5 β π_3 + 6 = 8 β 6 + 6 = 8$$
$$π_7 = π_6 β π_5 + 7 = 8 β 8 + 7 = 7$$
$$π_8 = π_7 β π_6 + 8 = 7 β 8 + 8 = 7$$
$$π_9 = π_8 β π_7 + 9 = 7 β 7 + 9 = 9$$
$$π_{10} = π_9 β π_8 + 10 = 9 β 7 + 10 = 12$$
$$π_{11} = π_{10} β π_9 + 11 = 12 β 9 + 11 = 14$$
$$π_{12} = π_{11} β π_{10} + 12 = 14 β 12 + 12 = 14$$
β¦
Enam suku berikutnya mengikuti pola yang sama dengan $$6$$ suku sebelumnya. $$2020\ πππ\ 6 = 4$$, artinya $$π_{2020}$$ mengikuti pola yang sama dengan $$π_4, π_{10}, β¦$$

Jadi nilai dari $$π_{2020} = 2020 + 2 = 2022$$

9. Pick 4 of the 9 balls which are marked from 1 to 9. Find the probability that the sum of the numbers on any two balls is not 10.

(A) $$\frac{36}{126}$$
(B) $$\frac{40}{126}$$
(C) $$\frac{45}{126}$$
(D) $$\frac{48}{126}$$

Dua bola yang berjumlah $$10$$ yaitu $$(1, 9), (2, 8), (3, 7), (4, 6)$$
Peluang ada $$2$$ bola yang berjumlah $$10$$ adalah

$$\frac{4{7\choose 2} β 6}{9\choose 4}=\frac{78}{126}$$

(ket: kurang 6 karena ada 6 himpunan yang terhitung dua kali di $$4{7\choose 2}$$)

Jadi peluang tidak ada $$2$$ bola yang jumlahnya $$10$$ adalah $$1β\frac{78}{126}=\frac{48}{126}$$

10. As in the picture, fill $$0ο½9$$ in the number sentences below. Each contains only a 1-digit number, and a 2-digit number should be filled in a linked or . Find $$AοΌBοΌCοΌDοΌEοΌF$$.

(A) 48
(B) 46
(C) 44
(D) 41