# 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

$$\frac{1}{7}＋\frac{1}{9}$$?

(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 $$(𝑥^2 − 𝑥 − 1)^{5𝑥^2+9𝑥−2} = 1$$?

• Syarat 1
$$𝑥^2 − 𝑥 − 1 = 1$$
$$𝑥^2 − 𝑥 − 2 = 0$$
$$(𝑥 − 2)(𝑥 + 1) = 0$$
$$𝑥 = 2 ∨ 𝑥 = −1$$
• Syarat 2
$$5𝑥^2 + 9𝑥 − 2 = 0$$
$$(5𝑥 − 1)(𝑥 + 2) = 0$$
$$𝑥 =\frac{1}{5}(𝑇𝑀) ∨ 𝑥 = −2$$
Nilai $$𝑥 = −2$$ memenuhi karena $$𝑥^2 − 𝑥 − 1 ≠ 0$$
• Syarat 3
$$𝑥^2 − 𝑥 − 1 = −1$$
$$𝑥^2 − 𝑥 = 0$$
$$𝑥(𝑥 − 1) = 0$$
$$𝑥 = 0 ∨ 𝑥 = 1$$
nilai $$𝑥 = 0 ∨ 𝑥 = 1$$ memenuhi karena $$5𝑥^2 + 9𝑥 − 2$$ bernilai genap.

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

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

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

5. Look at the picture. $$𝑃$$ is a point in $$Δ𝐴𝐵𝐶$$. Draw $$3$$ lines which pass through $$P$$ and are parallel to each side of $$Δ𝐴𝐵𝐶$$. If the areas of the small triangles $$𝑡_1, 𝑡_2$$ and $$𝑡_3$$ are $$4, 9$$, and $$49$$, respectively, find the area of $$Δ𝐴𝐵𝐶$$.

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

Perhatikan kesebangunan $$Δ𝐷𝑃𝐸$$ dan
$$Δ𝐼𝑃𝐹$$

$$(\frac{𝑛}{𝑚})^2=\frac{4}{49}⇒\frac{𝑛}{𝑚}=\frac{2}{7}$$

Perhatikan kesebangunan $$Δ𝐼𝐷𝐶$$ dan $$Δ𝐼𝑃𝐹$$

$$(\frac{𝑛}{𝑚 + 𝑛})^2=\frac{4}{49 + 4 + 𝑋}$$

$$⇒(\frac{2}{9})^2=\frac{4}{53 + 𝑋}$$

$$⇒ 81 = 53 + 𝑋 ⇒ 𝑋 = 28$$

Selanjutnya, Perhatikan kesebangunan $$Δ𝐷𝑃𝐸$$ dan $$Δ𝐻𝐺𝑃$$

$$(\frac{𝑟}{𝑚})^2=\frac{9}{49}⇒\frac{𝑟}{𝑚}=\frac{3}{7}$$

Perhatikan kesebangunan $$Δ𝐻𝐵𝐸$$ dan $$Δ𝐻𝐺𝑃$$

$$(\frac{𝑟}{𝑚 + 𝑟})^2=\frac{9}{49 + 9 + 𝑌}$$

$$⇒ (\frac{3}{10})^2=\frac{9}{58 + 𝑌}$$

$$⇒ 100 = 58 + 𝑌 ⇒ 𝑌 = 42$$

Perhatikan kesebangunan $$Δ𝐺𝐻𝑃$$ dan $$Δ𝐼𝑃𝐹$$

$$(\frac{𝑎}{𝑏})^2=\frac{4}{9}⇒\frac{𝑎}{𝑏}=\frac{2}{3}$$

Perhatikan kesebangunan $$Δ𝐴𝐺𝐹$$ dan $$Δ𝐼𝑃𝐹$$

$$(\frac{𝑎}{𝑎 + 𝑏})^2=\frac{4}{9 + 4 + 𝑍}$$

$$⇒ (\frac{2}{5})^2=\frac{4}{13 + 𝑍}$$

$$⇒ 25 = 13 + 𝑍 ⇒ 𝑍 = 12$$

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

6. Given that
$$\frac{1}{\log_x 3}+\frac{1}{\log_y 3} ≥8$$. If the smallest value of $$3^x + 3^y$$ is $$a×3^b$$ , and $$HCF(a, 3)＝1$$, find $$3a＋b$$.

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

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

$$\frac{1}{10!0!}+\frac{1}{8!2!}+\frac{1}{6!4!}+\frac{1}{4!6!}+\frac{1}{2!8!}+\frac{1}{0!10!}=\frac{𝑘}{10!}$$,

find $$𝑘$$.

$$\frac{1}{10!0!}+\frac{1}{8!2!}+\frac{1}{6!4!}+\frac{1}{4!6!}+\frac{1}{2!8!}+\frac{1}{0!10!}=\frac{𝑘}{10!}$$

$$⇒\frac{10!}{10!0!}+\frac{10!}{8!2!}+\frac{10!}{6!4!}+\frac{10!}{4!6!}+\frac{10!}{2!8!}+\frac{10!} {0!10!}=k$$

$$⇒1 + 45 + 210 + 210 + 45 + 1 = 𝑘$$
$$⇒𝑘 = 512$$

8. A sequence $${𝑎_𝑛}$$ satisfies $$𝑎_1 = 𝑎_2 = 1$$ and $$𝑎_𝑛 = 𝑎_{𝑛−1} − 𝑎_{𝑛−2} + 𝑛$$ (when $$𝑛 ≥ 3$$). Find $$𝑎_{2020}$$

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

$$𝑎_1 = 1$$
$$𝑎_2 = 1$$
$$𝑎_3 = 𝑎_2 − 𝑎_1 + 3 = 1 − 1 + 3 = 3$$
$$𝑎_4 = 𝑎_3 − 𝑎_2 + 4 = 3 − 1 + 4 = 6$$
$$𝑎_5 = 𝑎_4 − 𝑎_3 + 5 = 6 − 3 + 5 = 8$$
$$𝑎_6 = 𝑎_5 − 𝑎_3 + 6 = 8 − 6 + 6 = 8$$
$$𝑎_7 = 𝑎_6 − 𝑎_5 + 7 = 8 − 8 + 7 = 7$$
$$𝑎_8 = 𝑎_7 − 𝑎_6 + 8 = 7 − 8 + 8 = 7$$
$$𝑎_9 = 𝑎_8 − 𝑎_7 + 9 = 7 − 7 + 9 = 9$$
$$𝑎_{10} = 𝑎_9 − 𝑎_8 + 10 = 9 − 7 + 10 = 12$$
$$𝑎_{11} = 𝑎_{10} − 𝑎_9 + 11 = 12 − 9 + 11 = 14$$
$$𝑎_{12} = 𝑎_{11} − 𝑎_{10} + 12 = 14 − 12 + 12 = 14$$

Enam suku berikutnya mengikuti pola yang sama dengan $$6$$ suku sebelumnya. $$2020\ 𝑚𝑜𝑑\ 6 = 4$$, artinya $$𝑎_{2020}$$ mengikuti pola yang sama dengan $$𝑎_4, 𝑎_{10}, …$$

Jadi nilai dari $$𝑎_{2020} = 2020 + 2 = 2022$$

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

$$\frac{4{7\choose 2} − 6}{9\choose 4}=\frac{78}{126}$$

(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 $$1−\frac{78}{126}=\frac{48}{126}$$

10. As in the picture, fill $$0～9$$ 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 $$A＋B＋C＋D＋E＋F$$.

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