# Problems And Solutions SEAMO PAPER C 2021

The Southeast Asia Mathematical Olympiad (SEAMO) is an international Math Olympiad competition that originated in Singapore and was founded by Mr. Terry Chew in 2016 in 8 Southeast Asian Countries. Since then, it is growing its popularity around the world. In 2019 it was recognized by 18 countries. In 2020 total number of participating countries increased to 22, including students from Indonesia, Brazil, China, Newzealand, and Taiwan students enrolled in SEAMO 2022-23.

Problem and Solution SEAMO 2021 paper C. Soal ini bersumber dari seamo-official.org

1. Find the missing number.

(A) 20
(B) 18
(C) 24
(D) 52
(E) 48

• (3 + 2) × (3 − 2) = 5
• (5 + 3) × (5 − 3) = 16
• (6 + 4) × (6 − 4) = 20

2. 3-digit numbers are formed by selecting 3 numbers 0, 2, 3, 5 and 8, each digit used exactly once each time. What is the probability that the number is even?
(A) $$\frac{1}{4}$$
(B) $$\frac{2}{5}$$
(C) $$\frac{4}{7}$$
(D) $$\frac{2}{3}$$
(E) $$\frac{5}{8}$$

S = {banyaknya kemungkinan bilangan 3 digit yang terbentuk}
$$𝑛(𝑆) = 4 × 4 × 3 = 48$$
A = {banyaknya bilangan genap 3 angka yang terbentuk dari angka 0,2,3,5 dan 8 dan tiap angka digunakan sekali}
* Angka satuan 0
$$3×4×1$$
* Angka puluhan 0
$$3×1×2$$
* Tanpa memuat angka 0
$$2×3×2$$
Jadi peluang A adalah $$\frac{𝑛(𝐴)}{𝑛(𝑆)}=\frac{12+6+12}{48}=\frac{30}{48}=\frac{5}{8}$$

3. $$𝐵𝐷𝐸𝐹$$ is a square embedded in a rightangled triangle $$𝐴𝐵𝐶$$. Given that $$𝐴𝐸 = 10$$ 𝑐𝑚 and $$𝐸𝐶 = 15$$ 𝑐𝑚 , find, in 𝑐𝑚² , the area of the shaded region.

(A) 65
(B) 70
(C) 75
(D) 80
(E) 120

Karena $$Δ𝐸𝐷𝐶 ≈ Δ𝐴𝐵𝐶$$, maka
$$\frac{𝑥}{𝐴𝐵}=\frac{15}{25}=\frac{3}{5}⟹ 𝐴𝐵 =\frac{5}{3}𝑥$$

$$𝐴𝐹 = 𝐴𝐵 − 𝐹𝐵 =\frac{5}{3}𝑥 − 𝑥 =\frac{2}{3}𝑥$$

Dengan menggunakan rumus pythagoras pada segitiga $$AFE$$
$$(\frac{2}{3}x)^2 + 𝑥^2 = 10^2$$
$$⇒\frac{4}{9}𝑥^2 +\frac{9}{9}x^2= 100$$
$$⇒\frac{13}{9}𝑥^2 = 100 ⇒𝑥^2 =\frac{900}{13}$$

Karena $$Δ𝐸𝐷𝐶 ≈ Δ𝐴𝐹𝐸$$, maka
$$\frac{𝑦}{𝑥}=\frac{15}{10}=\frac{3}{2}⇒𝑦 =\frac{3}{2}𝑥$$

\begin{align} Luas\; arsiran & = [AFE] + [ EDC]\\ &=\frac{1}{2}(𝑥)(\frac{2}{3}𝑥) +\frac{1}{2}(𝑥)(\frac{3}{2}𝑥)\\ &=\frac{1}{3}𝑥^2 +\frac{3}{4}𝑥^2\\ &=\frac{13}{12}𝑥^2\\ &=\frac{13}{12}(\frac{900}{13})\\ &= 75\;𝑐𝑚^2\\ \end{align}

4. Given that $$𝑚 ⊝ 𝑛 = 𝑚^2 − 𝑛^2$$.
Evaluate $$(2021 ⊝ 2020) ⊝ (2020 ⊝ 2019)$$.
(A) 16016
(B) 16154
(C) 16158
(D) 16160
(E) 16162

$$𝑚 ⊝ 𝑛 = 𝑚^2 − 𝑛^2 = (𝑚 + 𝑛)(𝑚 − 𝑛)$$
$$(2021 ⊝ 2020) ⊝ (2020 ⊝ 2019)$$
$$= (2021 + 2020)(2021 − 2020) ⊝ (2020 + 2019)(2020 − 2019)$$
$$= (4041)(1) ⊝ (4039)(1)$$
$$= 4041 ⊝ 4039$$
$$= (4041 + 4039)(4041 − 4039)$$
$$= (8080)(2) = 16160$$

5. The profit from the sale of an item is $180 if it is sold at a discounted price of 10%. The loss is$240 if it is sold at a discount of 20%. What is the cost price of that item?
(A) $3400 (B)$3600
(C) $3800 (D)$4000
(E) \$4200

Misalkan Modal = $$M$$ dan harga jual = $$J$$
$$90\%J – M = 180$$
$$M – 80\%J = 240$$
_______________________ +
$$10\%J = 420$$
$$J = 4200$$
Subtitusi nilai $$J$$ ke persamaan $$M – 80%J = 240$$ diperoleh
$$M = 240 + 0.8(4200)=3600$$

6. In the figure shown, ∠𝐴 = 72°. The bisector of ∠𝐴𝐵𝐶 and the bisector of ∠𝐴𝐶𝐸 intersect at 𝐷. Find ∠𝐷.

(A) 18°
(B) 24°
(C) 30°
(D) 32°
(E) 36°

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