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Wardaya College Departemen Matematika

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A B

C

a

a a

Φ

D

E

SOLUSI SOAL OLIMPIADE MATEMATIKA Persiapan Olimpiade Sains Provinsi dan Nasional Tingkat SMA

Departemen Matematika - Wardaya College

MMXVIII-VIII

1.

Dimisalkan; AB = AC = BC = a

BD = p AD = a - p

CE = q AE = a – q

ED + BC = BD + CE

ED + a = p + q

ED = p + q – a

Menurut aturan Cosinus:

𝐸𝐷2 = 𝐴𝐸2 + 𝐴𝐷2 − 2. 𝐴𝐸. 𝐴𝐷. cos 60

(𝑝 + 𝑞 − 𝑎)2 = (𝑎 − 𝑞)2 + (𝑎 − 𝑝)2 − (𝑎 − 𝑞)(𝑎 − 𝑝)

𝑎 =3𝑝𝑞

𝑝 + 𝑞

Sehingga;

Jadi;

𝐴𝐷 = 𝑎 − 𝑝 =𝑝(2𝑞 − 𝑝)

𝑝 + 𝑞

𝐴𝐸 = 𝑎 − 𝑞 =𝑞(2𝑝 − 𝑞)

𝑝 + 𝑞

𝐴𝐷

𝐷𝐵+

𝐴𝐸

𝐸𝐶=

𝑝(2𝑞 − 𝑝)

𝑝(𝑝 + 𝑞)+

𝑞(2𝑝 − 𝑞)

𝑞(𝑝 + 𝑞)

𝐴𝐷

𝐷𝐵+

𝐴𝐸

𝐸𝐶=

𝑝 + 𝑞

𝑝 + 𝑞= 1



2. Misalkan titik P (a, a)

𝑥2 + 𝑦2 = 1 𝑎2 + 𝑎2 = 1

𝑎 = −1

√2

Maka: 𝑃 (−1

√2 , −

1

√2)

Luas BOC = 𝜋

4

Luas ΔPOC = Luas ΔPOB = 2 × 1

2 × 1 ×

1

√2

= 1

2√2

Luas arsiran = 𝜋

4+

1

2√2

3. C = 3B ; A = 1800 – 4B

Misal: jari-jari lingkaran luar ABC, R = 1

𝑎 = 2 sin 4𝐵

𝑐 = 2 sin 3𝐵 𝑐 =10

3𝑎

10 sin 4𝐵 = 3 sin 3𝐵

40 sin 𝐵 cos 𝐵 cos 2𝐵 = −12𝑠𝑖𝑛3𝐵 + 9 sin 𝐵

sin 𝐵 ≠ 0

Misal: 𝑥 = cos 𝐵

80𝑥3 − 12𝑥2 − 40𝑥 + 3 = 0

(4𝑥 − 3)(20𝑥2 + 12𝑥 − 1) = 0

𝑥 =3

4

atau 20𝑥2 + 12𝑥 − 1 = 0

(𝑇𝑀) 180 − 4𝐵 > 0

𝐵 < 45



𝑥 = cos 𝐵 =3

4

cos 𝐶 = cos 3𝐵 = −9

16

cos 𝐴 = − cos 4𝐵 =31

32

cos 𝐴 + cos 𝐵 + cos 𝐶 =3

4+ (−

9

16) +

31

32=

37

32

4. 𝑛2 − 11𝑛 + 63 = 𝑘2

4𝑛2 − 44𝑛 + 252 = 4𝑘2

(2𝑛 − 11)2 + 131 = (2𝑘)2

(2𝑘)2 − (2𝑛 − 11)2 = 131

(2𝑘 + 2𝑛 − 11)(2𝑘 − 2𝑛 + 11) = 131

131 adalah bilangan prima

Kasus 1: 2𝑘 + 2𝑛 − 11 = 131

2𝑘 − 2𝑛 + 11 = 1

4𝑛 − 22 = 130 𝑛 = 38

Kasus 2: 2𝑘 + 2𝑛 − 11 = 1

2𝑘 − 2𝑛 + 11 = 131

4𝑛 − 22 = −130 𝑛 = −27

𝑛 = 38, −27



5. AM – GM

1 + 4𝑥2 ≥ 2√4𝑥2 = 4𝑥

𝑦 ≤4𝑥2

4𝑥= 𝑥

𝐷𝑒𝑛𝑔𝑎𝑛 𝑐𝑎𝑟𝑎 𝑦𝑎𝑛𝑔 𝑠𝑎𝑚𝑎: 𝑧 ≤4𝑦2

4𝑦= 𝑦

𝑥 ≤4𝑧2

4𝑧= 𝑧

6. ∠𝐴𝑋𝑌 = 900 (𝑘𝑎𝑟𝑒𝑛𝑎 𝐴𝑋2 + 𝑋𝑌2 = 𝐴𝑌2)

∠𝑌𝑋𝐶 + ∠𝐴𝑋𝐵 = 900

∠𝑋𝐴𝐵 + ∠𝐴𝑋𝐵 = 900

∠𝑌𝑋𝐶 = ∠𝑋𝐴𝐵

∴ 𝛥𝐴𝐵𝑋 ~ 𝛥𝑋𝐶𝑌

𝐴𝐵

4=

𝑋𝐶

3 → 𝑋𝐶 =

3

4𝐴𝐵

𝐵𝑋 = 𝐵𝐶 − 𝑋𝐶 = 𝐴𝐵 −3

4𝐴𝐵 =

1

4𝐴𝐵

𝑃𝐻𝑌𝑇𝐴𝐺𝑂𝑅𝐴𝑆: 𝐴𝐵2 + (1

4𝐴𝐵2) = 42

𝐴𝐵 =16

17√17

(1

2,1

2,1

2) 𝑑𝑎𝑛 (0,0,0)



7. (𝑥√𝑥)𝑥

= 𝑥3

2𝑥

𝑥𝑥√𝑥 = (𝑥√𝑥)𝑥

𝑥𝑥√𝑥 = 𝑥3

2𝑥

𝑥√𝑥 =3

2𝑥

𝑥 =9

4 dan 1x , 0x tidak memenuhi

8.

𝑇𝑖𝑡𝑖𝑘 𝐴: 𝑏 = 2𝑎2 + 4𝑎 − 2

𝑇𝑖𝑡𝑖𝑘 𝐵: − 𝑏 = (−2𝑎)2 + (−4𝑎) − 2

0 = 4𝑎2 − 4

𝑎 = ± 1

Untuk a = 1 ; b = 1

Untuk a = -1 ; b = -1

𝐴𝐵̅̅ ̅̅ = √(1 + 1)2 + (4 + 4)2 = √68 = 2√17

9. 𝑃𝑒𝑟𝑝𝑜𝑡𝑜𝑛𝑔𝑎𝑛 𝑘𝑒𝑑𝑢𝑎 𝑘𝑢𝑟𝑣𝑎 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎ℎ: 𝑥 = ±√3

2 𝑑𝑎𝑛 𝑦 =

3

2

0

A (a,b)

B (-a,-b)



𝑉 = 2 × 2𝜋 ∫ (2√1 − 𝑥2 − 1) 𝑑𝑥

32

0

𝑉 =

10. √3𝑥2 − 18𝑥 + 52 = 3(𝑥 − 3)2 + 25

√2𝑥2 − 12𝑥 + 162 = 2(𝑥 − 3)2 + 144

√−𝑥2 + 6𝑥 + 280 = −(𝑥 − 3)2 + 289

𝑅𝑢𝑎𝑠 𝐾𝑖𝑟𝑖 √25 + √144 = 17

𝑅𝑢𝑎𝑠 𝐾𝑎𝑛𝑎𝑛 √289 = 17

∴ 𝑥 = 3

11. Bola-bola yang ada di dalam kotak:

6 bola bernomor ganjil dan 5 bola bernomor genap

Jumlah Ganjil: 1 ganjil dan 5 genap

3 ganjil dan 3 genap

5 ganjil dan 2 genap

6C1 . 5C5 + 6C3 . 5C3 + 6C5 . 5C1 = 236

11C6 = 432

PELUANG = 236

432=

118

231

12. 𝑥2 + 2𝑥(𝑦 + 1) + 3(𝑦 + 1)2 + 1 = (𝑥 + 𝑦 + 1)2 + 2(𝑦 + 1)2 + 1

𝑁𝑖𝑙𝑎𝑖 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑘𝑒𝑡𝑖𝑘𝑎: 𝑥 + 𝑦 + 1 = 0 𝑥 = 0



𝑦 + 1 = 0 𝑦 = −1

∴ 𝑁𝑖𝑙𝑎𝑖 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = 1

13. 3

5sin 𝜃 +

4

5cos 𝜃 = 1 ; 𝑘𝑒𝑑𝑢𝑎 𝑟𝑢𝑎𝑠 𝑑𝑖𝑘𝑢𝑎𝑑𝑟𝑎𝑡𝑘𝑎𝑛 𝑚𝑒𝑛𝑗𝑎𝑑𝑖:

9

25𝑠𝑖𝑛2𝜃 +

16

25𝑐𝑜𝑠2𝜃 +

24

25sin 𝜃 cos 𝜃 = 1

9


16

25𝑐𝑜𝑠2𝜃 +

24

25sin 𝜃 cos 𝜃 = 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃

16


9

25𝑐𝑜𝑠2𝜃 −

24

25sin 𝜃 cos 𝜃 = 0

(4

5sin 𝜃 −

3

5cos 𝜃)

2

= 0

tan 𝜃 =3

4

14. Dimisalkan koordinat titik lattice adalah (m, n)

𝑥2 + 𝑦2 = 25

𝑚2 + 𝑛2 = 25

m dan n adalah bilangan bulat;

𝑚2 ≥ 0

𝑛2 ∈ {0, 1,4,9,16,25}

𝑈𝑛𝑡𝑢𝑘 𝑛2 = 0 → 𝑚2 = 25 ; 𝑛 = 0 𝑑𝑎𝑛 𝑚 = ±5

(0,5) & (0, −5)



𝑈𝑛𝑡𝑢𝑘 𝑛2 = 1 → 𝑚2 = 24 ; 𝑛 = ±1 𝑑𝑎𝑛 𝑚 = ±√24 ∴ 𝑏𝑢𝑘𝑎𝑛 𝑡𝑖𝑡𝑖𝑘 𝑙𝑎𝑡𝑡𝑖𝑐𝑒

𝑈𝑛𝑡𝑢𝑘 𝑛2 = 4 → 𝑚2 = 20 ; 𝑛 = ±2 𝑑𝑎𝑛 𝑚 = ±√20 ∴ 𝑏𝑢𝑘𝑎𝑛 𝑡𝑖𝑡𝑖𝑘 𝑙𝑎𝑡𝑡𝑖𝑐𝑒

𝑈𝑛𝑡𝑢𝑘 𝑛2 = 9 → 𝑚2 = 16 ; 𝑛 = ±3 𝑑𝑎𝑛 𝑚 = ±4

(3,4); (3, −4); (−3,4); (−3, −4)

𝑈𝑛𝑡𝑢𝑘 𝑛2 = 16 → 𝑚2 = 9 ; 𝑛 = ±4 𝑑𝑎𝑛 𝑚 = ±3

(4,3); (4, −3); (−4,3); (−4, −3)

𝑈𝑛𝑡𝑢𝑘 𝑛2 = 25 → 𝑚2 = 0 ; 𝑛 = ±5 𝑑𝑎𝑛 𝑚 = 0

(5,0) & (−5,0)

∴ 𝐴𝑑𝑎 12 𝑡𝑖𝑡𝑖𝑘 𝑙𝑎𝑡𝑡𝑖𝑐𝑒

15. n = 48

16. ∏ (𝑘 + 1)2010

𝑘=2 ∏ (𝑘 − 1)2010𝑘=2

∏ (𝑘)2010𝑘=2 ∏ (𝑘)2010

𝑘=2

=∏ (𝑘)2011

𝑘=3 ∏ (𝑘)2009𝑘=1

∏ (𝑘)2010𝑘=2 ∏ (𝑘)2010

𝑘=2

=2011

2.2010=

2011

4020

17. a = 133333

𝑎 + (𝑎 + 1) + ⋯ + (𝑎 + 400000) = 400001𝑎 + 200000(400001)

= (400001)(333333) = 13333533333

18. 𝑥log 𝑥 + 10𝑙𝑜𝑔 –log 𝑥 − 1 = 10

𝑎 = 10 𝑎 = 1

𝑥log 𝑥 = 10 𝑥log 𝑥 = 1



𝑙𝑜𝑔2𝑥 = 10 𝑙𝑜𝑔2𝑥 = 0

log 𝑥 = 1 & log 𝑥 = −1 log 𝑥 = 0

𝑥 = 10 & 𝑥 =1

10 𝑥 = 1

𝑥1 ∙ 𝑥2 ∙ 𝑥3 = 10 ∙1

10∙ 1 = 1

19. 2

2+

3

4=

7

4

20. Misal: 𝑡𝑎𝑛−1 1

5= 𝑥 𝑡𝑎𝑛−1 1

99= 𝑧

𝑡𝑎𝑛−1 1

70= 𝑦

tan 4𝑥 =2 tan 2𝑥

1 − 𝑡𝑎𝑛22𝑥

=2 (

2 tan 𝑥1 − 𝑡𝑎𝑛2𝑥

)

1 − (2 tan 𝑥

1 − 𝑡𝑎𝑛2𝑥)

2

=4 tan 𝑥(1 − 𝑡𝑎𝑛2𝑥)

(1 − 𝑡𝑎𝑛2𝑥)2 − 4𝑡𝑎𝑛2𝑥

=4 (

15

) (1 −1

25)

(1 −1

25)

2

− 4(25)

tan 4𝑥 =120

119

tan(𝑦 − 𝑧) =tan 𝑦 − tan 𝑧

1 + tan 𝑦 ∙ tan 𝑧=

170

−1

99

1 +1

70 ∙1

99

=1

239

tan(4𝑥 − 𝑦 + 𝑧) = tan[4𝑥 − (𝑦 − 𝑧)]

=tan 4𝑥 − tan(𝑦 − 𝑧)

1 + tan 4𝑥 ∙ tan(𝑦 − 𝑧)



=

120119 −

1239

1 +120119 ∙

1239

= 1

21. 𝑥2 + (7𝑥 + 5)2 = 1

𝑥2 + 49𝑥2 + 70𝑥 + 25 − 1 = 0

50𝑥2 + 70𝑥 + 24 = 0

25𝑥2 + 35𝑥 + 12 = 0

(5𝑥 + 4)(5𝑥 + 3) = 0

𝑥 = −4

5 𝑑𝑎𝑛 𝑥 = −

3

5

(−4

5 , −

3

5) (−

3

5 ,

4

5)

𝐴𝐵 = √2

𝐴𝑂 = 𝐵𝑂 = 1

∠𝐴𝑂𝐵 = 90° (Phytagoras)

22. (𝑖) 𝑥2 − 3𝑥 + 1 = 1

𝑥(𝑥 − 3) = 0 → 𝑥 = 0 𝑑𝑎𝑛 𝑥 = 3

(𝑖𝑖) 𝑥2 − 3𝑥 + 1 = −1

(𝑥 − 1)(𝑥 − 2) = 0 → 𝑥 = 1 𝑑𝑎𝑛 𝑥 = 2

𝑥 + 1 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑒𝑛𝑎𝑝; 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑎𝑑𝑎𝑙𝑎ℎ 𝑥 = 1

(𝑖𝑖𝑖) 𝑥 + 1 = 0 → 𝑥 = −1

23. 𝑥1 − 𝑎

𝑏,𝑥2 − 𝑎

𝑏, … ,

𝑥𝑛 − 𝑎

𝑏

�̅�′ =�̅� − 𝑎

𝑏 𝐽𝑎𝑛𝑔𝑘𝑎𝑢𝑎𝑛′ =

𝐽

𝑏



�̅�′ =12 − 𝑎

𝑏 3 =

6

𝑏 → 𝑏 = 2

2 =12 − 𝑎

2

𝑎 = 8

∴ 𝑎 + 𝑏 = 8 + 2 = 10

24. 𝑥 = 2 → 𝑓(2) + 2𝑓(2010) = 2010 |× 2|

𝑥 = 2010 → 𝑓(2010) + 2𝑓(2) = 2 |× 1|

2𝑓(2) + 4𝑓(2010) = 4020

2𝑓(2) + 𝑓(2010) = 2

3𝑓(2010) = 4018

𝑓(2010) =4018

3

25. |�̅� + �̅�|2

= |�̅�|2 + |�̅�|2

+ |�̅�||�̅�| cos 𝜃

8 = 16 + 4 + 2 ∙ 4 ∙ 2 ∙ cos 𝜃

cos 𝜃 = −3

4 → sin 𝜃 =

√7

4

|𝑎 × 𝑏| = |�̅�||�̅�| sin 𝜃

= 4 ∙ 2 ∙√7

4

|𝑎 × 𝑏| = 2√7

26. |𝑥| = {−𝑥, 𝑥 < 0

𝑥, 𝑥 ≥ 0



∫𝑥2

|𝑥|

5

−4

𝑑𝑥 = ∫𝑥2

−𝑥

0

−4

𝑑𝑥 + ∫𝑥2

𝑥

5

0

𝑑𝑥

= ∫ −𝑥

0

−4

𝑑𝑥 + ∫ 𝑥

5

0

𝑑𝑥

= −1

2𝑥2|

0

−4 +

1

2𝑥2|

5

0

= −(0 − 8) + (1

2× 25 − 0)

= 8 +25

2=

41

2

27. 𝑑

𝑑𝑥sin 𝑥 = cos 𝑥

𝑑2

𝑑𝑥2sin 𝑥 = − sin 𝑥

𝑑3

𝑑𝑥3 sin 𝑥 = − cos 𝑥

𝑑4

𝑑𝑥4 sin 𝑥 = sin 𝑥

Asumsi : 2 . 1 = - sin

2 . 2 = sin

2 . 3 = - sin

2 . 4 = sin

2010 = 2 . 1005 = - sin

28. 𝑎 + 𝑏 = 16 − 𝑐

𝑎2 + 𝑏2 = 160 − 𝑐2



2𝑎𝑏 = (𝑎 + 𝑏)2 − (𝑎2 + 𝑏2)

2𝑎𝑏 = (16 − 𝑐)2 − (160 − 𝑐)

2𝑎𝑏 = 2𝑐2 − 32𝑐 + 96

𝑎𝑏 = 𝑐2 − 16𝑐 + 48

𝑎𝑐𝑏 = 𝑐3 − 32𝑐2 + 48𝑐

Agar didapat nilai maksimum, maka turunan pertama harus = 0

3𝑐2 − 64𝑐 + 48 = 0

𝑐 =16±4√7

3

𝑢𝑛𝑡𝑢𝑘 𝑐 = 16+4√73

𝑚𝑎𝑘𝑎 𝑛𝑖𝑙𝑎𝑖 𝑎𝑏𝑐 < 0

𝑚𝑎𝑘𝑎 𝑑𝑖𝑎𝑚𝑏𝑖𝑙 𝑐 = 16−4√73

∴ 𝑎𝑏𝑐 = 12827

(7√7 − 10)

29. (1 + 𝑖)2 = 2𝑖 (1 + 𝑖)3 = 2𝑖 − 2

𝑓(1 + 𝑖) = 3(1 + 𝑖)3 − 2(1 + 𝑖)2 + (1 + 𝑖) − 3

𝑓(1 + 𝑖) = 3(2𝑖 − 2) − 2(2𝑖) + (1 + 𝑖) − 3

𝑓(1 + 𝑖) = −8 + 3𝑖

30. 318 − 218 = (39 − 29)(39 + 29)

= (33 − 23)(36+33 ∙ 23+26)(33 + 23)(36−33 ∙ 23+26)

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