solusi soal olimpia mat matika - wardayacollege.com · menurut aturan cosinus: ... wardaya college...
TRANSCRIPT
Wardaya College Departemen Matematika
021-29336036 / 0816950875 1 www.antonwardaya.com
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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 2 www.antonwardaya.com
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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 3 www.antonwardaya.com
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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 4 www.antonwardaya.com
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)
Wardaya College Departemen Matematika
021-29336036 / 0816950875 5 www.antonwardaya.com
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)
Wardaya College Departemen Matematika
021-29336036 / 0816950875 6 www.antonwardaya.com
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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 7 www.antonwardaya.com
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
25𝑠𝑖𝑛2𝜃 +
16
25𝑐𝑜𝑠2𝜃 +
24
25sin 𝜃 cos 𝜃 = 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃
16
25𝑠𝑖𝑛2𝜃 +
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)
Wardaya College Departemen Matematika
021-29336036 / 0816950875 8 www.antonwardaya.com
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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 9 www.antonwardaya.com
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)
Wardaya College Departemen Matematika
021-29336036 / 0816950875 10 www.antonwardaya.com
=
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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 11 www.antonwardaya.com
�̅�′ =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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 12 www.antonwardaya.com
∫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
Wardaya College Departemen Matematika
021-29336036 / 0816950875 13 www.antonwardaya.com
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)