KTO Matematika

Kontes Terbuka Olimpiade.org Desember 2015: Bagian A

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luas ABCD=LuaS ABE  + l.ADCE                         =5 luas ABE                           2/5 AB x CD= AB x BE                    jadi BE =40

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No.6 ada dua bangun yakni.segitiga ABC dan Segitiga ADC.pencarian sisi pake pytagoras.AB=3 .Bc= 4 dan Ac=5 dan CD= 13 serta AD= 12 maka luasnya( AB x BC plus AC x AD)/2=6 + 30=36

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Aku dapat ide untuk ngerjain soal nomor $14$. Cari dulu nilai $n$ terkecil sehingga $99^2=3^{4}11^{2}|10^{n}-1$. Pakai LTE, nanti dapatnya $n=2\times 9\times 11=198$. Terus tinggal dibagi. 

Edited by AlifAqsha

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Untuk yang nomor 10,

P(x) = ax³ + bx² + cx + d

P(0) = 0+0+0+d = 10

Maka d = 10.

P(1) = a+b+c+d = 20

Maka a+b+c=10 ....(1)

P(2) = 8a + 4b + 2c + d = 30

8a + 4b + 2c = 20 (bagi 2)

4a + 2b + c = 10 ....(2)

Persamaan (2) kurang (1)

(4a + 2b +c)- (a+b+c) = 0

3a + b = 0

Maka b = -3a.

Substitusi ke (1)

a + (-3a) + c = 10

Didapat c = 10 + 2a..

Berarti, P(x) = ax³ + (-3)x² + (10+2a)x + 10

Sejauh ini, baru itu yg saya dapat.. Ada yang bisa bantu lanjutkan?

Kalo coba-coba, saya dpt a=1, b=-3, c=12, d=10, jadi

P(x) = x³ - 3x² + 12x + 10

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Spoiler

 

$10a+b= \frac{9}{2} (10b+a)$

$10a+b = \frac{9}{2} \cdot 10b + \frac{9}{2}a$

$10a+b = 45b + 4\frac{1}{2}a$

$5\frac{1}{2}a = 44b$

$a = 44b \cdot \frac{2}{11} $

$a = 8b$

Karena batasannya harus $0 <a, b < 10$ , maka nilai $a = 1$ dan $b = 8$

Spoiler

 

Edited by Wildan Bagus W

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