A Math Olympiad Question | Can You Solve This? | System of Equations

[@taniacsibi6879]

Demonstrația mi-a plăcut. Cind s-a ajuns la eg. Xy+ xz+zy=x+y +z. Deoarece x,y,z€z+ rezulta ca x=y=z=1 E=3

[@abcekkdo3749]

E=3

[@Fjfurufjdfjd]

χ=y=z=1 ; χ^2+y^2+ z^2=3 (λυση προφανης)

[@Fjfurufjdfjd]

Το συστημα ειναι ισοδυναμο με χz+y=2z χy+z=2χ yz+χ=2y προκυπτει η σχεση χ+y+z= χz+zy+yχ

[@Fjfurufjdfjd]

Για μενα η πιο δυσκολη ασκηση. Δεν μπορεσα να τη λυσω. Ξοδεψα ενα πακετο κολλες Α4

[@key_board_x]

(1/y) + (1/xz) = 2/xy
(1/z) + (1/xy) = 2/yz
(1/x) + (1/yz) = 2/xz
Let: a = 1/x
Let: b = 1/y
Let: c = 1/z
(1): b + ac = 2ab
(2): c + ab = 2bc
(3): a + bc = 2ac
From (1)
b + ac = 2ab
c = (2ab - b)/a ← equation (3)
From (2)
c + ab = 2bc
c - 2bc = - ab
c.(1 - 2b) = - ab
c = ab/(2b - 1) → recall (3)
ab/(2b - 1) = (2ab - b)/a
a²b = (2b - 1).(2ab - b)
a²b = 4ab² - 2b² - 2ab + b
a²b - 4ab² + 2b² + 2ab - b = 0
b.(a² - 4ab + 2b + 2a - 1) = 0
First case: b = 0 → the system becomes:
b + ac = 2ab → 0 + ac = 0 → ac = 0 → a = 0 or c = 0
c + ab = 2bc → c + 0 = 0 → c = 0
a + bc = 2ac → a + 0 = 2ac → a - 2ac = 0 → a.(1 - 2c) = 0 → and if c = 0 → a = 0
So, this case is not possible.
Second case: (a² - 4ab + 2b + 2a - 1) = 0
a² - 4ab + 2b + 2a - 1 = 0
- 4ab + 2b = - a² - 2a + 1
b.(- 4a + 2) = - a² - 2a + 1
b = (a² + 2a - 1)/(4a - 2) ← equation (4)
From (1): b + ac = 2ab
b - 2ab = - ac
b.(1 - 2a) = - ac
b = ac/(2a - 1) → recall (4)
ac/(2a - 1) = (a² + 2a - 1)/(4a - 2)
ac.(4a - 2) = (2a - 1).(a² + 2a - 1)
4a²c - 2ac = 2a³ + 4a² - 2a - a² - 2a + 1
4a²c - 2ac = 2a³ + 3a² - 4a + 1
c.(4a² - 2a) = 2a³ + 3a² - 4a + 1
c = (2a³ + 3a² - 4a + 1)/(4a² - 2a) ← equation (4)
From (3): a + bc = 2ac
bc = 2ac - a
b = (2ac - a)/c → recall (4)
(2ac - a)/c = (a² + 2a - 1)/(4a - 2)
(2ac - a).(4a - 2) = c.(a² + 2a - 1)
8a²c - 4ac - 4a² + 2a = a²c + 2ac - c
7a²c - 6ac + c = 4a² - 2a
c.(7a² - 6a + 1) = 4a² - 2a
c = (4a² - 2a)/(7a² - 6a + 1) → recall (5)
(4a² - 2a)/(7a² - 6a + 1) = (2a³ + 3a² - 4a + 1)/(4a² - 2a)
(4a² - 2a).(4a² - 2a) = (7a² - 6a + 1).(2a³ + 3a² - 4a + 1)
16a⁴ - 8a³ - 8a³ + 4a² = 14a⁵ + 21a⁴ - 28a³ + 7a² - 12a⁴ - 18a³ + 24a² - 6a + 2a³ + 3a² - 4a + 1
- 14a⁵ + 7a⁴ + 28a³ - 30a² + 10a - 1 = 0
14a⁵ - 7a⁴ - 28a³ + 30a² - 10a + 1 = 0 ← memorize this equation as (6)
14a⁵ - 7a⁴ - 28a³ + 30a² - 10a + 1 = 0 → before to continue, generally, I try to find an obvious root (- 2 or - 1 or 0 or 1 or 2)
14a⁵ - 7a⁴ - 28a³ + 30a² - 10a + 1 = 0 → you can see that (1) is an obvious root, so you can factorize (a - 1)
(a - 1).(14a⁴ + αa³ + βa² + γa - 1) = 0 → you expand
14a⁵ + αa⁴ + βa³ + γa² - a - 14a⁴ - αa³ - βa² - γa + 1 = 0 → you group
14a⁵ + a⁴.(α - 14) + a³.(β - α) + a².(γ - β) - a.(1 + γ) + 1 = 0 → you identify
(α - 14) = - 7 → α = 7
(β - α) = - 28 → β = - 28 + α → β = - 21
(γ - β) = 30 → γ = 30 + β → γ = 9
(1 + γ) = 10 → γ = 9, of course because above
(a - 1).(14a⁴ + αa³ + βa² + γa - 1) = 0 → it becomes
(a - 1).(14a⁴ + 7a³ - 21a² + 9a - 1) = 0 → you can see that (1/2) is an obvious root, so you can factorize [a - (1/2)]
[a - (1/2)].(14a³ + αa² + βa + 2) = 0 → you expand
14a⁴ + αa³ + βa² + 2a - 7a³ - (1/2).αa² - (1/2).βa - 1 = 0 → you group
14a⁴ + a³.(α - 7) + a².[β - (1/2).α] + a.[2 - (1/2).β] - 1 = 0 → you identify
(α - 7) = 7 → α = 14
[β - (1/2).α] = - 21 → β = - 21 + (1/2).α → β = - 14
[2 - (1/2).β] = 9 → (1/2).β = - 7 → β = - 14, of course because above
[a - (1/2)].(14a³ + αa² + βa + 2) = 0 → it becomes
[a - (1/2)].(14a³ + 14a² - 14a + 2) = 0 → don't forget the first factor (a - 1)
(a - 2).[a - (1/2)].(14a³ + 14a² - 14a + 2) = 0 → you can multiply by 2 both sides
We can stop here, because we have 2 values for a (1) & (1/2). Let's keep (1), because it's easier.
When a = 1, the system becomes
(1): b + ac = 2ab → (1'): b + c = 2b
(2): c + ab = 2bc → (2'): c + b = 2bc
(3): a + bc = 2ac → (3'): 1 + bc = 2c
Let's calculate (1') + (2')
(b + c) + (c + b) = 2b + 2bc
2b + 2c = 2b + 2bc
2c = 2bc
c = bc → we know that a, b & c cannot be zero
c/c = b ← so we can see that: b = 1
When b = 1, the system becomes
(1'): b + c = 2b → (1''): 1 + c = 2 → c = 1
(2'): c + b = 2bc → (2''): c + b = 2bc → c + 1 = 2c → c = 1
(3'): 1 + bc = 2c → (3''): 1 + bc = 2c → 1 + c = 2c → c = 1
Conclusion: a = b = c
Recall: a = 1/x = 1 → x = 1
Recall: b = 1/y = 1 → y = 1
Recall: c = 1/z = 1 → z = 1
→ x² + y² + z² = 1² + 1² + 1² = 3
To go further, when a = 1/2, the system becomes
(1): b + ac = 2ab → b + (c/2) = b → c/2 = 0 → c = 0
(2): c + ab = 2bc → c + (b/2) = 2bc → c - 2bc = - b/2 → c.(1 - 2b) = - b/2 → c = b/2.(2b - 1) ← not possible because above
(3): a + bc = 2ac → (1/2) + bc = c → c - bc = 1/2 → c.(1 - b) = 1/2 → c = (1 - b)/2 ← not possible because above
So, we can see that other value for a (excepted 1) doesn't solve the system.

[@Quest3669]

?= 3

[@RajeshKumar-wu7ox]

X=Y=z=1 . E=3

[@ЛюбовьМаркова-б9г]

То же самое!!!

[@SidneiMV]

xz + y = 2z
xy + z = 2x
yz + x = 2y
xy + xz + yz = x + y + z
E = x² + y² + z²
(x + y + z)² = x² + y² + z² + 2(xy + xz + yz)
E = x² + y² + z² = (x + y + z)² - 2(x + y + z)
..... I don't know how to continue .....