A USAMO Problem | Video Response To @PrimeNewtons

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❤️❤️❤️ This video is a video response to @PrimeNewtons
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❤️ x + log(x) = 11

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PLAYLISTS 🎵 :

[MA-bmjz]

You can consider the polynomial which has x y z as roots, and use viete's formulas from there, this is the standard method when dealing with symetric sistems of equations

[pwmiles]

Denote
P1=x+y+z=3
P2=x^2+y^2+z^2=3
P3=x^3+y^3+z^3=3

Some elementary symmetric polynomials are
E1=x+y+z=P1
E2=xy+yz+zx
E3=xyz

You can use two Newton-Girard identities
P1^2 = P2 + 2E2
P3 - 3E3 = P1(P2 - E2)

Substituting known values
9=3+2E2, E2=3
3-3E3=0, E3=1

-E1, E2, -E3 are the Vieta formulas for coefficients of x^2, x and 1 in a cubic with roots x, y and z
x^3-3x^2+3x-1=0
(x-1)^3=0
So the roots x, y and z are all 1

[nasrullahhusnan]

Just by inspection (x, y, z)=(1, 1, 1)
Algebraicly we can use Newton, -Girard formula.

[armacham]

I solved it a little differently.

I started with "suppose x = y" and then replace y with x in all 3 equations, from there it's not hard to solve (use the first equation to solve for z, then square both sides, substitute that into the second equation, and you see that x=1, (1, 1, 1) is the only solution under that circumstance

From there, all remaining solutions would have to have x != y != z !=x. We already found all the solutions when any of the values are equal so we know they can't be equal. WLOG we can say:


x is smallest, then y, then z

so x = a
y = a + b
z = a + b + c
where a can be any real number but b and c must be positive

substitute that into the first two equations and from there it's not too hard to solve. You will see that b and c must both be equal to zero. but neither of them can be equal to zero, it's a contradiction, so there are no solutions other than the one we found earlier (1, 1, 1)

[Altair]

Just substract 2 times equation 1 from equation 2, so you have:
x²-2x + y²-2y + z²-2z = -3

Add 1+1+1 on each side:
x²-2x+1 + y²-2y+1 + z²-2z+1 = 0

Factorize:
(x-1)² + (y-1)² + (z-1)² = 0

Each term is greater or equal than 0, so as the sum is 0 they must all be 0, so x=y=z=1.

[VictorPensioner]

Take only first two equations

(1) x + y + z = 3
(2) x² + y² + z² = 3

From (1)
(3) z = 3 - x - y
Substitute (3) in (2)
(4) x² + y² + (3 - x - y)² = 3
Find function minimum
f(x) = x² + y² + (3 - x - y)²

∂f/∂x = 0
∂f/∂y = 0
or
(5) 2x - 2(3 - x - y) = 0
(6) 2y - 2(3 - x - y) = 0
Subtract (5) - (6):
2x - 2y = 0 => x = y
and
2x - 2(3 - x - x) = 0
2x - 6 + 4x = 0 => x = 1
Therefore
min f(x) = 1² + 1² + (3 - 1 - 1)² = 3 when x = 1 and y = 1
and
x = 1,
y = 1
is solution of equation (4) and
x = 1,
y = 1,
z = 1
is solution of system (1) - (2).
P.S. The equation
x³ + y³ + z³ = 3
is redudant

[roberttelarket]

If you proctored in 1973 then you must be at least 70!

[coreyyanofsky]

i did it the rather hard way

from
(x + y + z)² = 9
we get
xy + xz + yz = 3

then
(x + y + z)³ = 27
x³ + y³ + z³ + 3(x + y)(xy + xz + yz + z²) = 27
3 + 3(x + y)(3 + z²) = 27
(x + y)(3 + z²) = 8
(x + y + z)(3 + z²) = 8 + 3z + z³
9 + 3z² = 8 + 3z + z³
z³ - 3z² + 3z - 1 = 0
(z - 1)³ = 0
z = 1

by symmetry the solution is x = y = z = 1

[williamspostoronnim]

Я ждал комплексных корней, а все сошлось на единицах, что и сразу было ясно на глазок.

[vladimirkaplun]

Third equation is redundant, Just a smokescreen. Without any calculation sphere x^2+y^2+z^2=3 tangents plane x+y+z=3 at x=y=z=1 -visible at a glance.If you want calculations - see @Altair705 below.

[SidneiMV]

x = u + 1
y = v + 1
z = w + 1

(u + 1) + (v + 1) + (w + 1) = 3
u + v + w = 0

(u + 1)² + (v + 1)² + (w + 1)² = 3
u² + v² + w² + 2(u + v + w) + 3 = 3
u² + v² + w² = 0

(u + 1)³ + (v + 1)³ + (w + 1)³ = 3
u³ + v³ + w³ + 3(u² + v² + w²)
+ 3(u + v + w) + 3 = 3
u³ + v³ + w³ = 0

B = u + v + w = 0
C = uv + uw + vw
D = uvw

t³ - 0t² + Ct - D = 0
t³ = -Ct + D
u³ = -Cu + D
v³ = -Cv + D
w³ = -Cw + D
0 = 0 + 3D => D = 0

t³ + Ct = 0

(u + v + w)² = u² + v² + z² + 2(uv + uw + vw)
C = uv + uw + vw = 0
t³ + Ct = 0
t³ = 0
(u, v, w) = (0, 0, 0)
*(x, y, z) = (1, 1, 1)*