A Nice Polynomial System | Math Olympiads

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[takeoverurmemes]

I like to visualize problems like these using Desmos, and what you see is when you treat z like a constant, you graphically get a diagonal line and a circle from these equations. The only time they ever intersect (meaning we have a valid answer) is exactly when the line is tangent to the circle, which happens when z = 10. The point at which they touch are at (10, 10). Any other value for z, and either the circle becomes too small to intersect with the line, or its equation has no real solutions, showing that this is the only answer (at least visually, which isn't rigorous, but I find it interesting nonetheless).

[Roq-stone]

I was side shot by this technique. Didn’t think of doing that way.

[angelishify]

x+y+z=10 je enačba ravnine z normalo (1, 1, 1), ki je od izhodišča (0, 0, 0) oddaljena za vektor (a, a, a): x-a+y-a+z-a=30-3a=0. a=10 in d=a√3=10√3.
x^2+y^2+z^2=300 je enačba sfere z radijem r=10√3.
Ker je d=r, se ravnina in sfera dotikata v točki (a, a, a)=(10, 10, 10). To je edina skupna točka in rešitev sistema.

[carlfisher]

I took the ratio of one over the other and got the answer.

[angelishify]

x+y+z=10: ∂z/∂x=-1 ter ∂z/∂y=-1
minimiziramo razdaljo f(x, y)=x^2+y^2+z(x, y)^2:
∂ f(x, y)/∂x=2x-2z=0 in ∂ f(x, y)/∂y=2y-2z=0.
sledi: x=y=z. Če je to rešitev sistema (kar pa je; za x=y=z=10), je to tudi edina rešitev.

[leif]

The multiplying through by 2 really comes out of nowhere so why don't you solve without that? Thanks for sharing.

[vitowidjojo]

So, no other solutions?
I was thinking that because they are 3 variable equation.
Unless there is another restriction i didn't know...

[giuseppemalaguti]

Si svolge....si arriva ad un discriminante<=0...x=y=Z=10