UKMT Special (Problem 1616)

Find all real numbers x,y,zx, y, z which satisfy the simultaneous equations

x24y+7=0x^2 - 4y + 7 = 0

y26z+14=0y^2 - 6z + 14 = 0

z22x7=0z^2 - 2x - 7 = 0

[UKMT BMO Round 11 20122012, Q33]

Note by Yajat Shamji
8 months, 2 weeks ago

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You have until next Friday, 3:003:00pm!

Yajat Shamji - 8 months, 2 weeks ago

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Step 1) Add all of them and factor the term.

   x^2 - 2x + y^{2} - 4y + z^{2} - 6z + 14 = 0

   x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = 0

   (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 0

Step 2) If any of the squared term is not 0, it must be a positive real number.

If there is at least one squared term that is not 0, the equation will be false because the right hand sight must be greater than 0.

So, each the square term must be equal to 0.

   (x - 1)^2 = 0    =>    x - 1  = 0    =>    x=1

   (y - 2)^2 = 0    =>    y - 2 = 0    =>    y=2

   (z - 3)^2 = 0    =>    z - 3 = 0    =>    z=3

(x , y , z)=(1 , 2 , 3) is the only real solution.

I apologize for the bad representation.

Quest Keeper - 8 months, 2 weeks ago

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Correct method and solution!

Yajat Shamji - 8 months, 2 weeks ago

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