2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 12 Medium

Which of the following conditions is sufficient to guarantee that integers x, y, and z satisfy the equation x(x-y)+y(y-z)+z(z-x)=1 ?(2021 AMC Fall 10B, Question #12)

  • A.

    x>y and y=z

  • B.

    x=y-1 and y=z-1

  • C.

    x=z+1 and y=x+1

  • D.

    x=z and y-1=x

  • E.

    x+y+z=1

Answer:D

Solution 1:

It is obvious x, y, and z are symmetrical. We are going to solve the problem by Completing the Square. \begin{aligned} &x^{2}+y^{2}+z^{2}-x y-y z-z x=1 \\ &2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x=2 \\ &(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2 \end{aligned} Because x, y, z are integers, (x-y)^{2},(y-z)^{2}, and (z-x)^{2} can only equal 0,1,1. So one variable must equal another, and the third variable is 1 different from those 2 equal variables. So the answer is D.

Solution 2:

Plugging in every choice, we see that choice (D) works. We have y=x+1, z=x, so x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1 . Our answer is (D).

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