Difference between revisions of "1970 Canadian MO Problems/Problem 1"

Revision as of 05:58, 11 May 2012

Problem

Find all number triples $(x,y,z)$ such that when any of these numbers is added to the product of the other two, the result is $2$ .

Solution

We have:

$\begin{matrix} x + yz &=& 2\\ y + xz &=& 2\\ z + yx &=& 2 \end{matrix}$

From the first equation minus the second, we get :

$\begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x+y) &=& 0\end{matrix}$

So either $z=1$ or $x=y$ . For $z=1$ , since the equations were symmetric, we have one solution of $(x,y,z)=(1,1,1)$ . From $x=y$ , substituting it into the original three derived equations, we have: $\begin{matrix} x+xz &=& 2\\ z + x^2 &=& 2 \end{matrix}$ We then get $z = 2-x^2$ Substituting this into $x+xz = 2$ , $\begin{matrix} x + x(2-x^2) &=& 2\\ x^3 - 3x + 2 &=& 0\\ (x-1)^2(x+2) = 0 \end{matrix}$ Thus, either $x = 1$ , or $x = -2$ . Since the equations were symmetric, we then get the full solution set of:

$(x,y,z) = {(1,1,1), (-2,-2,-2)}$

@@ Line 23: / Line 23: @@
 <cmath> (x,y,z) = {(1,1,1), (-2,-2,-2)} </cmath>
-{{Old CanadaMO box|num-b=1|num-a=2|year=1970}}
+== See Also ==
+{{Old CanadaMO box|before=First Question|num-a=2|year=1970}}
+[[Category:Olympiad Algebra Problems]]

1970 Canadian MO (Problems)
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 2

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Difference between revisions of "1970 Canadian MO Problems/Problem 1"

Revision as of 05:58, 11 May 2012

Problem

Solution

See Also