Difference between revisions of "1970 Canadian MO Problems/Problem 1"
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From the first equation minus the second, we get : | From the first equation minus the second, we get : | ||
− | <cmath> \begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x | + | <cmath> \begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x-y) &=& 0\end{matrix} </cmath> |
So either <math>z=1</math> or <math>x=y</math>. For <math>z=1</math>, since the equations were symmetric, we have one solution of <math> (x,y,z)=(1,1,1)</math>. | So either <math>z=1</math> or <math>x=y</math>. For <math>z=1</math>, since the equations were symmetric, we have one solution of <math> (x,y,z)=(1,1,1)</math>. |
Latest revision as of 06:54, 30 December 2012
Problem
Find all number triples such that when any of these numbers is added to the product of the other two, the result is
.
Solution
We have:
From the first equation minus the second, we get :
So either or
. For
, since the equations were symmetric, we have one solution of
.
From
, substituting it into the original three derived equations, we have:
We then get
Substituting this into
,
Thus, either
, or
. Since the equations were symmetric, we then get the full solution set of:
See Also
1970 Canadian MO (Problems) | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 2 |