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Difference between revisions of "1965 IMO Problems/Problem 4"
Line 102: Line 102:
 
the solutions <math>(-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)</math>.
 
the solutions <math>(-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)</math>.
  
−
Case 3.  
+
Case 3. Plug in <math>x_2x_3, x_2x_4</math> in (4) and (3), and get
−
 
+
<math>x_4 + x_1 = 2</math> and <math>x_3 + x_1 = 2</math>.  Now plug <math>x_2, x_3, x_4</math>
−
 
+
in (1), and get <math>x_1 + (2 - x_1)^2x_1 = 2</math>.  This equation becomes
−
 
+
<math>(x_1 - 1)^2(x_1 - 2) = 0</math>.  <math>x_1 = 1</math> yields <math>(1, 1, 1, 1)</math> which
 +
we already know, and <math>x_2 = 2</math> yields <math>(2, 0, 0, 2)</math>, which is not
 +
a possible solution.
  
 +
Case 4.  plugging <math>x_2x_3, x_2x_4, x_3x_4</math> into (4), (3), (2), we
 +
get <math>x_4 + x_1 = 2, x_3 + x_1 = 2, x_2 + x_1 = 2</math>.  Plugging
 +
<math>x_2, x_3, x_4</math> into (1), we get <math>x_1 + (2 - x_1)^3 = 2</math>.  The
 +
solutions of this equation are <math>1, 2, 3</math>.  Note that <math>x_1 = 1</math>
 +
and <math>x_1 = 3</math> yield solutions we already know, and <math>x_1 = 2</math>
 +
is impossible.
  
 +
Thus, the five solutions to the problem are <math>(1, 1, 1, 1),
 +
(3, -1, -1, -1), (-1, 3, -1, -1), (-1, -1, 3, -1), (-1, -1, -1, 3)</math>.
  
 
(Solution by pf02, November 2024)  
 
(Solution by pf02, November 2024)  
−
 
−
TO BE CONTINUED.  I AM SAVING MID WAY SO I DON'T LOSE WORK DONE SO FAR.
 
  
  
 
== See Also ==  
 
== See Also ==  
 
{{IMO box|year=1965|num-b=3|num-a=5}}
 
{{IMO box|year=1965|num-b=3|num-a=5}}

Revision as of 19:27, 3 November 2024

Problem

Find all sets of four real numbers $x_1$, $x_2$, $x_3$, $x_4$ such that the sum of any one and the product of the other three is equal to $2$.


Solution

Let $P = x_1x_2x_3x_4$ be the product of the four real numbers.

Then, for $i = 1,2,3,4$ we have: $x_i + \prod_{j \neq i}x_j = 2$.

Multiplying by $x_i$ yields:

$x^2_i + P = 2x_i \Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \Longleftrightarrow x_i = 1 \pm t$ where $t = \pm \sqrt{1-P} \in \mathbb{R}$.

If $t=0$, then we have $(x_1,x_2,x_3,x_4)=(1,1,1,1)$ which is a solution.

So assume that $t \neq 0$. WLOG, let at least two of $x_i$ equal $1+t$, and $x_1 \ge x_2 \ge x_3 \ge x_4$ OR $x_1 \le x_2 \le x_3 \le x_4$.

Case I: $x_1 = x_2 = x_3 = x_4 = 1+t$

Then we have:

$(1+t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+4t = 0 \Longleftrightarrow t(t^2+3t+4) = 0$

Which has no non-zero solutions for $t$.

Case II: $x_1 = x_2 = x_3 = 1+t$ AND $x_4 = 1-t$

Then we have:

$(1-t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+2t = 0$ $\Longleftrightarrow t(t+1)(t+2) = 0 \Longleftrightarrow t \in \{0,-1,-2\}$

AND

$(1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0$ $\Longleftrightarrow -t(t-1)(t+2) = 0 \Longleftrightarrow t \in \{0,1,-2\}$

So, we have $t = -2$ as the only non-zero solution, and thus, $(x_1,x_2,x_3,x_4) = (-1,-1,-1,3)$ and all permutations are solutions.

Case III: $x_1 = x_2 = 1+t$ AND $x_3 = x_4 = 1-t$

Then we have:

$(1-t)+(1-t)(1+t)^2 = 2 \Longleftrightarrow -t^3-t^2 = 0$ $\Longleftrightarrow -t^2(t+1) = 0 \Longleftrightarrow t \in \{0,-1\}$

AND

$(1+t)+(1+t)(1-t)^2 = 2 \Longleftrightarrow t^3-t^2 = 0$ $\Longleftrightarrow t^2(t-1) = 0 \Longleftrightarrow t \in \{0,1\}$

Thus, there are no non-zero solutions for $t$ in this case.

Therefore, the solutions are: $(1,1,1,1)$; $(3,-1,-1,-1)$; $(-1,3,-1,-1)$; $(-1,-1,3,-1)$; $(-1,-1,-1,3)$.


Solution 2

We have to solve the system of equations

$x_1 + x_2x_3x_4 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ x_2 + x_1x_3x_4 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\ x_3 + x_1x_2x_4 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\ x_4 + x_1x_2x_3 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

Subtract (2) from (1) and factor. We get

$(x_1 - x_2)(1 - x_3x_4) = 0$,

which implies $x_1 - x_2 = 0$ or $1 - x_3x_4 = 0$.

Similarly, subtracting (3) and then (4) from (1) and factoring, we get

$(x_1 - x_3)(1 - x_2x_4) = 0 \\ (x_1 - x_4)(1 - x_2x_3) = 0$

They imply $x_1 - x_3 = 0$ or $1 - x_2x_4 = 0$, and $x_1 - x_4 = 0$ or $1 - x_2x_3 = 0$.

We will consider four possibilities:

1. $x_2 = x_1, x_3 = x_1, x_4 = x_1$

2. $x_2 = x_1, x_3 = x_1$ and $x_2x_3 = 1$

3. $x_2 = x_1$ and $x_2x_3 = 1, x_2x_4 = 1$

4. $x_2x_3 = 1, x_2x_4 = 1, x_3x_4 = 1$

Note that in fact, there are four more possibilities, but they just correspond to permutations in $x_1, x_2, x_3, x_4$ of cases 2. and 3., so there is no harm in not dealing with them explicitly.

Case 1. Plug $x_2, x_3, x_4$ in equation (1). We get $x_1 + x_1^3 = 2$. This is an equation of degree $3$ whose only real root is $x_1 = 1$. We get the solution $x_1 = x_2 = x_3 = x_4 = 1$.

Case 2. Plug $x_2, x_3$ into $x_2x_3 = 1$, and get $x_1^2 = 1$. We get $x_1 = 1$ or $x_1 = -1$. The first solution, $x_1 = 1$, yields $x_2 = x_3 = 1$, and using (4), $x_4 = 1$. The second solution, $x_1 = -1$, yields $x_2 = x_3 = -1$, and using (4), $x_4 = 3$. We get the solution $x_1 = x_2 = x_3 = -1, x_4 = 3.$ Because of the permutations of $x_1, x_2, x_3, x_4$, we also get the solutions $(-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)$.

Case 3. Plug in $x_2x_3, x_2x_4$ in (4) and (3), and get $x_4 + x_1 = 2$ and $x_3 + x_1 = 2$. Now plug $x_2, x_3, x_4$ in (1), and get $x_1 + (2 - x_1)^2x_1 = 2$. This equation becomes $(x_1 - 1)^2(x_1 - 2) = 0$. $x_1 = 1$ yields $(1, 1, 1, 1)$ which we already know, and $x_2 = 2$ yields $(2, 0, 0, 2)$, which is not a possible solution.

Case 4. plugging $x_2x_3, x_2x_4, x_3x_4$ into (4), (3), (2), we get $x_4 + x_1 = 2, x_3 + x_1 = 2, x_2 + x_1 = 2$. Plugging $x_2, x_3, x_4$ into (1), we get $x_1 + (2 - x_1)^3 = 2$. The solutions of this equation are $1, 2, 3$. Note that $x_1 = 1$ and $x_1 = 3$ yield solutions we already know, and $x_1 = 2$ is impossible.

Thus, the five solutions to the problem are $(1, 1, 1, 1), (3, -1, -1, -1), (-1, 3, -1, -1), (-1, -1, 3, -1), (-1, -1, -1, 3)$.

(Solution by pf02, November 2024)


See Also

1965 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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