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1965 IMO Problems/Problem 4: Difference between revisions

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== Problem ==
== Problem ==
Find all sets of four real numbers <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, <math>x_4</math> such that the sum of any one and the product of the other three is equal to <math>2</math>.
Find all sets of four real numbers <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, <math>x_4</math> such that the sum of any one and the product of the other three is equal to <math>2</math>.


== Solution ==  
 
== Solution ==
 
Let <math>P = x_1x_2x_3x_4</math> be the product of the four real numbers.  
Let <math>P = x_1x_2x_3x_4</math> be the product of the four real numbers.  


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


Multiplying by <math>x_i</math> yields:  
Multiplying by <math>x_i</math> yields:  
Line 15: Line 17:


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


Case I: <math>x_1 = x_2 = x_3 = x_4 = 1+t</math>  
Case I: <math>x_1 = x_2 = x_3 = x_4 = 1+t</math>  
Line 24: Line 25:


Which has no non-zero solutions for <math>t</math>.  
Which has no non-zero solutions for <math>t</math>.  


Case II: <math>x_1 = x_2 = x_3 = 1+t</math> AND <math>x_4 = 1-t</math>
Case II: <math>x_1 = x_2 = x_3 = 1+t</math> AND <math>x_4 = 1-t</math>
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So, we have <math>t = -2</math> as the only non-zero solution, and thus, <math>(x_1,x_2,x_3,x_4) = (-1,-1,-1,3)</math> and all permutations are solutions.
So, we have <math>t = -2</math> as the only non-zero solution, and thus, <math>(x_1,x_2,x_3,x_4) = (-1,-1,-1,3)</math> and all permutations are solutions.


Case III: <math>x_1 = x_2 = 1+t</math> AND <math>x_3 = x_4 = 1-t</math>
Case III: <math>x_1 = x_2 = 1+t</math> AND <math>x_3 = x_4 = 1-t</math>
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Thus, there are no non-zero solutions for <math>t</math> in this case.  
Thus, there are no non-zero solutions for <math>t</math> in this case.  


Therefore, the solutions are: <math>(1,1,1,1)</math>; <math>(3,-1,-1,-1)</math>; <math>(-1,3,-1,-1)</math>; <math>(-1,-1,3,-1)</math>; <math>(-1,-1,-1,3)</math>.
== Solution 2 ==
We have to solve the system of equations
<math>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)</math>
Subtract (2) from (1) and factor.  We get
<math>(x_1 - x_2)(1 - x_3x_4) = 0</math>,
which implies <math>x_1 = x_2</math> or <math>x_3x_4 = 1</math>.
Similarly, subtracting (3) and then (4) from (1) and factoring, we get
<math>(x_1 - x_3)(1 - x_2x_4) = 0 \\
(x_1 - x_4)(1 - x_2x_3) = 0</math>
They imply <math>x_1 = x_3</math> or <math>x_2x_4 = 1</math>, and <math>x_1 = x_4</math> or <math>x_2x_3 = 1</math>.
We will consider four possibilities:
1. <math>x_1 = x_2 = x_3 = x_4</math>
2. <math>x_1 = x_2 = x_3</math> and <math>x_2x_3 = 1</math>
3. <math>x_1 = x_2</math> and <math>x_2x_3 = 1, x_2x_4 = 1</math>
4. <math>x_2x_3 = 1, x_2x_4 = 1, x_3x_4 = 1</math>
Note that in fact, there are four more possibilities, but they just
correspond to permutations of the indexes <math>1, 2, 3, 4</math> of <math>x</math>, so
there is no harm in not dealing with them explicitly.
(Solution by pf02, November 2024)
TO BE CONTINUED.  I AM SAVING MID WAY SO I DON'T LOSE WORK DONE SO FAR.


Therefore, the solutions are: <math>(1,1,1,1)</math>; <math>(3,-1,-1,-1)</math>; <math>(-1,3,-1,-1)</math>; <math>(-1,-1,3,-1)</math>; <math>(-1,-1,-1,3)</math>.


== 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 19:28, 2 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 $x_1 = x_2$ or $x_3x_4 = 1$.

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 $x_1 = x_3$ or $x_2x_4 = 1$, and $x_1 = x_4$ or $x_2x_3 = 1$.

We will consider four possibilities:

1. $x_1 = x_2 = x_3 = x_4$

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

3. $x_1 = x_2$ 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 of the indexes $1, 2, 3, 4$ of $x$, so there is no harm in not dealing with them explicitly.



(Solution by pf02, November 2024)

TO BE CONTINUED. I AM SAVING MID WAY SO I DON'T LOSE WORK DONE SO FAR.


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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