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Difference between revisions of "1965 IMO Problems/Problem 4"

(Created page with '== 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 …')
 
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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>.
  
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== Solution ==
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== Solution ==  
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{{solution}}
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Let <math>P = x_1x_2x_3x_4</math> be the product of the four real numbers.
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Then, for <math>i = 1,2,3,4</math> we have: <math>x_i + \prod_{j \neq i}x_j = 2</math>.
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Multiplying by <math>x_i</math> yields:
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<math>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</math> where <math>t = \pm \sqrt{1-P} \in \mathbb{R}</math>. 
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If <math>t=0</math>, then we have <math>(x_1,x_2,x_3,x_4)=(1,1,1,1)</math> which is a solution.
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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>.
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Case I: <math>x_1 = x_2 = x_3 = x_4 = 1+t</math>
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Then we have:
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<math>(1+t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+4t = 0 \Longleftrightarrow t(t^2+3t+4) = 0</math>
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Which has no non-zero solutions for <math>t</math>.
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Case II: <math>x_1 = x_2 = x_3 = 1+t</math> AND <math>x_4 = 1-t</math>
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Then we have:
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<math>(1-t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+2t = 0</math> <math>\Longleftrightarrow t(t+1)(t+2) = 0 \Longleftrightarrow t \in \{0,-1,-2\}</math>
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AND
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<math>(1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0</math> <math>\Longleftrightarrow -t(t-1)(t+2) = 0 \Longleftrightarrow t \in \{0,1,-2\}</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.
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Case III: <math>x_1 = x_2 = 1+t</math> AND <math>x_3 = x_4 = 1-t</math>
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Then we have:
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<math>(1-t)+(1-t)(1+t)^2 = 2 \Longleftrightarrow -t^3-t^2 = 0</math> <math>\Longleftrightarrow -t^2(t+1) = 0 \Longleftrightarrow t \in \{0,-1\}</math>
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AND
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<math>(1+t)+(1+t)(1-t)^2 = 2 \Longleftrightarrow t^3-t^2 = 0</math> <math>\Longleftrightarrow t^2(t-1) = 0 \Longleftrightarrow t \in \{0,1\}</math>
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Thus, there are no non-zero solutions for <math>t</math> in this case.
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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>.

Revision as of 16:52, 16 July 2009

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: $(1,1,1,1)$; $(3,-1,-1,-1)$; $(-1,3,-1,-1)$; $(-1,-1,3,-1)$; $(-1,-1,-1,3)$.

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