1965 IMO Problems/Problem 4: Difference between revisions
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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>. | ||
== Solution == | == Solution == | ||
{{solution}} | 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>. | |||
Multiplying by <math>x_i</math> yields: | |||
<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>. | |||
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. | |||
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> | |||
Then we have: | |||
<math>(1+t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+4t = 0 \Longleftrightarrow t(t^2+3t+4) = 0</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> | |||
Then we have: | |||
<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> | |||
AND | |||
<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> | |||
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> | |||
Then we have: | |||
<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> | |||
AND | |||
<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> | |||
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>. | |||
Revision as of 15:52, 16 July 2009
Problem
Find all sets of four real numbers 
, 
, 
, 
such that the sum of any one and the product of the other three is equal to 
.
Solution
Let 
be the product of the four real numbers.
Then, for 
we have: 
.
Multiplying by 
yields:
where 
.
If 
, then we have 
which is a solution.
So assume that 
. WLOG, let at least two of equal 
, and 
OR 
.
Case I: 
Then we have:
Which has no non-zero solutions for 
.
Case II: 
AND 
Then we have:
AND
So, we have 
as the only non-zero solution, and thus, 
and all permutations are solutions.
Case III: 
AND 
Then we have:
AND
Thus, there are no non-zero solutions for in this case.
Therefore, the solutions are: 
; 
; 
; 
; 
.
