Difference between revisions of "1966 IMO Problems/Problem 5"

Latest revision as of 18:12, 10 November 2024

Problem

Solve the system of equations

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\ |a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\ |a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3 - a_4| x_4 = 1 \\ |a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1$

where $a_1, a_2, a_3, a_4$ are four different real numbers.

Solution

Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:

$- x1 + x2 + x3 + x4 = 0.$

Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:

$- x1 - x2 - x3 + x4 = 0.$

Hence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:

$- x1 - x2 + x3 + x4 = 0.$

Hence $x2 = x3 = 0$ , and $x1 = x4 = 1/(a1 - a4)$ .

Remarks (added by pf02, September 2024)

The solution above is in the realm of flawed and incorrect. It is flawed because you can not claim to have solved a system of equations by having solved a particular case. It is correct in solving the particular case, but it is incorrect in stating that the result obtained is a solution to the system in general.

Below I will give a complete solution to the problem. The first few lines will be a repetition of the "solution" above, and I will repeat them for the sake of completeness and of a more tidy writing.

Solution 2

There are 24 possibilities when we count the ordering of $a_1, a_2, a_3, a_4$ , and each ordering gives a different system of equations. Let us consider one of them, like in the "solution" above.

Assume $a_1 > a_2 > a_3 > a_4$ . In this case, the system is

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_1 - a_2) x_2 + (a_1 - a_3) x_3 + (a_1 - a_4) x_4 = 1 \\ (a_1 - a_2) x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_2 - a_3) x_3 + (a_2 - a_4) x_4 = 1 \\ (a_1 - a_3) x_1 + (a_2 - a_3) x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_3 - a_4) x_4 = 1 \\ (a_1 - a_4) x_1 + (a_2 - a_4) x_2 + (a_3 - a_4) x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1$

Subtract the second equation from the first, and divide by $(a_1 - a_2)$ . Also, subtract the fourth equation from the third, and divide by $(a_3 - a_4)$ . We obtain

$-x_1 + x_2 + x_3 + x_4 = 0 \\ -x_1 - x_2 - x_3 + x_4 = 0$

It follows that $x_1 - x_4 = 0$ and $x_2 + x_3 = 0$ .

Subtract the third equation from the second, and divide by $(a_2 - a_3)$ . We obtain

$-x_1 - x_2 + x_3 + x_4 = 0$

Since $x_1 - x_4 = 0$ , it follows that $x_2 - x_3 = 0$ . Combining with $x_2 + x_3 = 0$ , we get $x_2 = x_3 = 0$ . Replacing these in the first equation of the system, we get $x_4 = \frac{1}{a_1 - a_4}$ , so we also have $x_1 = \frac{1}{a_1 - a_4}$ .

Now we have two ways of proceeding. We could consider each of the other 23 cases, and solve it by a similar method. The task is made easy if we notice that each case is obtained from the first case by a permutation of indices, so it can be viewed as a change of notation. With some care, we can just write the solution in each case. For example, in the case $a_2 > a_1 > a_3 > a_4$ , we will obtain $x_1 = x_3 = 0$ and $x_2 = x_4 = \frac{1}{a_2 - a_4}$ .

We will proceed differently, but we will use the same idea. Let $m, n, p, q$ be the indices such that $a_m > a_n > a_p > a_q$ . Written in a compact way, our system becomes

$\sum_{\substack{i = 1 \\ i \ne j}}^4 |a_j - a_i| x_i = 1, \ \ \ \ j = 1, 2, 3, 4$ .

Make the following change of notation: $a_m \rightarrow b_1 \\ a_n \rightarrow b_2 \\ a_p \rightarrow b_3 \\ a_q \rightarrow b_4$

and

$x_m \rightarrow y_1 \\ x_n \rightarrow y_2 \\ x_p \rightarrow y_3 \\ x_q \rightarrow y_4$

In the new notation we have $b_1 > b_2 > b_3 > b_4$ and the system becomes

$\sum_{\substack{k = 1 \\ k \ne l}}^4 |b_l - b_k| y_k = 1, \ \ \ \ l = 1, 2, 3, 4$ .

This is exactly the system we solved above, just with a new notation ( $b, y$ instead of $a, x$ ). So the solutions are $y_2 = y_3 = 0, y_1 = y_4 = \frac{1}{b_1 - b_4}$ .

Returning to our original notation, we have $x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}$ .

In conclusion, here is a compact way of giving the solution to the system: let $m$ be the index of the largest of the $a_i$ 's, and q = the index of the smallest of the $a_i$ 's, and let $n, p$ be the other two indices. Then $x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}$ .

[Solution by pf02, September 2024]

Revision as of 17:00, 23 September 2024 (view source) Pf02 (talk \| contribs) m ← Older edit		Latest revision as of 18:12, 10 November 2024 (view source) Pf02 (talk \| contribs) m
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	<math>x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}</math>.		<math>x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}</math>.

−	(Solution by pf02, September 2024)	+	[Solution by pf02, September 2024]

1966 IMO (Problems) • Resources
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