Difference between revisions of "2018 IMO Problems/Problem 2"
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+ | ==Problem== | ||
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Find all numbers <math>n \ge 3</math> for which there exists real numbers <math>a_1, a_2, ..., a_{n+2}</math> satisfying <math>a_{n+1} = a_1, a_{n+2} = a_2</math> and | Find all numbers <math>n \ge 3</math> for which there exists real numbers <math>a_1, a_2, ..., a_{n+2}</math> satisfying <math>a_{n+1} = a_1, a_{n+2} = a_2</math> and | ||
<cmath>a_{i}a_{i+1} + 1 = a_{i+2}</cmath> | <cmath>a_{i}a_{i+1} + 1 = a_{i+2}</cmath> | ||
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So <math>a_2 = – 1 \implies a_1 = 2, a_3 = – 1.</math> | So <math>a_2 = – 1 \implies a_1 = 2, a_3 = – 1.</math> | ||
− | <i><b>Case | + | <i><b>Case 1a</b></i> |
Let <math>n = 3k, k={1,2,...}.</math> | Let <math>n = 3k, k={1,2,...}.</math> | ||
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<cmath>(a_1 – a_2)^2 + (a_2 – a_3)^2 + ... + (a_4 – a_1)^2 = 0 \implies a_1 = a_2 = a_3 = a_4 .</cmath> | <cmath>(a_1 – a_2)^2 + (a_2 – a_3)^2 + ... + (a_4 – a_1)^2 = 0 \implies a_1 = a_2 = a_3 = a_4 .</cmath> | ||
Substituting into any of the initial equations, we obtain the equation <math>a_1^2 + 1 = a_1,</math> which does not have real roots. Hence, there are no such real numbers. | Substituting into any of the initial equations, we obtain the equation <math>a_1^2 + 1 = a_1,</math> which does not have real roots. Hence, there are no such real numbers. | ||
+ | |||
+ | <i><b>Case 2a</b></i> | ||
+ | |||
+ | Let <math>n = 5.</math> We get system of equations | ||
+ | <cmath>\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_4 \\a_3 a_4 + 1 = a_5\\a_4 a_5 + 1 = a_1 \\a_5 a_1 + 1 = a_2\end{cases}</cmath> | ||
+ | We repeat all steps of <i><b>Case 2</b></i> and get: there are no such real numbers. | ||
+ | |||
+ | <i><b>Case 2b </b></i> | ||
+ | |||
+ | Let <math>n = 3k \pm 1.</math> We repeat all steps of cases <math>2</math> and <math>2a</math> and get: there are no such real numbers. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2018|num-b=1|num-a=3}} |
Latest revision as of 01:44, 19 November 2023
Problem
Find all numbers for which there exists real numbers
satisfying
and
for
Solution
We find at least one series of real numbers for for each
and we prove that if
then the series does not exist.
Case 1
Let We get system of equations
We subtract the first equation from the second and get:
So
Case 1a
Let
Real numbers
satisfying
and
.
Case 2
Let We get system of equations
We multiply each equation by the number on the right-hand side and get:
We multiply each equation by a number that precedes a pair of product numbers in a given sequence
So we multiply the equation with product
by
, we multiply the equation with product
by
etc. We get:
We add all the equations of the first system, and all the equations of the second system. The sum of the left parts are the same! It includes the sum of all the numbers
and the sum of the triples of consecutive numbers
Hence, the sums of the right parts are equal, that is,
It is known that this expression is doubled
Substituting into any of the initial equations, we obtain the equation
which does not have real roots. Hence, there are no such real numbers.
Case 2a
Let We get system of equations
We repeat all steps of Case 2 and get: there are no such real numbers.
Case 2b
Let We repeat all steps of cases
and
and get: there are no such real numbers.
vladimir.shelomovskii@gmail.com, vvsss
See Also
2018 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |