# Difference between revisions of "1974 IMO Problems/Problem 5"

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<cmath>S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.</cmath> | <cmath>S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.</cmath> | ||

Note that <math>S</math> is a continuous function and that <math>f(m) = m + \frac{1}{2-m}</math> is a strictly increasing function. We can now decrease <math>m</math> and <math>l</math> to make <math>m</math> tend arbitrarily close to <math>1</math>. We see <math>\lim_{m\to1} m + \frac{1}{2-m} = 2</math>, meaning <math>S</math> can be brought arbitrarily close to <math>2</math>. | Note that <math>S</math> is a continuous function and that <math>f(m) = m + \frac{1}{2-m}</math> is a strictly increasing function. We can now decrease <math>m</math> and <math>l</math> to make <math>m</math> tend arbitrarily close to <math>1</math>. We see <math>\lim_{m\to1} m + \frac{1}{2-m} = 2</math>, meaning <math>S</math> can be brought arbitrarily close to <math>2</math>. | ||

− | < | + | <cmath> </cmath> |

~Imajinary | ~Imajinary |

## Revision as of 03:50, 7 November 2020

## Problem 5

Determine all possible values of where are arbitrary positive numbers.

## Solution

Note that We will now prove that can reach any range in between and .

Choose any positive number . For some variables such that and , let , , and . Plugging this back into the original fraction, we get The above equation can be further simplified to Note that is a continuous function and that is a strictly increasing function. We can now decrease and to make tend arbitrarily close to . We see , meaning can be brought arbitrarily close to . ~Imajinary