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

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+ | == Problem 5 == | ||

+ | Determine all possible values of <cmath>S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}</cmath> where <math>a, b, c, d,</math> are arbitrary positive numbers. | ||

+ | |||

+ | ==Solution== | ||

+ | Note that <cmath>2 = \frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d} > S > \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d} = 1.</cmath> We will now prove that <math>S</math> can reach any range in between <math>1</math> and <math>2</math>. | ||

+ | |||

+ | Choose any positive number <math>a</math>. For some variables such that <math>k, m, l > 0</math> and <math>k + m + l = 1</math>, let <math>b = ak</math>, <math>c = am</math>, and <math>d = al</math>. Plugging this back into the original fraction, we get | ||

+ | <cmath>S = \frac{a}{a+ak+al}+\frac{ak}{a+ak+am}+\frac{am}{ak+am+al}+\frac{al}{a+am+al} = \frac{1}{1+k+l}+\frac{k}{1+k+m}+\frac{m}{k+m+l}+\frac{l}{1+m+l}.</cmath> | ||

+ | The above equation can be further simplified to | ||

+ | <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>. | ||

+ | <math> </math> | ||

+ | ~Imajinary |

## Revision as of 03:49, 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 . $$ (Error compiling LaTeX. ! Missing $ inserted.) ~Imajinary