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

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== Problem 5 ==
  
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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.
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==Solution==
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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>.
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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
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<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>
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The above equation can be further simplified to
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<cmath>S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.</cmath>
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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>.
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<math> </math>
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~Imajinary

Revision as of 03:49, 7 November 2020

Problem 5

Determine all possible values of \[S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}\] where $a, b, c, d,$ are arbitrary positive numbers.

Solution

Note that \[2 = \frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d} > S > \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d} = 1.\] We will now prove that $S$ can reach any range in between $1$ and $2$.

Choose any positive number $a$. For some variables such that $k, m, l > 0$ and $k + m + l = 1$, let $b = ak$, $c = am$, and $d = al$. Plugging this back into the original fraction, we get \[S = \frac{a}{a+ak+al}+\frac{ak}{a+ak+am}+\frac{am}{ak+am+al}+\frac{al}{a+am+al} = \frac{1}{1+k+l}+\frac{k}{1+k+m}+\frac{m}{k+m+l}+\frac{l}{1+m+l}.\] The above equation can be further simplified to \[S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.\] Note that $S$ is a continuous function and that $f(m) = m + \frac{1}{2-m}$ is a strictly increasing function. We can now decrease $m$ and $l$ to make $m$ tend arbitrarily close to $1$. We see $\lim_{m\to1} m + \frac{1}{2-m} = 2$, meaning $S$ can be brought arbitrarily close to $2$. $$ (Error compiling LaTeX. ! Missing $ inserted.) ~Imajinary

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