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

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Determine, with proof, whether or not one can find 1975 points on the circumference
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== Problem 5 ==
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of a circle with unit radius such that the distance between any
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two of them is a rational number.
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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>k</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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Now, set <math>a = d = x</math> and <math>b = c = y</math> for some positive real numbers <math>x, y</math>. Then
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<cmath>S = \frac{2x}{2x+y} + \frac{2y}{2y+x} = \frac{2y^2 + 8xy + 2x^2}{2y^2 + 5xy + 2x^2}.</cmath>
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Notice that if we treat the numerator and denominator each as a quadratic in <math>y</math>, we will get <math>1 + \frac{g(x)}{2y^2 + 5xy + 2x^2}</math>, where <math>g(x)</math> has a degree lower than <math>2</math>. This means taking <math>\lim_{y\to\infty} 1 + \frac{g(x)}{2y^2 + 5xy + 2x^2} = 1</math>, which means <math>S</math> can be brought arbitrarily close to <math>1</math>. Therefore, we are done.
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<cmath> </cmath>
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~Imajinary
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== See Also == {{IMO box|year=1974|num-b=4|num-a=6}}

Latest revision as of 15:00, 29 January 2021

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 $k$ 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 $2$. Now, set $a = d = x$ and $b = c = y$ for some positive real numbers $x, y$. Then \[S = \frac{2x}{2x+y} + \frac{2y}{2y+x} = \frac{2y^2 + 8xy + 2x^2}{2y^2 + 5xy + 2x^2}.\] Notice that if we treat the numerator and denominator each as a quadratic in $y$, we will get $1 + \frac{g(x)}{2y^2 + 5xy + 2x^2}$, where $g(x)$ has a degree lower than $2$. This means taking $\lim_{y\to\infty} 1 + \frac{g(x)}{2y^2 + 5xy + 2x^2} = 1$, which means $S$ can be brought arbitrarily close to $1$. Therefore, we are done. \[\] ~Imajinary

See Also

1974 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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