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1974 IMO Problems/Problem 5: Difference between revisions

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==Solution==
==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>.
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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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.
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.
<cmath> </cmath>
<cmath> </cmath>
~Imajinary
~Imajinary
==Remarks (added by pf02, October 2025)==
1. Strictly speaking, the solution given above is incomplete.
It shows that <math>S</math> can be arbitrarily close to <math>1</math> and to <math>2</math>,
but it does not show that <math>S</math> can take <math>\mathbf{any}</math> value
in <math>(1, 2)</math>.  This is not difficult to prove, but it is not
obvious or trivial, so it should be mentioned.
Here is an argument:  Let <math>S_1</math> be given by <math>a_1, b_1, c_1, d_1</math>
and let <math>S_2</math> be given by <math>a_2, b_2, c_2, d_2</math>.  (Think of <math>S_1</math>
close to <math>1</math> and <math>S_2</math> close to <math>2</math>).  Let <math>a = (1 - t) a_1 + t a_2</math>,
and similarly for <math>b, c, d</math>.  Let <math>S(t)</math> be given by <math>a, b, c, d</math>.
Clearly <math>S(t)</math> is a continuous function of <math>t</math>, <math>S(0) = S_1</math> and
<math>S(1) = S_2</math>.  By the [[Intermediate Value Theorem]] for any
<math>S \in [S_1, S_2]</math> there is <math>t</math>, such that <math>S(t) = S</math>.
2. Motivation for the solution:  Take <math>a, b, c, d</math> of the same
magnitude, for example <math>a = b = c = d = 1</math>.  Then <math>S = 1.333\dots\ .</math>
Now take <math>a, b</math> being of higher magnitude than <math>c, d</math>, for example
<math>a = b = 1000</math>, <math>c = d = 1</math>.  We get <math>S = 1.001\dots\ .</math>  And, take
<math>a, c</math> of higher magnitude than <math>b, d</math>, for example <math>a = c = 1000</math>,
<math>b = d = 1</math>.  We get <math>S = 1.997\dots\ .</math>
Take a few more examples if desired.  This suggests <math>1 < S < 2</math>,
and that we can get values as close to <math>1</math> and <math>2</math> as we want.
3. I will give another solution below.  It is computationally more
involved, but conceptually simpler, and it uses no calculus.
==Solution 2==


== See Also == {{IMO box|year=1974|num-b=4|num-a=6}}
== See Also == {{IMO box|year=1974|num-b=4|num-a=6}}

Revision as of 12:55, 14 October 2025

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


Remarks (added by pf02, October 2025)

1. Strictly speaking, the solution given above is incomplete. It shows that $S$ can be arbitrarily close to $1$ and to $2$, but it does not show that $S$ can take $\mathbf{any}$ value in $(1, 2)$. This is not difficult to prove, but it is not obvious or trivial, so it should be mentioned.

Here is an argument: Let $S_1$ be given by $a_1, b_1, c_1, d_1$ and let $S_2$ be given by $a_2, b_2, c_2, d_2$. (Think of $S_1$ close to $1$ and $S_2$ close to $2$). Let $a = (1 - t) a_1 + t a_2$, and similarly for $b, c, d$. Let $S(t)$ be given by $a, b, c, d$. Clearly $S(t)$ is a continuous function of $t$, $S(0) = S_1$ and $S(1) = S_2$. By the Intermediate Value Theorem for any $S \in [S_1, S_2]$ there is $t$, such that $S(t) = S$.

2. Motivation for the solution: Take $a, b, c, d$ of the same magnitude, for example $a = b = c = d = 1$. Then $S = 1.333\dots\ .$ Now take $a, b$ being of higher magnitude than $c, d$, for example $a = b = 1000$, $c = d = 1$. We get $S = 1.001\dots\ .$ And, take $a, c$ of higher magnitude than $b, d$, for example $a = c = 1000$, $b = d = 1$. We get $S = 1.997\dots\ .$

Take a few more examples if desired. This suggests $1 < S < 2$, and that we can get values as close to $1$ and $2$ as we want.

3. I will give another solution below. It is computationally more involved, but conceptually simpler, and it uses no calculus.


Solution 2

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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