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Difference between revisions of "1974 IMO Problems/Problem 5"
 
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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==
 +
 +
Compute, but group terms to make the computations easier.
 +
Add 1st and 3rd terms, 2nd and 4th terms, and group <math>a, c</math>
 +
and <math>b, d</math>:
 +
 +
<math>S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d} =</math>
 +
 +
<math>\frac{2ac + (a + c)(b + d)}{ac + (a + c)(b + d) + (a + c)^2} +
 +
\frac{2bd + (a + c)(b + d)}{bd + (a + c)(b + d) + (b + d)^2}</math>
 +
 +
Showing that <math>S < 2</math> amounts to showing that
 +
 +
<math>4acbd + 3(ac + bd)(a + c)(b + d) + 2ac(b + d)^2 + 2bd(a + c)^2 +
 +
2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2] <</math>
 +
 +
<math>2acbd + 2(ac + bd)(a + c)(b + d) + 2ac(b + d)^2 + 2bd(a + c)^2 +
 +
4(a + c)^2(b + d)^2 + 2(a + c)(b + d)[(a + c)^2 + (b + d)^2]</math>
 +
 +
This reduces to
 +
 +
<math>-2acbd - (ac + bd)(a + c)(b + d) + 2(a + c)^2(b + d)^2 +
 +
(a + c)(b + d)[(a + c)^2 + (b + d)^2] > 0</math>
 +
 +
This is clearly true because the negative terms cancel with
 +
parts from the positive terms: <math>-2acbd</math> is cancelled by terms
 +
from <math>2(a + c)^2(b + d)^2</math> and <math>-(ac + bd)(a + c)(b + d)</math> is
 +
cancelled by terms from <math>(a + c)(b + d)[(a + c)^2 + (b + d)^2]</math>
 +
 +
Showing that <math>S > 1</math> amounts to showing that
 +
 +
<math>4acbd + 3(ac + bd)(a + c)(b + d) + 2ac(b + d)^2 + 2bd(a + c)^2 +
 +
2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2] ></math>the negative terms cancel with
 +
parts from the positive terms: <math>-2acbd</math> is cancelled by terms
 +
from <math>2(a + c)^2(b + d)^2</math> and <math>-(ac + bd)(a + c)(b + d)</math> is
 +
cancelled by terms from <math>(a + c)(b + d)[(a + c)^2 + (b + d)^2]</math>
 +
 +
<math>acbd + (ac + bd)(a + c)(b + d) + ac(b + d)^2 + bd(a + c)^2 +
 +
2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2]</math>
 +
 +
This reduces to
 +
 +
<math>3acbd + 2(ac + bd)(a + c)(b + d) + ac(b + d)^2 + bd(a + c)^2 > 0</math>
 +
 +
This is clearly true because all the terms are <math>> 0</math>.
 +
 +
Now show that any number <math>S \in (1, 2)</math> is the result of the sum
 +
of the fractions in the statement of the problem for some
 +
<math>a, b, c, d</math>.  First, look for <math>a = c, b = d</math> which would yield
 +
<math>S</math>.  We are looking for <math>a, b</math> such that
 +
<math>S = \frac{2a}{a + 2b} + \frac{2b}{2a + b}</math>.  After some easy
 +
computations this becomes
 +
 +
<math>(4 - 2S)\left(\frac{a}{b}\right)^2 - (5S - 4)\left(\frac{a}{b}\right) +
 +
(4 - 2S) = 0</math>
 +
 +
This yields <math>\frac{a}{b} = \frac{5S - 4 \pm \sqrt{9S^2 + 24S - 48}}{4 - 2S}</math>
 +
 +
Remembering that <math>S > 0</math>, we see that we obtain <math>\frac{a}{b}</math> real and
 +
positive when <math>S \in \left[\frac{4}{3}, 2\right)</math>.
 +
 +
Now, look for <math>a = b, c = d</math> which would yield
 +
<math>S</math>.  We are looking for <math>a, c</math> such that
 +
<math>S = \frac{2a}{2a + c} + \frac{2c}{a + 2c}</math>.  After some easy
 +
computations this becomes
 +
 +
<math>(2S - 2)\left(\frac{a}{c}\right)^2 - (8 - 5S)\left(\frac{a}{c}\right) +
 +
(2S - 2) = 0</math>
 +
 +
This yields <math>\frac{a}{c} = \frac{8 - 5S \pm \sqrt{9S^2 - 48S + 48}}{4S - 4}</math>
 +
 +
Remembering that <math>S > 0</math>, we see that we obtain <math>\frac{a}{c}</math> real and
 +
positive when <math>S \in \left(1, \frac{4}{3}\right]</math>.
 +
 +
Thus, any <math>S \in (1, 2)</math> is the result of the sum of the fractions in the
 +
statement of the problem for some <math>a, b, c, d</math>.
 +
 +
[Solution by pf02, October 2025]
 +
  
 
== See Also == {{IMO box|year=1974|num-b=4|num-a=6}}
 
== See Also == {{IMO box|year=1974|num-b=4|num-a=6}}

Latest revision as of 16:32, 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

Compute, but group terms to make the computations easier. Add 1st and 3rd terms, 2nd and 4th terms, and group $a, c$ and $b, d$:

$S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d} =$

$\frac{2ac + (a + c)(b + d)}{ac + (a + c)(b + d) + (a + c)^2} + \frac{2bd + (a + c)(b + d)}{bd + (a + c)(b + d) + (b + d)^2}$

Showing that $S < 2$ amounts to showing that

$4acbd + 3(ac + bd)(a + c)(b + d) + 2ac(b + d)^2 + 2bd(a + c)^2 + 2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2] <$

$2acbd + 2(ac + bd)(a + c)(b + d) + 2ac(b + d)^2 + 2bd(a + c)^2 + 4(a + c)^2(b + d)^2 + 2(a + c)(b + d)[(a + c)^2 + (b + d)^2]$

This reduces to

$-2acbd - (ac + bd)(a + c)(b + d) + 2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2] > 0$

This is clearly true because the negative terms cancel with parts from the positive terms: $-2acbd$ is cancelled by terms from $2(a + c)^2(b + d)^2$ and $-(ac + bd)(a + c)(b + d)$ is cancelled by terms from $(a + c)(b + d)[(a + c)^2 + (b + d)^2]$

Showing that $S > 1$ amounts to showing that

$4acbd + 3(ac + bd)(a + c)(b + d) + 2ac(b + d)^2 + 2bd(a + c)^2 + 2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2] >$the negative terms cancel with parts from the positive terms: $-2acbd$ is cancelled by terms from $2(a + c)^2(b + d)^2$ and $-(ac + bd)(a + c)(b + d)$ is cancelled by terms from $(a + c)(b + d)[(a + c)^2 + (b + d)^2]$

$acbd + (ac + bd)(a + c)(b + d) + ac(b + d)^2 + bd(a + c)^2 + 2(a + c)^2(b + d)^2 + (a + c)(b + d)[(a + c)^2 + (b + d)^2]$

This reduces to

$3acbd + 2(ac + bd)(a + c)(b + d) + ac(b + d)^2 + bd(a + c)^2 > 0$

This is clearly true because all the terms are $> 0$.

Now show that any number $S \in (1, 2)$ is the result of the sum of the fractions in the statement of the problem for some $a, b, c, d$. First, look for $a = c, b = d$ which would yield $S$. We are looking for $a, b$ such that $S = \frac{2a}{a + 2b} + \frac{2b}{2a + b}$. After some easy computations this becomes

$(4 - 2S)\left(\frac{a}{b}\right)^2 - (5S - 4)\left(\frac{a}{b}\right) + (4 - 2S) = 0$

This yields $\frac{a}{b} = \frac{5S - 4 \pm \sqrt{9S^2 + 24S - 48}}{4 - 2S}$

Remembering that $S > 0$, we see that we obtain $\frac{a}{b}$ real and positive when $S \in \left[\frac{4}{3}, 2\right)$.

Now, look for $a = b, c = d$ which would yield $S$. We are looking for $a, c$ such that $S = \frac{2a}{2a + c} + \frac{2c}{a + 2c}$. After some easy computations this becomes

$(2S - 2)\left(\frac{a}{c}\right)^2 - (8 - 5S)\left(\frac{a}{c}\right) + (2S - 2) = 0$

This yields $\frac{a}{c} = \frac{8 - 5S \pm \sqrt{9S^2 - 48S + 48}}{4S - 4}$

Remembering that $S > 0$, we see that we obtain $\frac{a}{c}$ real and positive when $S \in \left(1, \frac{4}{3}\right]$.

Thus, any $S \in (1, 2)$ is the result of the sum of the fractions in the statement of the problem for some $a, b, c, d$.

[Solution by pf02, October 2025]


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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