1974 IMO Problems/Problem 5

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]

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

Page

Toolbox

Search

1974 IMO Problems/Problem 5

Contents

Problem 5

Solution

Remarks (added by pf02, October 2025)

Solution 2

See Also