4. Assume that for positive real numbers $a, b, c, d$ the following holds:

$$
a b+c d=a c+b d=4 \quad \text { and } \quad a d+b c=5 .
$$

Find the smallest possible value of the sum $a+b+c+d$ and determine which quadruples $a, b, c, d$ achieve it.

The regional round of category A takes place

## on Tuesday, January 17, 2012

so that it starts in the morning and contestants have 4 hours of clean time to solve the problems. Contestants can earn 6 points for each problem, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. These details will be communicated to the students before the start of the competition.
4. Since the given equations contain mixed products of variables, it will be advantageous to examine the square of the sum $a+b+c+d$. By rearrangement and substitution of the given equations, we get

$$
\begin{aligned}
(a+b+c+d)^{2} & =a^{2}+b^{2}+c^{2}+d^{2}+2(ab+cd+ac+bd+ad+bc)= \\
& =a^{2}+b^{2}+c^{2}+d^{2}+2(4+4+5)=a^{2}+d^{2}+b^{2}+c^{2}+26 .
\end{aligned}
$$

Now we will use the known inequalities $a^{2}+d^{2} \geqq 2ad, b^{2}+c^{2} \geqq 2bc$, where equality holds if and only if $a=d$ and $b=c$. Based on this, from (1) we get

$$
(a+b+c+d)^{2} \geqq 2ad+2bc+26=2 \cdot 5+26=36 .
$$

Therefore, for such positive numbers $a, b, c, d$, we have $a+b+c+d \geqq \sqrt{36}=6$, with equality holding if and only if $a=d$ and $b=c$. Substituting into the original equations, we get the system

$$
2ab=4, \quad a^{2}+b^{2}=5 .
$$

This can be solved in several ways. For example, we can express

$$
(a+b)^{2}=a^{2}+b^{2}+2ab=5+4=9
$$

thus $a+b=3$ (since $a, b>0$). According to Vieta's formulas, $a, b$ are the roots of the quadratic equation $x^{2}-3x+2=0$, hence $\{a, b\}=\{1,2\}$. It is easy to verify that the quadruples $a=d=1$, $b=c=2$, or $a=d=2, b=c=1$ indeed satisfy the given equations and for them $a+b+c+d=6$.

Answer. The smallest possible value of the sum $a+b+c+d$ is 6, and it is achieved only by the quadruples $(1,2,2,1)$ and $(2,1,1,2)$.

Alternative Solution. From the equation $ab+cd=ac+bd$ it follows that $a(b-c)=d(b-c)$, so either $a=d$ or $b=c$. Considering the symmetry, we can only consider the quadruples $(a, b, c, d)$ where $d=a$, and find the smallest value of the sum $S=2a+b+c$ under the assumption that the positive numbers $a, b, c$ satisfy the equations $a(b+c)=4$ and $a^{2}+bc=5$.

According to Vieta's formulas, the positive numbers $b, c$ are the roots of the quadratic equation

$$
x^{2}-\frac{4}{a}x+(5-a^{2})=0
$$

This equation has two positive (not necessarily distinct) roots if and only if its discriminant

$$
D=\frac{16}{a^{2}}-4(5-a^{2})=\frac{4(a^{2}-1)(a^{2}-4)}{a^{2}}
$$

is non-negative and if, in addition to the inequality $a>0$, we also have $a^{2} \leq 1$ or $a^{2} \geq 4$. This means that $\min S=6$ and that $(1,2,2,1)$ and $(2,1,1,2)$ are the only quadruples providing the found minimum (since in both of them $b=c$, no other such quadruples - given the restriction of our considerations to the first of the possibilities $a=d, b=c$ - exist). From the expression for the difference $S-6$ in the form

$$
S-6=2a+b+c-6=2a+\frac{4}{a}-6=\frac{2(a-1)(a-2)}{a}
$$

we see that the desired inequality $S>6$ indeed holds for every $a \in (0,1) \cup (2, \sqrt{5})$.

For a complete solution, award 6 points, of which 4 points for estimating $a+b+c+d \geqq 6$ and 2 points for determining the quadruples for which equality holds.

If the solver uses the AG-inequalities $a^{2}+b^{2} \geqq 2ab, c^{2}+d^{2} \geqq 2cd$ in (1) and thus derives the weaker estimate $a+b+c+d \geqq \sqrt{34}$, award 2 points.

In the case of the second solution, award 1 point for deriving the equation $(a-d)(b-c)=0$, 1 point for transitioning to solving the system of two unknowns $b, c$ with parameter $a$, 2 points for finding the set of admissible values of $a$ (or at least deriving the necessary condition $a \leqq 1 \vee a \geqq 2$), 1 point for proving the inequality $2a+4/a \geqq 6$ for all admissible $a$, and 1 point for determining both sought quadruples.

If the student only guesses the result and finds the quadruples $(1,2,2,1)$ and $(2,1,1,2)$, award 1 point (this point should only be awarded if the student does not earn any other points in the problem).

[^0]:    1 Numbers having only twos and threes on the remaining $n-1$ places will be counted, as they already have the required at least one digit 1 in the units place.

    2 When filling in the table, we do not need to laboriously divide $10^{n}-1$ by seven. It is sufficient to use the fact that $10^{n+1}$ gives the same remainder when divided by seven as 10 times the remainder of $10^{n}$.