8.4. Determine all functions $f, g: R \rightarrow R$, knowing that

$$
f(x+4)+2 g(2 x+7)=\frac{x+3}{2} . \text { and } f\left(\frac{x+3}{2}\right)+g(x+2)=x+6
$$
Solution: Let $f\left(\frac{x+3}{2}\right)+g(x+2)=x+6, \forall x \in R$. Here we replace $x$ with $2x+5$. We get $f(x+4)+g(2x+7)=2x+11$.

Since $f(x+4)+2g(2x+7)=\frac{x+3}{2}$, by subtracting we obtain

$g(2x+7)=\frac{x+3}{2}-2x-11=\frac{-3x-19}{2}=-\frac{3}{2}x-\frac{19}{2}$.

Let $t=2x+7$, then $x=\frac{t-7}{2} \Rightarrow g(t)=-\frac{3}{2} \cdot \frac{t-7}{2}-\frac{19}{2}=-\frac{3}{4}t+\frac{21}{4}-\frac{19}{2} = -\frac{3}{4}t-\frac{17}{4}$, or $g(x)=-\frac{3}{4}x-\frac{17}{4}$.

Then from (1) and (2) we get $f(x+4)=2x+11-g(2x+7)=2x+11+\frac{3}{2}x+\frac{19}{2}=\frac{7}{2}x+\frac{41}{2}$.

Let $t=x+4$, then $x=t-4 \Rightarrow f(t)=\frac{7}{2}(t-4)+\frac{41}{2}=\frac{7}{2}t+\frac{13}{2}$, or

$f(x)=\frac{7}{2}x+\frac{13}{2}$. Answer: $f(x)=\frac{7}{2}x+\frac{13}{2}$ and $g(x)=-\frac{3}{4}x-\frac{17}{4}$.

## Moldova Mathematical Olympiad

March 04, 2017, 8th grade, first day

Official evaluation criteria for problems

| 8.1. Prove that if $x, y \in R$ and $x^{2}+y^{2}-2x+12y+33=0$, then $x>y$. |  |  |  |
| :---: | :---: | :---: | :---: |
| Step |  | Stages of the solution | Points awarded |
| 1. |  | Obtain the relation $(x-1)^{2}+(y+6)^{2}=4$ | 1 point |
| 2. |  | Conclude that $\left\{\begin{array}{l}(x-1)^{2} \leq 4 \\ (y+6)^{2} \leq 4\end{array}\right.$. | 1 point |
| 3. | Write | $y4$  $y>-4 \Rightarrow(y+6)^{2}>4$ with the conclusion that {f6c948b7d-9442-4282-9632-edbf6de988a9} | 2 points |
| 4. | Write | $\left\{\begin{array}{l}x4 \\ x>3 \Rightarrow(x-1)^{2}>4\end{array}\right.$ with the conclusion that $x \in[-1,3]$ | 2 points |
| 5. |  | From 3. and 4. it follows that $x>y$ | 1 point |
|  |  | Total points | 7 points |

Remark: Any other correct solution is awarded 7 points.

| 8.2. If $a, b, c, d$ are positive real numbers such that $ab=2$ and $cd=27$, find the minimum value of the expression $E=(a+1)(b+2)(c+3)(d+4)$. |  |  |
| :--- | :---: | :--- |
| Step | Solution with evaluation criteria | Points awarded |
| 1. | Applied $ab=2$ and obtained $(a+1)(b+2)=4+2a+b$ | 1 point |
| 2. | Applied $cd=27$ and obtained $(c+3)(d+4)=39+4c+3d$ | 1 point |


| 3. | Applied the inequality of means for $(a+1)(b+2):$  $(a+1)(b+2) \geq 2 \sqrt{a} \cdot 2 \sqrt{2b}=4 \sqrt{2ab}=4 \sqrt{4}=8$ | 1 point |
| :--- | :--- | :--- |
| 4. | Applied the inequality of means for $(c+3)(d+4):$  $(c+3)(c+4)=39+4c+3d \geq 39+2 \sqrt{4c \cdot 3d}=39+2 \sqrt{12 \cdot 27}=39+36=75$ | 1 point |
| 5. | Obtained $E \geq 8 \cdot 75=600$ | 1 point |
| 6. | Deduced that the minimum value is obtained for $a=1$ and $b=2$ | 1 point |
| 7. | Deduced that the minimum value is obtained for $c=4.5$ and $d=6$ | 7 points |

Remark: Any other correct solution is awarded 7 points.