Exercise 2. Let $\mathrm{ABC}$ be a triangle, $\mathrm{O}$ a point inside this triangle. The line parallel to $(\mathrm{BC})$ passing through $O$ intersects $[C A]$ at $D$, the line parallel to $(C A)$ passing through $O$ intersects $[A B]$ at $E$, the line parallel to $(A B)$ passing through $\mathrm{O}$ intersects $[\mathrm{BC}]$ at $\mathrm{F}$. What is the value of the following sum of ratios:

$$
\frac{\mathrm{BF}}{\mathrm{BC}}+\frac{\mathrm{AE}}{\mathrm{AB}}+\frac{C D}{\mathrm{AC}} ?
$$
## Solution to Exercise 2 First Solution:

Let $|A B C|$ denote the area of triangle $A B C$. We observe that

$$
\frac{\mathrm{BF}}{\mathrm{BC}}=\frac{|\mathrm{ABF}|}{|\mathrm{ABC}|}=\frac{|\mathrm{ABO}|}{|\mathrm{ABC}|}
$$

because triangles $A B F$ and $A B C$ with bases $B F$ and $B C$ have the same height, and $A B O$ and $A B F$ have the same base $A B$ and two parallel heights of the same length (since, if $\mathrm{O}^{\prime}, \mathrm{F}^{\prime}$ are the feet of the heights, $\mathrm{OO}^{\prime} \mathrm{F}^{\prime} \mathrm{F}$ is a rectangle).

We deduce that

$$
\frac{B F}{B C}+\frac{A E}{A B}+\frac{C D}{A C}=\frac{|A B O|}{|A B C|}+\frac{|A O C|}{|A B C|}+\frac{|O C B|}{|A B C|}=1 .
$$

Second Solution:

Here we will use a slightly more general result than Thales' theorem: the fact that two triangles with the same angles have proportional side lengths (such triangles are called similar).

Let $F^{\prime}$ be the intersection of $[A C]$ and (OF).

![](https://cdn.mathpix.com/cropped/2024_06_04_93e83a8cba80c1bd2fe8g-2.jpg?height=352&width=567&top_left_y=366&top_left_x=779)

For example, triangles $\mathrm{ABC}$ and $\mathrm{F}^{\prime} \mathrm{CF}$ in the exercise have parallel sides, so they have the same angles and are similar, hence $\frac{B F}{B C}=\frac{A F^{\prime}}{A C}$. (But this can also be shown using Thales' theorem.)

However, triangles $A B C$ and $F^{\prime} O D$ are also similar, so $\frac{F^{\prime} D}{A C}=\frac{F^{\prime} O}{A B}$.

Moreover, $A F O^{\prime} E$ is a parallelogram, so $\mathrm{F}^{\prime} \mathrm{O}=A E$, and finally:

$$
\frac{B F}{B C}+\frac{A E}{A B}+\frac{C D}{A C}=\frac{A F^{\prime}}{A C}+\frac{F^{\prime} D}{A C}+\frac{C D}{A C}=1
$$

Exercise 3. Consider a number $N$ that can be written in the form $30 x 070 y 03$, where $x$ and $y$ are digits between 0 and 9. For which values of $(\mathrm{x}, \mathrm{y})$ is the integer $N$ divisible by 37?

Solution to Exercise 3 We can write $N=300070003+1000000 x+100 y$. Since $100=37 \times 3-11$ and $1000=37 \times 30-110=37 \times 27+1$, we have $1000000=1000^{2}=37 \times (37 \times 27^{2}+2 \times 27)+1$ (by double distributivity).

Since 37 divides $N$, 37 also divides $300070003+x-11 y$. By performing the Euclidean division of 300070003 by 37, we find that $300070003=37 \times 8110000+3$.

In short, 37 divides $N$ if and only if it divides $3+x-11 y$, which is an integer between 12 (if $x=9$ and $y=0$) and -96 (if $x=0$ and $y=9$). Therefore, there are three possibilities:

- We have $3+x-11 y=0$, i.e., $y=1$ and $x=8$;
- We have $3+x-11 y=-37$, so $y=4$ and $x=4$;
- We have $3+x-11 y=-74$, so $y=7$ and $x=0$.