## Task 3 - 060823

$18 \%$ of a number is equal to $15 \%$ of another number.

Determine the ratio of the first to the second of these two numbers!
If $x$ is the first number and $y$ is the second, then according to the given information, the following equation holds:

$$
\frac{18 \cdot x}{100}=\frac{15 \cdot y}{100}
$$

thus $6 x=5 y$. From this, we obtain the proportion $x: y=5: 6$.

By reasoning backwards, one can see that under this condition, the requirement is indeed satisfied.[^12]

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0609.jpg?height=482&width=334&top_left_y=233&top_left_x=273)

Claim: $\overline{A B}+\overline{C D}=\overline{B C}+\overline{A D}$

Proof:

Let the points of tangency on the circle be $E, F, G$, and $H$. Since the lengths of the tangent segments from any point outside the circle to the circle are equal, we have:

$$
\overline{A E}=\overline{A H} \quad \overline{B E}=\overline{B F} \quad \overline{C G}=\overline{C F} \quad \overline{D G}=\overline{D H}
$$

From this, it follows that:

$$
\begin{aligned}
\overline{A E}+\overline{B E}+\overline{C G}+\overline{D G} & =\overline{A H}+\overline{B F}+\overline{C F}+\overline{D H}, \\
\overline{A B}+\overline{C D} & =\overline{B C}+\overline{A D}
\end{aligned}
$$

Solutions from the second round of 1966 taken from [5]

### 5.8.3 Third Round 1966, Class 8