We are traveling by airplane. Looking out of the window that is $20 \mathrm{~cm}$ from our eyes, we spot a boat in the direction of each of the lower corners of the $25 \mathrm{~cm} \times 40 \mathrm{~cm}$ window. We know that the airplane is flying at an altitude of $10,3 \mathrm{~km}$, and our eye level is at the horizontal midpoint of the window. How far apart are the two boats?
Solution. Let's make a diagram. Let $S$ be our eye, $H_{1}$ and $H_{2}$ the two ships, and $A B C D$ the window. Then $A B \| H_{1} H_{2}$. Let the perpendicular projection of $S$ onto $A B$ be $T$, onto $H_{1} H_{2}$ be $U$, onto the Ground be $Q$; and the perpendicular projection of $T$ onto $S Q$ be $P$. Then $P T \| Q U$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2b7e91a7c844ac0b3e28g-1.jpg?height=580&width=471&top_left_y=306&top_left_x=816)

We know that $S P=\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm}, P T=20 \mathrm{~cm}, A B=40 \mathrm{~cm}$, and $S Q=10.3 \mathrm{~km}$.

The triangles $S A B$ and $S H_{1} H_{2}$ are similar, since their corresponding sides are parallel. Thus, $\frac{H_{1} H_{2}}{A B}=\frac{S U}{S T}$, or $H_{1} H_{2}=\frac{S U \cdot A B}{S T}$. Similarly, from the similarity of the triangles $S T P$ and $S U Q$, $\frac{S U}{S T}=\frac{S Q}{S P}$, which, when substituted into the previous expression and the appropriate values are inserted:

$$
H_{1} H_{2}=\frac{S Q}{S P} \cdot A B=\frac{10.3 \mathrm{~km}}{\frac{40}{2} \mathrm{~cm}} \cdot 40 \mathrm{~cm}=10.3 \mathrm{~km}
$$

It can be seen that the result does not depend on how far we look out the window.

Note. The problem did not specify that the height of the window is $40 \mathrm{~cm}$, which most people assumed. Those who took the width to be $40 \mathrm{~cm}$ could also get 5 points.