The figure below illustrates the operation of a garage door, represented by the segment $X Y$. By moving the point $X$, the point $A$ slides along a vertical track, represented by the segment $B D$. Some of the measurements in the figure are $A C=B C=C Y=0.5 \, \text{m}$ and $A X=1 \, \text{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-12.jpg?height=283&width=211&top_left_y=1292&top_left_x=1025)

a) In the figure, the point $X$ is $0.2 \, \text{m}$ from the track $B D$. What is the distance from $C$ to the track?

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-12.jpg?height=260&width=203&top_left_y=1720&top_left_x=1029)

b) Show that the height of point $Y$ relative to the ground does not change with the movement of the door.

c) If the bumper of a car is $0.4 \, \text{m}$ high, as shown in the figure, what should be the minimum distance between the track and the bumper so that it is not hit when the door opens?

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-12.jpg?height=308&width=346&top_left_y=2196&top_left_x=958)
Solution

a) $1^{st}$ solution: In the figure below, we have $X S=0.2$ and we want to find $C R$. We notice that the angles indicated in the figure with vertices at $C$ and $X$ are equal, as they are determined by the parallel lines $C R$ and $X S$ and the transversal $X Y$. Therefore, the right triangles $A R C$ and $A S X$ are similar, and we have

$$
\frac{C R}{X S}=\frac{A C}{A X}
$$

that is,

$$
C R=X S \times \frac{A C}{A X}=0.2 \times \frac{0.5}{1}=0.1
$$

We can also argue as follows. The similarity ratio between the triangles $A R C$ and $A S X$ is equal to $\frac{A C}{A X}=\frac{0.5}{1}=0.5$; since the segments $C R$ and $X S$ are corresponding, it follows that the length of $C R$ is half the length of $A X$, that is, it is equal to $0.1 \, \text{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-39.jpg?height=394&width=257&top_left_y=2284&top_left_x=817)
$2^{nd}$ solution: Let $\alpha$ be the angle $\widehat{D A X}$, as in the figure below. Since $\widehat{D A X}$ and $\widehat{B A Y}$ are vertical angles, we also have $\widehat{B A Y}=\alpha$. In the right triangles $A S X$ and $A R C$, we have $X S=A X \sin \alpha=\sin \alpha$ and $C R=A C \sin \alpha=\frac{1}{2} \sin \alpha$. Therefore, $C R=\frac{1}{2} X S=0.1 \, \text{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-40.jpg?height=386&width=240&top_left_y=415&top_left_x=1019)

b) $1^{st}$ solution: Since $A C=B C=Y C$, the triangles $A C B$ and $B C Y$ are isosceles; we can then mark the angles $\alpha$ and $\beta$ as in the figure below. The sum of the angles in triangle $A B Y$ is $2 \alpha+2 \beta=180^{\circ}$; hence $\alpha+\beta=90^{\circ}$. Therefore, $B Y$ is perpendicular to the rail $B D$, that is, $B Y$ is horizontal regardless of the position of $Y$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-40.jpg?height=399&width=257&top_left_y=1057&top_left_x=1005)

$2^{nd}$ solution: Since $A C=B C=Y C$, we can draw a circle with center $C$ passing through $A, B$, and $Y$, as in the left figure. Since the points $A, C$, and $Y$ are collinear, the segment $A Y$ is a diameter of this circle. Therefore, the angle $\widehat{A B Y}$ is inscribed in the semicircle, so its measure is $90^{\circ}$. Thus, $B Y$ is horizontal regardless of the position of $Y$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-40.jpg?height=402&width=266&top_left_y=1712&top_left_x=998)

c) In the figure below, we want to calculate $D T$ when $X T=0.4$. To do this, we first note that when the door closes, $X Y$ coincides with $B D$; thus

$$
B D=X Y=X A+A C+C Y=1+0.5+0.5=2
$$

Since DTXS is a rectangle, we have $S D=X T=0.4$ and it follows that $B S=B D-S D=2-0.4=1.6$. On the other hand, the triangles $A S X$ and $A B Y$ are congruent; indeed, they are both right triangles, their angles at $X$ and $Y$ are equal (as in item $a$), and $A X=A Y$. Therefore, $A S=A B$ and since $B S=1.6$, it follows that $A S=0.8$. The Pythagorean theorem then tells us that

$$
S X=\sqrt{A X^{2}-A S^{2}}=\sqrt{1-0.64}=\sqrt{0.36}=0.6
$$

and we conclude that $D T=0.6 \, \text{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-41.jpg?height=423&width=277&top_left_y=234&top_left_x=798)