When a light ray strikes a plane mirror, it is reflected in such a way that it makes equal angles with the mirror, as illustrated in Figure 1. Figure 2 shows two mirrors forming an angle $\alpha$. A light ray, parallel to mirror I, strikes mirror II at point $A$ and is reflected three times, until it strikes mirror I perpendicularly at point $D$.
![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-14.jpg?height=320&width=1082&top_left_y=1459&top_left_x=587)

a) What is the measure of the angle $\alpha$?

b) Let $A B$ be perpendicular to mirror I, as in Figure 2. If $A B=10 \mathrm{~cm}$, what is the length of $C D$?
Solution

a) $1^{a}$ solution: We mark on the figure the relevant angles for the solution. We note in particular that at A the angle of incidence (and therefore the angle of reflection) is equal to $\alpha$; indeed, the light ray enters parallel to mirror I and the support line of mirror II is transversal to both. Since $\gamma$ is the exterior angle of triangle $A F C$, it follows that $\gamma=2 \alpha$. Similarly, since $\beta$ is the exterior angle of triangle $C E F$, we have $\beta=\alpha+\gamma=3 \alpha$. Finally, from the right triangle $C D E$ we have $180^{\circ}=\alpha+\beta+90^{\circ}=4 \alpha+90^{\circ}$, from which $4 \alpha=90^{\circ}$, or $\alpha=22.5^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-44.jpg?height=397&width=789&top_left_y=1886&top_left_x=722)

$2^{\text {a }}$ solution: Since the sum of the angles of triangle $A B F$ is $180^{\circ}$, it follows that $B \hat{A} F=90^{\circ}-\gamma$. And since the sum of the angles at vertex $A$ is also $180^{\circ}$, it follows that $2 \alpha+\left(90^{\circ}-\gamma\right)+90^{\circ}=180^{\circ}$, from which $\gamma=2 \alpha$. Considering now the triangle AFE, we have $\alpha+\beta+\left(180^{\circ}-2 \gamma\right)=180^{\circ}$, from which we get $\beta=2 \gamma-\alpha=3 \alpha$. Finally, the triangle $C D E$ tells us that $180^{\circ}=\alpha+\beta+90^{\circ}=4 \alpha+90^{\circ}$ and it follows that $4 \alpha=90^{\circ}$, or $\alpha=22.5^{\circ}$.

b) $1^{\mathrm{a}}$ solution: We observe that, since $\gamma=2 \alpha=45^{\circ}$, the triangle $D E F$ is isosceles, that is, $E D=D F$. The Pythagorean theorem tells us that

$$
E F^{2}=E D^{2}+D F^{2}=2 E D^{2}
$$

from which we get $E F=\sqrt{2} E D$. The same argument applied to the triangle $A B F$ shows that $A F=\sqrt{2} A B=$ $10 \sqrt{2}$.

We now note that the triangles $C D E$ and $A F E$ are similar, as they have the angles $\alpha$ and $\beta$ in common. Thus

$$
\frac{C D}{A F}=\frac{C D}{10 \sqrt{2}}=\frac{D E}{F E}=\frac{D E}{\sqrt{2} D E}=\frac{1}{\sqrt{2}}
$$

from which we get $C D=10$.

$2^{\mathrm{a}}$ solution: We reflect the line $C F$ using the line $C A$ as the axis of symmetry, obtaining the ray $C F^{\prime}$, where $F^{\prime}$ is the symmetric of $F$ (figure below).

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-45.jpg?height=423&width=757&top_left_y=668&top_left_x=547)

Note that

$$
C \hat{E} A=C \hat{E} F+\beta=C \hat{E} F^{\prime}+F^{\prime} \hat{E} A
$$

Since $C \hat{E} F=C \hat{E} F^{\prime}$, the above equation can only be valid if $F^{\prime} \hat{E} A=\beta$. This implies that the points $D, E$ and $F^{\prime}$ are collinear (see figure above); thus, $C D F^{\prime}$ is a triangle. Since $\alpha=22.5^{\circ}$ it follows that $D \hat{C} F^{\prime}=45^{\circ}$, from which $C D F^{\prime}$ is isosceles and then $C D=D F^{\prime}$. To finish, we note that $A B D F^{\prime}$ is a rectangle, and it follows that $D F^{\prime}=A B$. Thus $C D=A B=10$.

34 Common region - Solution

Observation: The general argument for solving this question is illustrated below. Triangle $A B C$ is one of the triangles resulting from the cut of the square, and $D$ is any point on side $A B$. Making $D E$ perpendicular to $A B$, the triangle $A D E$ is also a right triangle with equal sides, and its area is equal to half the area of the square $A D E F$; the area of the triangle $A D G$ is then equal to $\frac{1}{4}$ of the area of the square $A D E F$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-45.jpg?height=394&width=394&top_left_y=1779&top_left_x=731)

a) When $x=1$, the figure formed by the overlap of the larger triangles is a smaller triangle, indicated in gray in the figure below. The observation above shows that its area is a quarter of the area of a square of side 1, that is, $f(1)=\frac{1}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-45.jpg?height=291&width=334&top_left_y=2393&top_left_x=770)

When $x=3$, the figure formed by the overlap of the two triangles is a pentagon, as in the figure below. Since the triangles have legs of measure 2 and $A B=3$, we see that the legs overlap in a segment of measure 1. Thus, the pentagon is the union of a square of side 1 and a triangle identical to the one we considered at the beginning of this question. Thus, $f(3)=1+\frac{1}{4}=\frac{5}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-46.jpg?height=305&width=403&top_left_y=493&top_left_x=929)

b) For values of $x$ such that $0 \leq x \leq 2$, the figure formed by the overlap of the triangles is the gray triangle on the left in the figure below, from which $f(x)=\frac{x^{2}}{4}$ for $0 \leq x \leq 2$, as per the initial observation. When $21$ we conclude that this is the maximum value of $f$ in the interval $[0,4]$.