The following figure shows a square $ABCD$, and two points $P$ and $Q$ on the sides $\overline{BC}$ and $\overline{DA}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-13.jpg?height=462&width=479&top_left_y=504&top_left_x=860)

We now fold the square along the segment $\overline{PQ}$, bringing the vertex $B$ to the midpoint of the segment $\overline{CD}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-13.jpg?height=500&width=492&top_left_y=1151&top_left_x=862)

It is known that the side of the square measures 24.

a) Calculate the length of the segment $\overline{PC}$.

b) Calculate the length of the segment $\overline{AQ}$.

c) Calculate the length of the segment $\overline{PQ}$.
Solution

a) Let's call $x$ the length of the segment $\overline{P C}$. Since the side of the square measures 24, then $|P B|=24-x$ and also $\left|P B^{\prime}\right|=24-x$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-42.jpg?height=666&width=643&top_left_y=468&top_left_x=588)

Using the Pythagorean Theorem in the triangle $P C B^{\prime}$, we obtain the relation

$$
x^{2}+12^{2}=(24-x)^{2}
$$

Thus, we show that $|P C|=9$. Also observe that $\left|P B^{\prime}\right|=15$ (this will be used in the next items).

b) Let's call $\alpha^{\circ}$ the measure of the angle $P B^{\prime} C$. Then $\measuredangle R B^{\prime} D=(90-\alpha)^{\circ} \mathrm{e}$, and therefore, $\measuredangle D R B^{\prime}=\alpha^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-42.jpg?height=668&width=647&top_left_y=1617&top_left_x=586)

This shows that the triangles $P C B^{\prime}$ and $B^{\prime} D R$ are similar. Using the similarity relations, we see that $|R D|=12 \times 12 / 9=16$ and then $\left|R B^{\prime}\right|=20$ (by Pythagoras).

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-43.jpg?height=666&width=642&top_left_y=238&top_left_x=784)

Given that $\left|A^{\prime} B^{\prime}\right|=|A B|=24$, we obtain that $\left|A^{\prime} R\right|=4$. Moreover, we can see that $\measuredangle Q R A^{\prime}=\measuredangle D R B^{\prime}=\alpha^{\circ}$ and, therefore, the triangles $R D B^{\prime}$ and $R A^{\prime} Q$ are also similar. Using the similarity relation, we obtain that $\left|Q A^{\prime}\right|=4 \times 12 / 16=3$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-43.jpg?height=669&width=658&top_left_y=1400&top_left_x=781)

Finally, $|A Q|=\left|A^{\prime} Q\right|=3$.

c) To calculate $P Q$, it is sufficient to observe the quadrilateral $A^{\prime} Q P B^{\prime}$. We draw the segment $\overline{Q S}$ perpendicular to the segment $\overline{P B^{\prime}}$ such that $\left|Q A^{\prime}\right|=\left|S B^{\prime}\right|=3$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-44.jpg?height=666&width=579&top_left_y=235&top_left_x=615)

Using Pythagoras in the triangle $P S Q$ we have

$$
|P Q|^{2}=24^{2}+12^{2}
$$

Thus, $|P Q|=12 \sqrt{5}$.