## Task B-3.3.

A right triangle $A B C$ is given with a right angle at vertex $C$ and a hypotenuse of length $6 \mathrm{~cm}$. On the hypotenuse $\overline{A B}$, point $P$ is marked, and on the leg $\overline{B C}$, point $Q$ is marked such that $|B P|=|Q C|=1 \mathrm{~cm}$. If $\varangle B P Q=90^{\circ}$, determine $|P C|$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_803c5e1b47d2c202a3a9g-17.jpg?height=368&width=605&top_left_y=267&top_left_x=197)

Let $a, b$, and $c=6 \text{ cm}$ be the lengths of the sides of triangle $ABC$, and let $x$ be the length of the segment $\overline{PC}$. Triangles $ABC$ and $QBP$ are similar by the AA similarity criterion because both triangles are right-angled, and the angle at vertex $B$ is common to both.

Therefore, the lengths of the corresponding sides of these triangles are in the same ratio:

$\frac{1}{a-1}=\frac{a}{6}$.

It follows that $a^2 - a - 6 = 0$, where $a$ is a positive number because it represents the length of a side of the triangle.

By solving the quadratic equation, we conclude that $a = 3 \text{ cm}$.

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Since triangle $ABC$ is a right triangle, we have $\cos \beta = \frac{3}{6} = \frac{1}{2}$.

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Applying the cosine rule to triangle $PBC$:

$x^2 = |BP|^2 + |BC|^2 - 2 \cdot |BP| \cdot |BC| \cdot \cos \beta$.

$x^2 = 1^2 + 3^2 - 2 \cdot 1 \cdot 3 \cdot \frac{1}{2} = 1 + 9 - 3 = 7$

Finally, $|PC| = \sqrt{7} \text{ cm}$.

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