A circle has centre $O$ and diameter $A B=2 \sqrt{19}$. Points $C$ and $D$ are on the upper half of the circle. A line is drawn through $C$ and $D$, as shown. Points $P$ and $Q$ are on the line so that $A P$ and $B Q$ are both perpendicular to $P Q . Q B$ intersects the circle at $R$. If $C P=D Q=1$ and $2 A P=B Q$, what is the length of $A P$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_5629b903ac0448ea13d7g-2.jpg?height=342&width=360&top_left_y=2274&top_left_x=1357)
Since the circle has diameter $2 \sqrt{19}$, it has radius $\sqrt{19}$.

Let $A P=2 x$. Thus, $B Q=2 A P=4 x$.

Let $C D=2 y$ and let $M$ be the midpoint of $C D$. Thus, $C M=M D=y$.

Join $A$ to $R$. Since $A B$ is a diameter, then $\angle A R B=90^{\circ}$.

Since quadrilateral $P Q R A$ has right angles at $P, Q$ and $R$, then it must have four right angles, and so it is a rectangle.

Thus, $A R=P Q=P C+C D+D Q=1+2 y+1=2 y+2$ and $R Q=A P=2 x$.

Since $B Q=4 x$, then $R B=B Q-R Q=4 x-2 x=2 x$.

By the Pythagorean Theorem in $\triangle A R B$, we obtain

$$
\begin{aligned}
A R^{2}+R B^{2} & =A B^{2} \\
(2 y+2)^{2}+(2 x)^{2} & =(2 \sqrt{19})^{2} \\
(y+1)^{2}+x^{2} & =19
\end{aligned}
$$

Join $O$ to $M$.

![](https://cdn.mathpix.com/cropped/2024_04_17_118dc927cebd403196e3g-03.jpg?height=347&width=371&top_left_y=840&top_left_x=931)

Since $M$ is the midpoint of chord $C D$, then $O M$ is perpendicular to $C D$.

Since $A R$ is parallel to $C D$ (opposite sides in rectangle $P Q R A$ ), then $O M$ is also perpendicular to $A R$, crossing $A R$ at $T$.

Consider $\triangle A T O$ and $\triangle A R B$. These triangles are similar because they share a common angle at $A$ and each is right-angled (at $T$ and $R$, respectively).

Since $\frac{A O}{A B}=\frac{1}{2}$ (radius and diameter), then $\frac{T O}{R B}=\frac{1}{2}$ and so $T O=\frac{1}{2} \cdot R B=x$.

Therefore, $O M=M T+T O=2 x+x=3 x$. This is because $M T=P A$ since $M T$ is parallel to side $P A$ of rectangle $P Q R A$ and has its endpoints on opposite sides of the rectangle.

By the Pythagorean Theorem in $\triangle O M C$, we have

$$
\begin{aligned}
C M^{2}+O M^{2} & =C O^{2} \\
y^{2}+(3 x)^{2} & =(\sqrt{19})^{2} \quad \text { (since } C O \text { is a radius) } \\
y^{2}+9 x^{2} & =19
\end{aligned}
$$

Since $(y+1)^{2}+x^{2}=19$, then multiplying by 9 gives $9(y+1)^{2}+9 x^{2}=171$.

Subtracting $y^{2}+9 x^{2}=19$ gives

$$
\begin{aligned}
9\left(y^{2}+2 y+1\right)-y^{2} & =171-19 \\
8 y^{2}+18 y-143 & =0 \\
(4 y-13)(2 y+11) & =0
\end{aligned}
$$

Since $y>0$, then $y=\frac{13}{4}$. Therefore, $x^{2}=19-(y+1)^{2}=19-\left(\frac{17}{4}\right)^{2}=\frac{304}{16}-\frac{289}{16}=\frac{15}{16}$.

Since $x>0$, then $x=\frac{\sqrt{15}}{4}$. Finally, $A P=2 x=\frac{\sqrt{15}}{2}$.