The point $P$ is given on the diameter $AB$ of a unit circle. A chord passing through $P$ intersects the circle at points $C$ and $D$. What can the area of the quadrilateral $ABCD$ be?
Solution. Let the given circle be denoted by $k$, its center by $O$, and the distance $OP$ by $d$. We investigate how the maximum value in question depends on the value of $d$. Clearly, $0 \leq d < 1$. The triangles $ADB$ and $ODP$ share the altitude from $D$, so the ratio of their areas is $\frac{T_{ADB}}{T_{ODP}} = \frac{AB}{OP}$ (Figure 1), and similarly, we get $\frac{T_{ABC}}{T_{OPC}} = \frac{AB}{OP}$. Therefore,

$$
T_{ADBC} = T_{COD} \cdot \frac{AB}{OP} = T_{COD} \cdot \frac{2}{d}
$$

The area of quadrilateral $ADBC$ is thus maximized when the area of triangle $COD$, $\frac{1}{2} OC \cdot OD \cdot \sin \angle COD$, is as large as possible.

![](https://cdn.mathpix.com/cropped/2024_05_02_b3d2ad33fc92c37c309eg-1.jpg?height=458&width=487&top_left_y=855&top_left_x=800)

Figure 1

In this product, $OC = OD = 1$, so the area of triangle $COD$ is maximized when $\sin \angle COD$ is maximized. Let $\varphi = \angle OCP = \angle ODP$. Then $\sin \angle COD = \sin (180^\circ - 2\varphi) = \sin 2\varphi$, and we seek the maximum value of this expression as $C$ traverses the arc $AB$. If $\varphi_0$ denotes the maximum value of $\angle OCP$, then $2\varphi$ varies over the interval $\left(0, 2\varphi_0\right]$. The maximum value of $\sin 2\varphi$ thus depends on whether this interval contains a right angle, i.e., on the relationship between $\varphi_0$ and $45^\circ$. Let us determine the maximum value of $\varphi$ for a given position of $P$.

For this, consider the tangent circle $k'$ of segment $OP$ that touches the arc $AB$ (Figure 2). (Due to symmetry, it suffices to consider the upper half-plane.) If $C_0$ is the point of tangency, then any point $C$ on the arc $AB$ different from $C_0$ is an external point of the circle $k'$, and by a known theorem, $\varphi = \angle OCP < \varphi_0$. Therefore, the maximum value of $\varphi$ is $\varphi_0$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b3d2ad33fc92c37c309eg-2.jpg?height=431&width=487&top_left_y=478&top_left_x=800)

Figure 2

Now, $\sin 2\varphi$ reaches its maximum value of 1 when $2\varphi = 90^\circ$, i.e., when $\varphi = 45^\circ$. This is possible if and only if $d = OP \geq \frac{\sqrt{2}}{2}$. In this case, the maximum area of $ADBC$ is $\frac{1}{d}$. If $d < \frac{\sqrt{2}}{2}$, then $\sin 2\varphi$ reaches its maximum value when $\varphi = \varphi_0$, and the maximum area of $ADBC$ is $2 \sqrt{1 - d^2}$.

Summarizing our results:

$$
\max T_{ACBD} = \begin{cases} 2 \sqrt{1 - OP^2}, & \text{if } 0 \leq OP < \frac{\sqrt{2}}{2} \\ \frac{1}{OP}, & \text{if } \frac{\sqrt{2}}{2} \leq OP < 1 \end{cases}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_b3d2ad33fc92c37c309eg-2.jpg?height=431&width=487&top_left_y=478&top_left_x=800)

Figure 3

In the case where $OP \geq \frac{\sqrt{2}}{2}$, there are two positions of the chord $CD$ passing through $P$ for which $\angle COD = 90^\circ$, the corresponding points $C_1$ and $C_2$ being the points where the $45^\circ$-view circle of segment $OP$ intersects the arc $AB$ (Figure 3).