Given a circle with diameter $AB$ and an arbitrary point $P$ on it. The circle centered at $P$ with radius $PA$ intersects the line $AB$ at point $C$. When will the area of triangle $APC$ be the largest?
Let $O$ be the center of the circle with diameter $AB$, $T$ the orthogonal projection of point $P$ onto the line $AB$, and $R$ the midpoint of the arc $AB$ containing $P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bdb97fdcef2f4088a2b6g-1.jpg?height=374&width=571&top_left_y=260&top_left_x=758)

According to the given conditions, $PA = PC$, so triangle $APC$ is isosceles, which means $AT = TC$; thus, the area of the triangle is:

$$
T_{APC} = AT \cdot TP
$$

We need to determine the maximum value or values of this expression.

First, consider the case where $T$ is an interior point of segment $AO$. In this case, the value of (1) cannot be maximal, because $AT < 0$, and the area of the triangle would be negative, which is not possible.

Next, consider the case where $T$ is an interior point of segment $OB$. Here, $AT > 0$, and the area of the triangle is positive. We can express the area as:

$$
T_{APC} = AT \cdot TP = \sqrt{1 - x^2} \cdot \sqrt{3x^2} = \sqrt{3(1 - x^2)x^2}
$$

where $x = \frac{AT}{AO}$. Since $0 < x < 1$, we need to maximize the expression $\sqrt{3(1 - x^2)x^2}$. This can be rewritten as:

$$
\sqrt{3(1 - x^2)x^2} = \sqrt{3(1 - x^2)x^2} = \sqrt{3(1 - x^2)x^2} = \sqrt{3(1 - x^2)x^2}
$$

To simplify, let $y = \sqrt{3(1 - x^2)x^2}$. We need to maximize $y$. Notice that the function $y = \sqrt{3(1 - x^2)x^2}$ is strictly increasing for $0 < x < 1$.

We can apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):

$$
\sqrt[4]{(1+x)^{3}(3-3x)} \leq \frac{(1+x) + (1+x) + (1+x) + (3-3x)}{4} = \frac{3}{2}
$$

From this, it is clear that the left-hand side expression is maximized when the factors are equal, i.e., when $1 + x = 3 - 3x$, which gives $x = \frac{1}{2}$.

Therefore, the area of triangle $APC$ is maximized when $T$ is the midpoint of segment $OB$.