The segments $A B$ and $C D$ are parallel. Point $P$ is an internal point on segment $B C$, and the line $A P$ intersects the line $C D$ at point $E$. How should point $P$ be chosen so that the sum of the areas of triangles $A P B$ and $C P E$ is as small as possible?
In the diagram, the length of segment $AB$ is denoted by $a$, the distance between $P$ and $AB$ is denoted by $x$, and the distance between the two parallel lines is 1. The angles of triangles $ABP$ and $ECP$ are equal in pairs, so these triangles are similar. Therefore, $CE = a \frac{1-x}{x}$. Thus, the sum $T$ of the areas of the triangles is:

\[
T = \frac{a x}{2} + \frac{a (1-x)^2}{2 x} = \frac{a}{2} \cdot \frac{2 x^2 - 2 x + 1}{x}
\]

and since $P$ is an internal point, $0 < x < 1$. $T$ will be the smallest precisely when the fraction $\frac{2 x^2 - 2 x + 1}{x}$ is the smallest.

\[
\frac{2 x^2 - 2 x + 1}{x} = 2 x + \frac{1}{x} - 2 \geq 2 \sqrt{2 x \cdot \frac{1}{x}} - 2 = 2 \sqrt{2} - 2
\]

where from the arithmetic and geometric mean inequality, we obtained that the value of the fraction is certainly not less than $2 \sqrt{2} - 2$. It also follows that this minimum is achieved if $0 < x < 1$: if $2 x = \frac{1}{x}$, then $x = \frac{1}{\sqrt{2}}$. This condition uniquely determines the position of point $P$ on segment $BC$. Furthermore, although the problem does not ask for it, we can also provide the minimum value of $T$: $T_{\min} = a (\sqrt{2} - 1)$.

David Papp (Budapest, Szent István Gymnasium, 10th grade)

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