A rectangle with an isosceles right triangle inscribed over one of its sides, and a semicircle inscribed over the opposite side (both outward, with the rectangle's side as the hypotenuse or diameter), is called a $V$-shape. Which one has the largest area among $V$-shapes with the same perimeter?

 untranslated portion:
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Which one has the largest area among $V$-shapes with the same perimeter?
Let the lengths of the rectangle's sides be $x, y$ and write the semicircle and the triangle over the $x$ sides. Since we only consider $V$-shaped figures with the same perimeter $k$, only one of $x$ and $y$ can be chosen freely; let this be $x$, then

$$
2 y+\left(\sqrt{2}+\frac{\pi}{2}\right) x=k \text{, from which } \quad y=\frac{k}{2}-\frac{2 \sqrt{2}+\pi}{4} x
$$

Thus, the area is

$$
t=x y+\frac{x^{2}}{4}+\frac{x^{2} \pi}{8}=\frac{k x}{2}-\frac{(4 \sqrt{2}+\pi-2) x^{2}}{8}
$$

$x$ and $y$ cannot be negative, and the problem loses its meaning if $x=0$ (while $y=0$ is still acceptable), so considering (1), the area of $x$ is

$$
0 < x < \frac{2 k}{2 \sqrt{2}+\pi}
$$

Thus,

$$
t=-\frac{b}{8}\left(x^{2}-\frac{4 k x}{b}\right)=-\frac{b}{8}\left(x-\frac{2 k}{b}\right)^{2}+\frac{b}{8} \cdot \frac{4 k^{2}}{b^{2}}=\frac{k^{2}}{2 b}-\frac{b}{8}\left(x-\frac{2 k}{b}\right)^{2}
$$

and this is maximal when the subtracted term is 0, i.e., $x_{0}=2 k / b$. This is positive, and smaller than the right side of (2) because $b=2 \sqrt{2}+\pi+2(\sqrt{2}-1)>2 \sqrt{2}+\pi$, so the place of the maximum is within the domain of definition. The maximum area: $t_{\max }=k^{2} / 2 b=k^{2} /(8 \sqrt{2}+2 \pi-4)$.

The $x_{0}$ that gives the maximum area is directly proportional to $k$, so the ratio $x_{0} / k=2 / b$ is determined solely by the $\sqrt{2}$ and $\pi$ numbers associated with the $V$-shape. Let's calculate $y / x$ for the largest area $V$-shaped figure. From (1) with $x_{0}$,

$$
y_{0}=\frac{k}{2}-\frac{2 \sqrt{2}+\pi}{4} \cdot \frac{2 k}{b}=\frac{k}{2 b}(b-2 \sqrt{2}-\pi)=\frac{x_{0}}{4}(2 \sqrt{2}-2)=\frac{x_{0}}{2}(\sqrt{2}-1)
$$

thus

$$
\frac{y_{0}}{x_{0}}=(\sqrt{2}-1) / 2=0.2071 \ldots
$$

Gábor Szatmári (Budapest, Piarist g. IV. o. t.)

Remark. The simple ratio $y_{0} / x_{0}$ suggests that the shape of the largest area $V$-shaped figure can be constructed (but the figure itself cannot be constructed from the given $k$).