Is there a triangle that can be cut in half parallel to one of its sides so that the perimeters and areas of the two resulting parts are equal?
Let the sides of the triangle be denoted by $a, b, c$ in the usual way. If a line parallel to the side $a$ cuts off a smaller triangle with sides $x a, x b, x c$, then this smaller triangle is similar to the original triangle, so its area is $x^{2}$ times that of the original. This area is exactly equal to the area of the remaining part (trapezoid) if $x^{2}=\frac{1}{2}$, that is, $x=\frac{\sqrt{2}}{2}$. In this case, the perimeter of the smaller triangle is $x(a+b+c)$, and that of the trapezoid is $(1+x) a+(1-x)(b+c)$. The two perimeters are equal if and only if

$$
\begin{aligned}
x(a+b+c) & =(1+x) a+(1-x)(b+c) \\
x & =\frac{a+b+c}{2(b+c)} \\
\frac{\sqrt{2}}{2} & =\frac{a+b+c}{2(b+c)}, \quad \text { that is, } \\
a & =(\sqrt{2}-1)(b+c)
\end{aligned}
$$

Such a triangle clearly exists, since $0<\sqrt{2}-1<1$; for example, $a=\sqrt{2}-1, b=c=\frac{1}{2}$.

Klár Tímea (Gödöllő, Török Ignác Gymn., 8th grade) Sebestyén Krisztián (Bp., Berzsenyi D. Gymn., 9th grade) Tuska Gábor (Debrecen, Fazekas M. Gymn., 9th grade)