Mekkora a területe annak az egyenló szárú derékszögú háromszögnek, melynek kerülete adott $2 s$ ?

What is the area of an isosceles right triangle whose perimeter is given as $2 s$?
I. solution: First, let's calculate the common length $a$ of the legs of our triangle. With these, the hypotenuse is $a \sqrt{2}$, so from the equation

$$
a+a+a \sqrt{2}=a(2+\sqrt{2})=2 s
$$

we get

$$
a=\frac{2 s}{2+\sqrt{2}}=(2-\sqrt{2}) s=\sqrt{2}(\sqrt{2}-1) s
$$

and the area

$$
t=\frac{a^{2}}{2}=\left(\frac{a}{\sqrt{2}}\right)^{2}=[(\sqrt{2}-1) s]^{2}=(3-2 \sqrt{2}) s^{2}
$$

Téry László (Pécs, bányaipari techn. II. o. t.)

II. solution: The perimeter of an isosceles right triangle with a leg of $2 s$ is $2 s+2 s+2 \sqrt{2} s=2 s(2+\sqrt{2})$ and the area is $2 s^{2}$. The triangle under consideration is similar to this one, the ratio of their corresponding lengths is derived from the perimeters as $2 s : 2 s(2+\sqrt{2})=1:(2+\sqrt{2})=(\sqrt{2}-1): \sqrt{2}$. Using the theorem on the ratio of the areas of similar figures, we have

$$
t: 2 s^{2}=(\sqrt{2}-1)^{2}:(\sqrt{2})^{2}
$$

and thus again

$$
t=(\sqrt{2}-1)^{2} s^{2}
$$

is obtained.

Gáll Endre (Bp. XI., József A. g. I. o. t.)[^0]


[^0]:    ${ }^{1}$ It is easily provable that the ratio of the areas of two similar triangles, or more generally, two similar plane figures, is equal to the ratio of the squares of any corresponding lengths (e.g., sides, heights, perimeters, radii, etc.).