A ladder is $2 \frac{2}{3}$ m shorter than the height of the wall. If we lean it against the wall so that the foot of the ladder is at a distance of $\frac{3}{5}$ of the ladder's length from the wall, then the upper part of the ladder reaches $\frac{2}{5}$ of the wall's height. How long is the ladder?
I. Solution: Denoting the length of the ladder by $x$, the height of the wall is $x+2 \frac{2}{3}=\frac{3 x+8}{3}$. Assuming the wall is perpendicular to the ground (it is not certain that we are in a room, or that the wall is vertical; without this, the problem is indeterminate). In this case, the distances of the foot and the top of the ladder from the base of the wall are $\frac{3}{5} x$ and $\frac{2}{5} \cdot \frac{3 x+8}{3}=\frac{6 x+16}{15}$, respectively, forming a right-angled triangle whose hypotenuse is $x$. Applying the Pythagorean theorem to the "vertical" leg of this triangle (which also contains a constant with $x$):

$$
\left(\frac{6 x+16}{15}\right)^{2}=x^{2}-\left(\frac{3 x}{5}\right)^{2}=\frac{16 x^{2}}{25}=\left(\frac{4 x}{5}\right)^{2}
$$

Since we are dealing with distances, both sides of the equation are positive numbers squared, so the bases must also be equal:

$$
\frac{6 x+16}{15}=\frac{4 x}{5}
$$

From this, the length of the ladder is

$$
x=\frac{8}{3}=2 \frac{2}{3} \text { meters. }
$$

Indeed, in this case, the height of the wall is $x+2 \frac{2}{3}=\frac{16}{3}$ meters, the height of the top of the ladder is $\frac{2}{5} \cdot \frac{16}{3}=\frac{32}{15}$ meters, the foot of the ladder is $\frac{3}{5} \cdot \frac{8}{3}=\frac{8}{5}$ meters from the wall, and from these two projections, the length of the ladder is:

$$
\sqrt{\left(\frac{32}{15}\right)^{2}+\left(\frac{8}{5}\right)^{2}}=\sqrt{\frac{1024}{225}+\frac{64}{25}}=\sqrt{\frac{1600}{225}}=\frac{40}{15}=\frac{8}{3} \text { meters. }
$$

Julia Gonda (Mako, Jozsef A. g. I. o. t.)

II. Solution: Notice the $\frac{3}{5}$ data in the denominator and numerator of the most well-known Pythagorean triplet (3, $4,5$) hypotenuse number and one of its legs, and denote the length of the ladder by $5 x$; then the distance of the foot of the ladder from the wall is $3 x$ and the height of the top of the ladder is $4 x$. Since this is $\frac{2}{5}$ of the height of the wall, the height of the wall is $4 x: \frac{2}{5}=10 x$, exactly twice the length of the ladder. But this is also $2 \frac{2}{3} \mathrm{~m}$ more than the length of the ladder, so the sought length is also $2 \frac{2}{3}$ meters.

Tamas Biborka (Mako, Jozsef A. g. I. o. t.)