## Task 21/79

All triples $(x ; y ; z)$ of positive real numbers are to be determined that satisfy the two conditions:

$$
x+y+z=\pi \quad(1) \quad ; \quad \sin ^{2} x+\sin ^{2} y=\sin ^{2} z
$$
Given $x ; y ; z>0$ and (1), it follows that $0<x ; y ; z<\pi$. Therefore, the numbers $x ; y ; z$ can be (angles measured in radians) of angles in a plane triangle. If the opposite sides are denoted by $a ; b$ and $c$ respectively, then according to the Law of Sines,

$$
\sin x: \sin y=a: c \quad \text { and } \quad \sin y: \sin z=b: c
$$

from which we get

$$
\sin ^{2} x+\sin ^{2} y=\left(\frac{a}{c}\right)^{2} \sin ^{2} z+\left(\frac{b}{c}\right)^{2} \sin ^{2} z=\frac{a^{2}+b^{2}}{c^{2}} \sin ^{2} z
$$

From (2), it follows that

$$
\frac{a^{2}+b^{2}}{c^{2}}=1 \quad ; \quad a^{2}+b^{2}=c^{2}
$$

i.e., it is a right-angled triangle with the hypotenuse $c$. Then, however, $z=0.5 \pi, \sin z=1$ and due to (1) also $x+y=0.5 \pi$. Clearly, for all triples $(x ; y ; z)$ with $x+y=z=0.5 \pi$, the conditions (1) and (2) are also satisfied, so that all triples are found.