## Task 26/68

Under what conditions is the square of the circumcircle diameter in a triangle equal to the sum of the squares of two sides of the triangle?
We use the usual notation (triangle sides $a, b, c$; opposite angles $\alpha, \beta, \gamma$; circumdiameter $d$). The requirement is $s^{3}=a^{2}+b^{2}$. Because $a=d \cdot \sin \alpha$ and $b=d \cdot \sin \beta$, it follows that

$$
a^{2}+b^{2}=d^{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)
$$

By comparing coefficients, it follows that

$$
\sin ^{2} \alpha+\sin ^{2} \beta=1 ; \quad \sin ^{2} \beta=1-\sin ^{2} \alpha=\cos ^{2} \alpha ; \quad \sin \beta=|\cos \alpha|
$$

This relation is obviously satisfied for $0^{\circ}<\alpha ; \beta<180^{\circ}$ if and only if either

$$
\text { 1. } \alpha+\beta=90^{\circ} \text { or } 2 \text {. }|\alpha-\beta|=90^{\circ}
$$

is true. Solution 1 can be considered trivial (the triangle is right-angled, the circumcircle is the same as the Thales circle, its diameter is equal to the hypotenuse).

Result: In a triangle, the square of the circumdiameter is equal to the sum of the squares of two triangle sides if the sum or difference of two triangle angles is $90^{\circ}$. As the derivation shows, the condition is both necessary and sufficient. The theorem is thus reversible.