A2. For the lengths $a, b$, and $c$ of the sides of triangle $ABC$, it holds that $c^{2}=2 a b$ and $a^{2}+c^{2}=3 b^{2}$. The measures of the interior angles of triangle $ABC$ are
(A) $30^{\circ}, 60^{\circ}$, and $90^{\circ}$.
(B) $45^{\circ}, 60^{\circ}$, and $75^{\circ}$.
(C) $45^{\circ}, 45^{\circ}$, and $90^{\circ}$.
(D) $60^{\circ}, 60^{\circ}$, and $60^{\circ}$.
(E) It is impossible to determine.
A2. From $c^{2}=2 a b$ and $a^{2}+c^{2}=3 b^{2}$ it follows that $a^{2}+2 a b=3 b^{2}$ or $(a+b)^{2}=4 b^{2}$. From this we get $(a+b-2 b)(a+b+2 b)=0$. Since $a$ and $b$ are positive numbers, the only possibility is $a=b$. Then $c^{2}=2 a^{2}=a^{2}+b^{2}$. The triangle is thus isosceles and right-angled, so the measures of the interior angles are $45^{\circ}, 45^{\circ}$ and $90^{\circ}$.