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orl wrote: Show that the triangle whose angles $A$, $B$, $C$ satisfy the equality \[ \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2 \] is a rectangular triangle. The op02 problem Germany 6 = geo problem 56 states even more: Prove that a triangle ABC is right-angled if and only if $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$. The solution is simple once one knows the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 \left(\cos A \cos B \cos C + 1\right)$, which holds for every triangle ABC. This identity shows that, if we denote $S = \sin^2 A + \sin^2 B + \sin^2 C$, then we have S = 2 if and only if 2 (cos A cos B cos C + 1) = 2, i. e. if cos A cos B cos C = 0, i. e. if triangle ABC is right-angled. Now, obviously, $\cos^2 x = 1 - \sin^2 x$ yields $\cos^2 A + \cos^2 B + \cos^2 C = \left(1-\sin^2 A\right)+\left(1-\sin^2 B\right)+\left(1-\sin^2 C\right)$ $= 3-\left(\sin^2 A + \sin^2 B + \sin^2 C\right) = 3 - S$. Thus, the condition given in the problem, $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$, is equivalent to $\frac{S}{3-S} = 2$. But this is a linear equation for S with the only solution S = 2. Hence, we have $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$ if and only if S = 2. But we know that S = 2 holds if and only if triangle ABC is right-angled. Thus, we have $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$ if and only if the triangle ABC is right-angled, and the problem is solved. Darij Solution by campos: Multiplying out the fraction and adding $\cos^2A+\cos^2B+\cos^2C$ to both sides gives $3=3(\cos^2A+\cos^2B+\cos^2C) \Rightarrow 1=\cos^2A+\cos^2B+\cos^2C$. Finally, using the trigonometry identity $1=\cos^2A+\cos^2B+\cos^2C+2\cos{A}\cos{B}\cos{C}$ we have $\cos{A}\cos{B}\cos{C}=0$, so it must be a rectangular triangle.