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jasperE3 wrote: In a tetrahedron $ABCD$, all angles at vertex $D$ are equal to $\alpha$ and all dihedral angles between faces having $D$ as a vertex are equal to $\phi$. Prove that there exists a unique $\alpha$ for which $\phi=2\alpha$. See here page 3 in croatian. Translation by chatGPT: "Let E and F be points on the edges DA and DC, respectively, such that IDEI = IDFI = 1. The feet of the perpendiculars from points E and F to the line BD are point G. According to the definition of the angle between two planes, <p = ∠EGF. Let \( |GE| = |GF| = x \), \( |EF| = y \), then \( \sin \alpha = x \), \( \sin \frac{\alpha}{2} = \frac{y}{2} \), \( \sin \frac{\varphi}{2} = \frac{y}{2x} \). From here, \[ \sin \frac{\varphi}{2} = \frac{\sin \frac{\alpha}{2}}{\sin \alpha} = \frac{1}{2 \cos \frac{\alpha}{2}}. \]Since \( \varphi = 2\alpha \), then \( 2 \sin \alpha \cos \frac{\alpha}{2} = 1 \), from which we get \[ 4 \sin \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} = 1, \]\[ 4 \sin \frac{\alpha}{2} \left(1 - \sin^2 \frac{\alpha}{2} \right) = 1, \]\[ 4 \sin^3 \frac{\alpha}{2} - 4 \sin \frac{\alpha}{2} + 1 = 0. \]Let \( t = \sin \frac{\alpha}{2} \), the last equation turns into \( 4t^3 - 4t + 1 = 0. \) (1) Since \( 0 < \varphi < \pi \), then \( 0 < \alpha < \frac{\pi}{2} \), or \( 0 < t < \frac{\sqrt{2}}{2} \). Because of this, it is enough to show that equation (1) has exactly one solution in the interval \( \left(0, \frac{\sqrt{2}}{2} \right) \). Let us consider the function \( f(t) = 4t^3 - 4t + 1 \). It holds: \( f(0) = 1 > 0 \) and \( f\left( \frac{\sqrt{2}}{2} \right) = 2\sqrt{2} - 2\sqrt{2} + 1 = 1 - \sqrt{2} < 0 \). Since \( f(t) \) is a polynomial over the set of real numbers, from \( f(0) > 0 \) and \( f\left( \frac{\sqrt{2}}{2} \right) < 0 \), we conclude that \( f \) has (at least) one zero point in the interval \( \left(0, \frac{\sqrt{2}}{2} \right) \). It remains to be shown that this is the only zero point in that interval. From \( f(-2) = -32 + 8 + 1 = -23 < 0 \) and \( f(0) > 0 \), we see that the second zero point of the observed function is in the interval \((-2, 0)\). Likewise, because \( f\left( \frac{\sqrt{2}}{2} \right) < 0 \) and \( f(1) > 0 \), the third zero point of that function is in the interval \(\left( \frac{\sqrt{2}}{2}, 1 \right) \).

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