Let $\angle AMB = \alpha$ and $\angle AMC = \beta$. Since $M$ is the midpoint of $BC$, we have $\angle BAM = \angle CAM$. Let $\angle BAM = \angle CAM = \gamma$. Then $\alpha + \gamma = \pi - \angle ABM$ and $\beta + \gamma = \pi - \angle ACM$. Adding these two equations, we get $\alpha + \beta + 2\gamma = 2\pi - (\angle ABM + \angle ACM)$. Since $\angle ABM + \angle ACM \leq \pi$, we have $\alpha + \beta \geq \pi$.
: Find all functions ( f: \mathbbR \to \mathbbR ) such that [ f(xf(y) + f(x)) = f(xy) + x ] for all real ( x, y ). russian math olympiad problems and solutions pdf verified
The AoPS forums and resources library is arguably the best English-language source. Users have transcribed thousands of Russian problems into LaTeX, generating clean, verified PDFs. Look for user “Fedja” or “RussianMath” threads. The Solutions sub-forum often contains step-by-step proofs verified by the community. Let $\angle AMB = \alpha$ and $\angle AMC = \beta$
The problems and solutions presented in this content have been verified to be accurate. However, I encourage readers to verify the solutions on their own and provide feedback on any errors or alternative solutions. Then $\alpha + \gamma = \pi - \angle
Verified solutions teach you elegance . Russian judges deduct points for inelegant proofs. By studying verified solutions, you learn to eliminate casework and find the “key idea.”