INMO & RMO

$Problem$ #1 $$ $$ If $A$ and $B$ are different matrices satisfying \( A^3 = B^3 \) and \(A^2B = B^2A\), find \(det(A^2+B^2)\) $$ $$ Since $A$ and $B$ are different matrices \( A-B \neq O \), Now \((A^2+B^2)(A-B) = A^3-A^2B+B^2A-B^3\) $$ $$ =$O$ since \(A^3 = B^3 and A^2B = B^2A\) $$ $$ This shows that \((A^2+B^2)\) has a zero divisor, so it is not invertible hence \(det(A^2+B^2) = 0\) $$ $$ $Problem$ #2 $$ $$ Prove that \(\mathbf{\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^2} \right) = \frac{1}{2}}\) $$ $$ \(\prod_{n=2}^{N} \left( 1 - \frac{1}{n^2} \right) = \prod_{n=2}^{N} \left( 1 - \frac{1}{n} \right) \prod_{n=2}^{N} \left( 1 + \frac{1}{n} \right)\) $$ $$ \( = \prod_{n=2}^{N} \left( \frac{n-1}{n} \right) \prod_{n=2}^{N} \left( \frac{n+1}{n} \right) = \frac{1}{2}\frac{2}{3}\dots\frac{N-1}{N} \frac{3}{2}\frac{4}{3}\dots\frac{N+1}{N} = \frac{N+1}{2N}\) $$ $$ Letting \( N \to\) to \(\infty\), we see that \(\mathbf{\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^2} \right) = \frac{1+\frac{1}{N}}{2} = \frac{1}{2}} \) $$ $$ $Problem$ #3 $$ $$ Find all real numbers $x$ for which \(\sqrt{3-x}-\sqrt{x+1} > \frac{1}{2}\) $$ $$ Let $f(x)$ \(=\sqrt{3-x}-\sqrt{x+1}\). First note that $f(x)$ is defined for \( -1 \leq x \leq 3 \) $$ $$ \( f'(x) = \frac{-1}{2\sqrt{3-x}} - \frac{1}{\sqrt{x+1}} = - \big(\frac{1}{2\sqrt{3-x}} + \frac{1}{\sqrt{x+1}}\big) < 0 \Rightarrow f(x)\) is strictly decreasing $$ $$ Now \( f(-1) = 2 > \frac{1}{2}\) and \( f(3) = -2 < \frac{1}{2}\) Since $f(x)$ is continuous, $\exists$ at least one x $\in$ ${(-1,3)}$ suct that $f(x) = \frac{1}{2}$ $$ $$ \( f(x) = \frac{1}{2} \Rightarrow \sqrt{3-x}-\sqrt{x+1} = \frac{1}{2} \Rightarrow 64x^2-128x+33 = 0 \Rightarrow x = 1 \pm \frac{\sqrt{31}}{8} \)$$ $$ but \(x = 1 + \frac{\sqrt{31}}{8}\) does not satisfy \(\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}\) Check yourself! So the only solution is \(x = 1 - \frac{\sqrt{31}}{8}\) $$ $$ Since $f(x)$ is strictly decreasing, the given inequality is true for \( x \in {[-1,1 - \frac{\sqrt{31}}{8}\big)}\) $$ $$