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- olympiad proofs and inequalities
[hint: some of the inequalities from this document may be helpful to you!]
U697 from Mathematical Reflections 2025 Issue 2
{Problem U697} Let \(f : [a, b] \to \mathbb{R}\) be a continuous function. Prove the inequality: \begin{equation} \left( \int_a^b f(x) \, dx \right)^4 \leq (b-a)^2 \left( \int_a^b e^x f^2(x) \, dx \right) \left( \int_a^b \frac{f^2(x)}{e^x} \, dx \right). \end{equation}Spoilers
Define the functions: \begin{align*} u(x) &= f(x) \sqrt{e^x}, \\ v(x) &= f(x) \frac{1}{\sqrt{e^x}}. \end{align*} Then: \begin{align*} \int_a^b u(x) v(x) \, dx &= \int_a^b f(x) \sqrt{e^x} \cdot f(x) \frac{1}{\sqrt{e^x}} \, dx = \int_a^b f^2(x) \, dx. \end{align*} Also, \begin{align*} \|u\|^2 &= \int_a^b u^2(x) \, dx = \int_a^b f^2(x) e^x \, dx, \\ \|v\|^2 &= \int_a^b v^2(x) \, dx = \int_a^b \frac{f^2(x)}{e^x} \, dx. \end{align*} By the Cauchy-Schwarz inequality, \begin{equation} \left( \int_a^b f^2(x) \, dx \right)^2 \leq \left( \int_a^b f^2(x) e^x \, dx \right) \left( \int_a^b \frac{f^2(x)}{e^x} \, dx \right). \label{eq:cs1} \end{equation} Applying Cauchy-Schwarz again: \begin{equation} \left( \int_a^b f(x) \cdot 1 \, dx \right)^2 \leq \left( \int_a^b f^2(x) \, dx \right)(b-a). \end{equation} Squaring both sides: \begin{equation} \left( \int_a^b f(x) \, dx \right)^4 \leq (b-a)^2 \left( \int_a^b f^2(x) \, dx \right)^2. \label{eq:cs2} \end{equation} Combining equations 1 and 2 gives: \begin{align*} \left( \int_a^b f(x) \, dx \right)^4 &\leq (b-a)^2 \left( \int_a^b f^2(x) \, dx \right)^2 \\ &\leq (b-a)^2 \left( \int_a^b f^2(x) e^x \, dx \right) \left( \int_a^b \frac{f^2(x)}{e^x} \, dx \right). \end{align*}U698 from Mathematical Reflections 2025 Issue 2
{Problem U698} Let \( f(z) = u(x, y) + iv(x, y) \) be a holomorphic function on \( \mathbb{C} \), where \( z = x + iy \), and the real part of \( f \) is given by: \( u(x, y) = 2x^2 - 3xy - 2y^2. \)Spoilers
Since \( f \) is holomorphic, it satisfies the Cauchy-Riemann equations: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \] We now have: \[ \frac{\partial v}{\partial y} = 4x - 3y, \quad \frac{\partial v}{\partial x} = 3x + 4y. \] Integrating \( \frac{\partial v}{\partial y} \) with respect to \( y \): \[ v(x, y) = \int (4x - 3y)\,dy = 4xy - \frac{3}{2}y^2 + C(x). \] Differentiate with respect to \( x \) to determine \( C(x) \): \[ \frac{\partial v}{\partial x} = 4y + C'(x). \] Set this equal to \( \frac{\partial v}{\partial x} = 3x + 4y \), yielding: \[ C'(x) = 3x \Rightarrow C(x) = \frac{3}{2}x^2. \] Thus, \[ v(x, y) = 4xy - \frac{3}{2}y^2 + \frac{3}{2}x^2. \] For a holomorphic function, the derivative is: \[ f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}. \] From earlier: \[ \frac{\partial u}{\partial x} = 4x - 3y, \quad \frac{\partial v}{\partial x} = 3x + 4y, \] So, \[ f'(z) = (4x - 3y) + i(3x + 4y). \] Computing the modulus: \begin{align*} |f'(z)|^2 &= (4x - 3y)^2 + (3x + 4y)^2 \\ &= 16x^2 - 24xy + 9y^2 + 9x^2 + 24xy + 16y^2 \\ &= 25x^2 + 25y^2 = 25(x^2 + y^2). \end{align*} Therefore: \[ |f'(z)| = 5\sqrt{x^2 + y^2}. \] Since \( |z| = \sqrt{x^2 + y^2} \), we have: \[ \left| \frac{f'(z)}{z} \right| = \frac{|f'(z)|}{|z|} = \frac{5\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2}} = \boxed{5}. \]- geometry
Cutting a Circle--Problem 166 from Kordemsky's 1956 Moscow Puzzles
Problem: To cut a circle with 6 straight lines into the greatest possible number of parts. The following diagram shows a circle cut into 16, however this is not the maximum. The largest number of parts is \[ \frac{1}{2}(n^2 + n + 2) \], where n is the number of straight lines. Cut a circle into 22 parts with 6 straight lines, and try to achieve some symmetry.Spoilers
picture goes here
- Graph Theory
Problem 75: Edge-Coloring Geometric Graphs
For a set of n points in the plane in general position, draw a straight segment between every pair of points. What is the minimum number of colors that suffice to color the edges such that no two edges that cross have the same color? (With the general position assumption, all crossings are proper crossings.)Spoilers
Math section:
i love math puzzles. i have included a few of my favorites, with answers (spoilers) though i suggest you try them out yourself first.
hiiii.
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