## Solution of problem T10/363 in M&Y

$\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\geq \left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\\ \forall a,b,c\in [1,+\infty).$

Solution of a my student.

## Solution of Problem 2, Grade 9, RNO 2007

Let $ABC$ be an acute angled triangle and point $M$ chosen differently from $A,B,C$. Prove that $M$ is the orthocenter of triangle $ABC$ if and only if

$\dfrac{BC}{MA}\cdot\overrightarrow{MA}+\dfrac{CA}{MB}\cdot\overrightarrow{MB}+\dfrac{AB}{MC}\cdot\overrightarrow{MC}= \overrightarrow{0}.(*)$

Solution of my students.

## Solution of problem 11306 in AMM

Let $a,b,$ and $c$ be the lengths of the sides of a nondegenerate triangle, let $p=(a+b+c)/2$, and let $r$ and $R$ be the inradius and circumradius of the triangle, respectively. Show that $\dfrac{a}{2}\cdot \dfrac{4r-R}{R}\leq\sqrt{(p-b)(p-c)}\leq \dfrac{a}{2},$

and determine the cases of equality.

My solution.

## Problems in Mathematics and Youth Magazine, 2007 , Issue 9

For Lower Secondary Schools

1. The first $100$ positive integer numbers are written consecutively in a certain order. Call the resulting number $A$. Is $A$ a multiple of $2007$?

## Solution of problem T12/363 in M&Y

Let $f:\mathbb{N}\to\mathbb{R}$ be a function such that $f(1)=\dfrac{2007}{6}$ and $\dfrac{f(1)}{1}+\dfrac{f(2)}{2}+\cdots+\dfrac{f(n)}{n}=\dfrac{n+1}{2}\cdot f(n)\forall n\in\mathbb{N}$. Find the limit $\lim_{n\to\infty} (2008+n)f(n)$.

My solution.

## Hàm lồi và cực trị của biểu thức dạng xtanA+ytanB+ztanC

Posted by Nguyen Song Minh

Edited by Nguyen Trung Tuan

In this post $I = (a,b)\subset \mathbb{R}$.
Theorem. If $f: I\to\mathbb{R}$ such that $f''(x) > 0\forall x\in I$ then $f(c)\geq (c - d)f'(d) + f(d)$. Equal hold iff $c = d$. Where $c,d\in I$.

## Cau truc de thi mon Toan (du kien) nam 2008 cua Bo GD&DT

Dinh lan mot thoi gian, nhung viec nay qua quan trong nen minh da tro lai. 😀 Day la cau truc de thi mon Toan nam 2008,  nhung chi la ”du kien” thoi day! 😛