Revision as of 00:19, 8 April 2013 by Atmath2011 (talk | contribs) (Test)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

For any information, please contact me at


USAJMO Problem 1

Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP = AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle BPS = \angle PRS$, and $\angle CQR = \angle QSR$. Prove that $P$, $Q$, $R$, $S$ are concyclic (in other words, these four points lie on a circle).

Problem 2

Find all integers $n \ge 3$ such that among any $n$ positive real numbers $a_1$, $a_2$, $\dots$, $a_n$ with \[\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n),\] there exist three that are the side lengths of an acute triangle.

Problem 3

Let $a$, $b$, $c$ be positive real numbers. Prove that \[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).\]


We proceed to prove that \[\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2\]

(then the inequality in question is just the cyclic sum of both sides, since \[\sum_{cyc} (-\frac{1}{36} a^2 + \frac{25}{36} b^2) = \frac{24}{36}\sum_{cyc} a^2 = \frac{2}{3} (a^2+b^2+c^2)\] )

Indeed, by AP-GP, we have

\[41 (a^3 + b^3+b^3) \ge 41 \cdot 3 ab^2\]


\[b^3 + a^2b \ge 2 ab^2\]

Summing up, we have

\[41a^3 + 83b^3 + a^2 b \ge 125 ab^2\]

which is equivalent to:

\[36(a^3 + 3b^3) \ge (5a + b)(-a^2 + 25b^2)\]

Dividing $36(5a+b)$ from both sides, the desired inequality is proved.

Problem 4

Let $\alpha$ be an irrational number with $0 < \alpha < 1$, and draw a circle in the plane whose circumference has length 1. Given any integer $n \ge 3$, define a sequence of points $P_1$, $P_2$, $\dots$, $P_n$ as follows. First select any point $P_1$ on the circle, and for $2 \le k \le n$ define $P_k$ as the point on the circle for which the length of arc $P_{k - 1} P_k$ is $\alpha$, when travelling counterclockwise around the circle from $P_{k - 1}$ to $P_k$. Supose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$. Prove that $a + b \le n$.


Use mathematical induction. For $n=3$ it is true because one point can't be closest to $P_3$ in both ways, and that $1+2\le 3$. Suppose that for some $n$, the nearest adjacent points $P_a$ and $P_b$ on either side of $P_n$ satisfy $a+b \le n$. Then consider the nearest adjacent points $P_c$ and $P_d$ on either side of $P_{n+1}$. It is by the assumption of the nearness we can see that either $(c,d)=(a+1,b+1)$ still holds, or $P_1$ jumps into the interior of the arc $P_{a+1}P_{n}P_{b+1}$, so that $c$ or $d$ equals two $1$. Let's consider the following two cases.

(i) Suppose $a+b=n$.

Since the length of the arc $P_nP_a$ is $\{(a-n)\alpha\}$ (where $\{x\}$ equals to $x$ subtracted by the greatest integer not exceeding $x$) and length of the arc $P_bP_n$ is $\{(n-b)\alpha\} = \{a\alpha\}$, we now consider a point $P_0$ which is defined by $P_1$ traveling clockwise on the circle such that the length of arc $P_0P_1$ is $\alpha$. We claim that $P_0$ is in the interior of the arc $P_bP_nP_a$. Algebraically, it is equivalent to either $\{0-n\alpha\} < \{a\alpha-n\alpha\}$ or $\{n\alpha -0 \} < \{n\alpha - b\alpha\} = \{a\alpha\}$.

Suppose the latter fails, i.e. $\{n\alpha\} \ge \{a\alpha\}$. Then suppose $n\alpha = m_1 + r_1$ and $a\alpha = m_2 + r_2$, where $m_1$, $m_2$ are integers and $0< r_2\le r_1 <1$ ($r_2$ is not zero because $a\alpha$ is irrational). We now have \[\{0-n\alpha\} = \{-m_1-1 + (1 -r_1)\}=1-r_1\] and \[\{a\alpha -n\alpha\} = \{m_2-m_1-1 + (1+r_2-r_1)\} = 1+r_2-r_1>1-r_1\]

Therefore $P_0$ is either closer to $P_n$ than $P_a$ on the $P_a$ side, or closer to $P_n$ than $P_b$ on the $P_b$ side. In other words, $P_1$ is the closest adjacent point of $P_{n+1}$ on the $P_{a+1}$ side, or the closest adjacent point of $P_{n+1}$ on the $P_{b+1}$ side. Hence $P_c$ or $P_d$ is $P_1$, therefore $c+d \le n+1$.

(ii) Suppose $a+b\le n-1$ Then either $c+d = (a+1)+(b+1) \le n+1$ when $c=a+1$ and $d=b+1$, or $c+d \le 1+n$ when one of $P_c$ or $P_d$ is $P_1$.

In either case, $c+d\le n+1$ is true.

Problem 5

For distinct positive integers $a$, $b < 2012$, define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$.

Problem 6

Let $P$ be a point in the plane of triangle $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear.



Problem 1

给定三角形 $ABC$, 点$P$, $Q$ 分别落在边 $\overline{AB}$$\overline{AC}$上,使得 $AP = AQ$. 令 $S$, $R$$\overline{BC}$ 上相异的两点,使得 $S$ 落在 $B$$R$ 之间,而且有 $\angle BPS = \angle PRS$, $\angle CQR = \angle QSR$. 证明 $P$, $Q$, $R$, $S$ 四点共圆 (即,存在一个圆使得这四点同时落在其上。)

Problem 2

找出所有满足如下条件的正整数 $n \ge 3$: 对于任意的$n$个正实数 $a_1$, $a_2$, $\dots$, $a_n$,只要 \[\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n),\] 就存在其中的三个数,它们能构成锐角三角形的三边长。

Problem 3

找所有同时满足如下两个等式的整数 $a$,$b$,$c$,$d$:



Problem 4

找出所有满足如下条件的函数 $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ (其中$\mathbb{Z}^+$ 为正整数集) 使得 $f(n!) = f(n)!$ 对任意正整数$n$成立,而且对任意相异正整数$m,n$$f(m) - f(n)$能被 $m - n$ 整除。

Problem 5

Two people play a game with a bar of chocolate made of 60 pieces, in a 6 × 10 rectangle. The first person breaks off a part of the chocolate bar along the grooves dividing the pieces, and discards (or eats) the part he broke off. Then the second person breaks off a part of the remaining part and discards her part. The game continues until one piece is left. The winner is the one who leaves the other with the single piece (i.e. is the last to move). Which person has a perfect winning strategy?

Problem 6

Let $x, y, z$ be positive real numbers such that $x+y+z = 1$. Prove that \[\frac{3x+1}{y+z}+\frac{3y+1}{z+x}+\frac{3z+1}{x+y}\ge\frac{4}{2x+y+z}+\frac{4}{x+2y+z}+\frac{4}{x+y+2z},\] with equality if and only if $x=y=z$.