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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 $\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).$

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 $a + b \le n$ .

Solution outline

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 $c=a+1$ , $d=b+1$ , or one of $c$ 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 either $P_0$ is on the interior of the arc $P_nP_a$ or on the interior of the arc $P_bP_n$ . Algebraically, it is equivalent to either $\{0-n\alpha\} < \{(a-n)\alpha\}$ or $\{n\alpha -0 \} < \{a\alpha\}$ . Suppose the former fails, i.e. $\{0-n\alpha\} \ge \{(a-n)\alpha\}$ . Then suppose $-n\alpha = m_1 + r_1$ and $(a-n)\alpha = m_2 + r_2$ , where $m_1$ , $m_2$ are integers and $1> r_1 \ge r_2 \ge 0$ . We now have $\{n\alpha\} = \{-m_1-1 + (1 -r_1)\}=1-r_1$ and $a\alpha = \{m_2-m_1-1 + (1-r_2+r_1)\} = 1-r_2+r_1>1-r_1$ Therefore $P_0$ is either closer to $P_n$ on the $P_a$ side, or closer to $P_n$ on the $P_b$ 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.

@@ Line 22: / Line 22: @@
 Let <math>\alpha</math> be an irrational number with <math>0 < \alpha < 1</math>, and draw a circle in the plane whose circumference has length 1.  Given any integer <math>n \ge 3</math>, define a sequence of points <math>P_1</math>, <math>P_2</math>, <math>\dots</math>, <math>P_n</math> as follows.  First select any point <math>P_1</math> on the circle, and for <math>2 \le k \le n</math> define <math>P_k</math> as the point on the circle for which the length of arc <math>P_{k - 1} P_k</math> is <math>\alpha</math>, when travelling counterclockwise around the circle from <math>P_{k - 1}</math> to <math>P_k</math>.  Supose that <math>P_a</math> and <math>P_b</math> are the nearest adjacent points on either side of <math>P_n</math>.  Prove that <math>a + b \le n</math>.
+=== Solution outline ===
+Use mathematical induction. For <math>n=3</math> it is true because one point can't be closest to <math>P_3</math> in both ways, and that <math>1+2\le 3</math>. Suppose that for some <math>n</math>, the nearest adjacent points <math>P_a</math> and <math>P_b</math> on either side of <math>P_n</math> satisfy  <math>a+b \le n</math>. Then consider the nearest adjacent points <math>P_c</math> and <math>P_d</math> on either side of <math>P_{n+1}</math>. It is by the assumption of the nearness we can see that either <math>c=a+1</math>, <math>d=b+1</math>, or one of <math>c</math> or <math>d</math> equals two <math>1</math>. Let's consider the following two cases.
+(i) Suppose <math>a+b=n</math>.
+Since the length of the arc <math>P_nP_a</math> is <math>\{(a-n)\alpha\}</math> (where <math>\{x\}</math> equals to <math>x</math> subtracted by the greatest integer not exceeding <math>x</math>) and length of the arc <math>P_bP_n</math> is <math>\{(n-b)\alpha\} = \{a\alpha\}</math>, we now consider a point <math>P_0</math> which is defined by <math>P_1</math> traveling clockwise on the circle such that the length of arc <math>P_0P_1</math> is <math>\alpha</math>. We claim that either <math>P_0</math> is on the interior of the arc <math>P_nP_a</math> or on the interior of the arc <math>P_bP_n</math>. Algebraically, it is equivalent to either <math> \{0-n\alpha\} < \{(a-n)\alpha\}</math> or <math>\{n\alpha -0 \} < \{a\alpha\}</math>. Suppose the former fails, i.e. <math> \{0-n\alpha\} \ge \{(a-n)\alpha\}</math>. Then suppose <math>-n\alpha = m_1 + r_1</math> and <math>(a-n)\alpha = m_2 + r_2</math>, where <math>m_1</math>, <math>m_2</math> are integers and <math>1> r_1 \ge r_2 \ge 0</math>. We now have
+<cmath>\{n\alpha\} = \{-m_1-1 + (1 -r_1)\}=1-r_1</cmath> and <cmath>a\alpha = \{m_2-m_1-1 + (1-r_2+r_1)\} = 1-r_2+r_1>1-r_1</cmath>
+Therefore <math>P_0</math> is either closer to <math>P_n</math> on the <math>P_a</math> side, or closer to <math>P_n</math> on the <math>P_b</math> side. Hence <math>P_c</math> or <math>P_d</math> is <math>P_1</math>, therefore <math>c+d \le n+1</math>
+(ii) Suppose <math>a+b\le n-1</math>
+Then either <math>c+d = (a+1)+(b+1) \le n+1</math> when <math>c=a+1</math> and <math>d=b+1</math>, or <math>c+d \le 1+n</math> when one of <math>P_c</math> or <math>P_d</math> is <math>P_1</math>.
+In either case, <math>c+d\le n+1</math> is true.
 == Problem 5==

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