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{\centerline {\bf THE 1997 ASIAN PACIFIC MATHEMATICAL OLYMPIAD}}
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{\it Time allowed: 4 hours}
{\it NO calculators are to be used.}
{\it Each question is worth seven points.}
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{\bf Question 1}
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Given $$S = 1 + \frac{1}{1 + \frac{1}{3}} + \frac{1}{1 + \frac{1}{3} + \frac{1}
{6}} + \cdots + \frac{1}{1 + \frac{1}{3} + \frac{1}{6} + \cdots + \frac{1}
{1993006}} \ ,$$ where the denominators contain partial sums of the sequence of
reciprocals of triangular numbers (i.e. $k=n(n+1)/2$ for $n = 1$, 2, \ldots,
1996). Prove that $S>1001$.
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{\bf Question 2}
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Find an integer $n$, where $100 \leq n \leq 1997$, such that $$\frac{2^n+2}
{n}$$ is also an integer.
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{\bf Question 3}
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Let $ABC$ be a triangle inscribed in a circle and let $$l_a = \frac{m_a}{M_a} \
, \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ ,$$ where $m_a$,
$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle)
and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until
they meet the circle. Prove that $$\frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B}
+ \frac{l_c}{\sin^2 C} \geq 3,$$ and that equality holds iff $ABC$ is an
equilateral triangle.
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{\bf Question 4}
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Triangle $A_1 A_2 A_3$ has a right angle at $A_3$. A sequence of points is now
defined by the following iterative process, where $n$ is a positive integer.
From $A_n$ ($n \geq 3$), a perpendicular line is drawn to meet $A_{n-2}
A_{n-1}$ at $A_{n+1}$.
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(a) Prove that if this process is continued indefinitely, then one and only one
point $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$, $n \geq 3$.
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(b) Let $A_1$ and $A_3$ be fixed points. By considering all possible locations
of $A_2$ on the plane, find the locus of $P$.
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{\bf Question 5}
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Suppose that $n$ people $A_1$, $A_2$, $\ldots$, $A_n$, ($n \geq 3$) are seated
in a circle and that $A_i$ has $a_i$ objects such that $$a_1 + a_2 + \cdots +
a_n = nN,$$ where $N$ is a positive integer. In order that each person has the
same number of objects, each person $A_i$ is to give or to receive a certain
number of objects to or from its two neighbours $A_{i-1} $ and $A_{i+1}$.
(Here $A_{n+1}$ means $A_1$ and $A_n$ means $A_0$.) How should this
redistribution be performed so that the total number of objects transferred is
minimum?
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