Difference between revisions of "1997 USAMO Problems"

Latest revision as of 13:31, 12 April 2023

Day 1

Problem 1

Let $p_1,p_2,p_3,...$ be the prime numbers listed in increasing order, and let $x_0$ be a real number between $0$ and $1$ . For positive integer $k$ , define

$x_{k}=\begin{cases}0&\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&\text{ if }x_{k-1}\ne0\end{cases}$

where $\{x\}$ denotes the fractional part of $x$ . (The fractional part of $x$ is given by $x-\lfloor{x}\rfloor$ where $\lfloor{x}\rfloor$ is the greatest integer less than or equal to $x$ .) Find, with proof, all $x_0$ satisfying $0<x_0<1$ for which the sequence $x_0,x_1,x_2,...$ eventually becomes $0$ .

Solution

Problem 2

Let $ABC$ be a triangle, and draw isosceles triangles $BCD, CAE, ABF$ externally to $ABC$ , with $BC, CA, AB$ as their respective bases. Prove that the lines through $A,B,C$ perpendicular to the lines $\overleftrightarrow{EF},\overleftrightarrow{FD},\overleftrightarrow{DE}$ , respectively, are concurrent.

Solution

Problem 3

Prove that for any integer $n$ , there exists a unique polynomial $Q$ with coefficients in $\{0,1,...,9\}$ such that $Q(-2)=Q(-5)=n$ .

Solution

Day 2

Problem 4

To clip a convex $n$ -gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$ . In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$ -gon. A regular hexagon $P_6$ of area $1$ is clipped to obtain a heptagon $P_7$ . Then $P_7$ is clipped (in one of the seven possible ways) to obtain an octagon $P_8$ , and so on. Prove that no matter how the clippings are done, the area of $P_n$ is greater than $\frac{1}{3}$ , for all $n\ge6$ .

Solution

Problem 5

Prove that, for all positive real numbers $a, b, c,$

$\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}$ .

Solution

Problem 6

Suppose the sequence of nonnegative integers $a_1,a_2,...,a_{1997}$ satisfies

$a_i+a_j \le a_{i+j} \le a_i+a_j+1$

for all $i, j \ge 1$ with $i+j \le 1997$ . Show that there exists a real number $x$ such that $a_n=\lfloor{nx}\rfloor$ (the greatest integer $\le x$ ) for all $1 \le n \le 1997$ .

Solution

@@ Line 28: / Line 28: @@
 Prove that, for all positive real numbers <math>a, b, c,</math>
-<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.
+<math>\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}</math>.
 [[1997 USAMO Problems/Problem 5|Solution]]
@@ Line 37: / Line 37: @@
 <math>a_i+a_j \le a_{i+j} \le a_i+a_j+1</math>
-for all <math>i, j \ge 1</math> with <math>i+j \le 1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\lenx</math>) for all <math>1 \le n \le 1997</math>.
+for all <math>i, j \ge 1</math> with <math>i+j \le 1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\le x</math>) for all <math>1 \le n \le 1997</math>.
 [[1997 USAMO Problems/Problem 6|Solution]]

1997 USAMO (Problems • Resources)
Preceded by 1996 USAMO	Followed by 1998 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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Difference between revisions of "1997 USAMO Problems"

Latest revision as of 13:31, 12 April 2023

Contents

Day 1

Problem 1

Problem 2

Problem 3

Day 2

Problem 4

Problem 5

Problem 6

See Also